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How to prove the identity `sin^2x + cos^2x = 1` ?

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mathexpert | Student, Undergraduate | Honors

Posted October 29, 2008 at 9:20 PM via web

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How to prove the identity `sin^2x + cos^2x = 1` ?

88 Answers | Add Yours

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jeyaram | Student, Undergraduate | Valedictorian

Posted September 13, 2009 at 10:14 PM (Answer #1)

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sin^2X + cos^2X = 1

L.H.S

=sin^2X + cos^2X

=sinXsinX+cosXcosX

=cos(X-X)

=cos0

=1

R.H.S

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teasley | High School Teacher | Honors

Posted October 30, 2008 at 8:44 AM (Answer #2)

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Keep in mind that sin t = y/r, cos t = x/r, and r^2 = x^2+y^2.  By substitution, sin^2x = y^2/r^2 and cos^2x = x^2/r^2.  So the left hand side of the identitiy becomes (y^2/r^2) + (x^2/r^2).  Since these already have a common denominator, we can write them as one fraction:  (y^2+x^2)/r^2 = r^2/r^2 = 1. 

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beckden | High School Teacher | (Level 1) Educator

Posted June 21, 2011 at 3:09 PM (Answer #4)

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Another method is knowing to take the derivative of

f(x) = sin^2(x) + cos^2(x)

f '(x) = 2 sin(x) cos(x) + 2 cos(x) (-sin(x))

= 2 sin(x) cos(x) - 2 cos(x) sin(x)

= 0

Since the derivative is zero everywhere the function must be a constant.

Take f(0) = sin^2(0) + cos^2(0) = 0 + 1 = 1

So

sin^2(x) + cos^2(x) = 1 everywhere.

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sefua | Student, Undergraduate | Honors

Posted May 27, 2011 at 9:37 PM (Answer #3)

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from pythagoras theorem.....

r^2=x^2+y^2

divide through by r^2..

i.e..r^2/r^2=x^2/r^2+y^2/r^2

:. it implies that.. 1=x^2/r^2+y^2/r^2

1=(x/r)^2+(y/r)^2

but sinΘ=y/r,cosΘ=x/r

by substitution...

1=sin^2Θ+cos^2Θ

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revolution | College Teacher | Valedictorian

Posted July 4, 2010 at 9:42 PM (Answer #6)

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You can prove it by using pythagoras' theorem using the following equation:

r^2=y^2+x^2

Then divide throughout by r^2

y^2/r^2 + x^2/y^2 = 1

then subsituting sin Θ = y/r, cos Θ = x/r

sin^2x + Cos^2x = 1 (shown)

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arthurdong | Student, Grade 11 | eNoter

Posted June 7, 2011 at 4:06 PM (Answer #7)

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In a right triangle, let the opposite side of angle x is a, the other right-angle side is b, and the hypontenuse is r.

So, we can get: sin^2x+cos^2x=(a^2)/r^2+(b^2)/r^2=(a^2+b^2)/r^2=r^2/r^2=1

 

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dogsg | High School Teacher | (Level 2) Adjunct Educator

Posted November 11, 2008 at 8:47 AM (Answer #5)

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An alternate approach to proving this identity involves using the "unit circle" (radius = 1).  Since the radius is also the hypotenuse of the right triangle formed by the angle "x" within the circle, the sine is y and the cosine is x.  By the Pythagorean Theorem, x^2 + y^2 = 1^2 ... or x^2 + y^2 = 1

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tutoringcraft | High School Teacher | Honors

Posted June 4, 2011 at 1:31 PM (Answer #8)

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This is not a proof, but it sure is compelling evidence:

Enter y=(sin(x))^2+(cos(x))^2 into a graphing calculator and look at the result for -2pi < x < 2pi.  Yep, it's the constant function y=1.

Or think about it without technology.  What would the graph of f(x)=(sin(x))^2 look like?  All points with y-values of 0 or 1 would not change, and points with y=-1 would keep their x-values but get y-values of +1.  So this graph would be zero at even multiples of pi/2 and 1 at odd multiples of pi/2.  A similar analysis of g(x)=(cos(x))^2 gives a graph that is 1 where f(x) is 0, and 0 where f(x) is 1.  So y=f(x)+g(x) is clearly 1 in all of those places.

Again, it's not a proof, but it's good to have a graphical look at this important trig identity.

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zarghamenotes | Student, Grade 10 | Honors

Posted July 20, 2011 at 1:23 AM (Answer #9)

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sin theta=p/h

cos theta=b/h

Therefore,

sin^2 theta=p^2/h^2

cos^2 theta=b^2/h^2

Now,

sin^2 theta + cos^2 theta=(p^2/h^2) + (b^2/h^2)

=(p^2 + b^2)/h^2

By pythagoras' theorem,

p^2 + b^2=h^2

therefore,

(p^2 + b^2)/h^2 = h^2/h^2

=1

Hence, proved

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jdukowitz | Honors

Posted August 16, 2011 at 11:31 PM (Answer #10)

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Consider a right triangle with sides a, b and c, with c as hypotenuse and Θ as the angle between lines a and b. We know from triginometry that: sinΘ = opposite / hypotenuse (a/c)
cosΘ = adjacent / hypotenuse (b/c) We know from the Pythagorean theorum that a² + b² = c²   Divide both sides by c² 1/c² (a² + b²) = 1/c² ( c²) Then a²/c² + b²/c² = c²/c² (a/c)² + (b/c)² = 1 sin²Θ + cos²Θ = 1  

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verushkabrijrajh | Student, Undergraduate | eNoter

Posted August 20, 2011 at 3:02 AM (Answer #11)

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Remember: sin is y/r  & cos is x/r

So:

LHS= sin ²x + cos ²x

= (y/r  ) ² (x/r) ²

=y²/r²  + x²/r²

= (y² + x²)/ r²                                        r² = x² + y² (theorem of Pythagoras)

= r²/r²

=1

= RHS

 

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crazynitin1998 | Student, Grade 9 | Salutatorian

Posted June 23, 2011 at 8:19 PM (Answer #12)

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Keep in mind that sin t = y/r, cos t = x/r, and r^2 = x^2+y^2.  By substitution, sin^2x = y^2/r^2 and cos^2x = x^2/r^2.  So the left hand side of the identitiy becomes (y^2/r^2) + (x^2/r^2).  Since these already have a common denominator, we can write them as one fraction:  (y^2+x^2)/r^2 = r^2/r^2 = 1.

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tilakkumar | Student, Grade 11 | Honors

Posted July 5, 2011 at 10:26 PM (Answer #13)

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if sinx=p/randcosx=q/r then,

sin^2x=p^2/r^2

cos^2x=q^2/r^2

sin^2x+cos^2x=p^2/r^2+q^2/r^2=p^2+q^2/r^2=r^2/r^2=1

in right angled triangle p^2+q^2=r^2

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tilakkumar | Student, Grade 11 | Honors

Posted July 5, 2011 at 10:28 PM (Answer #14)

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if sinx=p/randcosx=q/r then,

sin^2x=p^2/r^2

cos^2x=q^2/r^2

sin^2x+cos^2x=p^2/r^2+q^2/r^2=p^2+q^2/r^2=r^2/r^2=1

in right angled triangle p^2+q^2=r^2

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sureshpandey | Student, Undergraduate | Honors

Posted August 30, 2011 at 10:21 PM (Answer #15)

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sin2x + cos2x =sinx sinx + cosx cosx =cos x cos x +sinx sinx =cos(x-x) =cos0 =1
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egotripper | Student, Undergraduate | eNoter

Posted September 6, 2011 at 7:10 AM (Answer #16)

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Proof by Euler's formula:

e^(i*x) = cos(x) + i*sin(x)

 

-2*sin(x) = i*e^(i*x) -i* e^(-i*x)

sin(x) = 1/(2*i) * (e^(i*x)-e^-(i*x))

2*cos(x) = e^(i*x)+e^(-i*x)

cos(x) = (1/2) * (e^(i*x)+e^-(i*x))

 

so cos(x)^2 = (1/4)*(e^(2*i*x)+e^-(2*i*x)+2*e^(0))

and sin(x)^2 = -(1/4)*(e^(2*i*x)+e^-(2*i*x)-2*e^(0))

adding these 2 you can get

cos(x)^2 + sin(x)^2 = (2/4) * e^(0) + (2/4) * e^(0) = 1

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vattamattam | College Teacher | eNoter

Posted September 28, 2011 at 1:18 AM (Answer #19)

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  1. Case(1) x is an acute angle. 
    In triangle ABC, let B be a right angle, x be the measure of angle A, AB = c, BC = a, CA = b.
    Then sin x = a/b, cos x = c/b.
    sin^2 x + cos^2 x = a^2/b^2 + c^2/b^2 =(a^2+c^2)\b^2= b^2/b^2 = 1, by Pythagoras theorem.
  2. Case(2) x is any angle.
    An angle whose initial side coincides with OX, in the xy-plane, is said to be in the standard position. The circle with center at the origin and radius 1, is called the unit circle. 
    Let x be the measure of an angle (in the positive or negative direction) in the standard position. Suppose its terminal side cuts the unit circle at P. Then (cos x, sin x) are the coordinates of P. Since P is on the unit circle,
    cos^2 x + sin^2 x = 1.

arddirection

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reachanukriti | Student, Grade 10 | Honors

Posted October 24, 2011 at 5:26 PM (Answer #20)

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in right triangle ABC with angle  B=90

sin A=BC/AC and cos A=AB/AC

sin^2A+cos^2A= (BC/AC)^2 + (AB/AC)^2

= (BC^2+AB^2)/AC^2,        as, AB^2+BC^2=AC^2

=AC^2/AC^2=1

hence proved sin^2A+cos^2A=1

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rimi9 | Student, Grade 11 | eNoter

Posted November 5, 2011 at 9:36 PM (Answer #17)

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L.H.S

=sin^2X + cos^2X

=sinXsinX+cosXcosX

=cos(X-X)

=cos0

=1

R.H.S

hence proved :)

 

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uranium9 | Student, Grade 11 | eNoter

Posted November 29, 2011 at 11:45 AM (Answer #21)

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In a right angle triangle,

sinx=perpendicular(p)/hypotenuse(h)

cosx=base(b)/hypotenuse(h)

Now,

L.H.S=Sin^2x+Cos^2x

=P^2/h^2 + b^2/h^2

=(p^2+b^2)/h^2

=h^2/h^2   (p^2+b^2 =h^2)

=1

proved

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crazynitin1998 | Student, Grade 9 | Salutatorian

Posted January 27, 2012 at 11:06 PM (Answer #22)

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sin^2X + cos^2X = 1

L.H.S

=sin^2X + cos^2X

=sinXsinX+cosXcosX

=cos(X-X)

=cos0

=1

R.H.S

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bishalnaskar | Student, Undergraduate | Salutatorian

Posted February 24, 2012 at 3:57 PM (Answer #23)

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sin^2X + cos^2X = 1

L.H.S

=sin^2X + cos^2X

=sinXsinX+cosXcosX

=cos(X-X)

=cos0

=1= R.H.S hence proved!!

 

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mohanchandru17 | eNoter

Posted July 5, 2011 at 3:08 AM (Answer #18)

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sin^2x=1-cos2a/2,cos^2x=1+cos2a/2

=(1-cos2a/2)+(1+cos2a/2)

=2/2

=1

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sushrut52 | Student, Undergraduate | eNoter

Posted September 3, 2011 at 9:29 PM (Answer #25)

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Simple!!!

Consider a Right angled triangle ABC,

By Pythagorean theorem,

AB^2+BC^2=AC^2

Divide by AC^2..,

AB^2/AC^2+BC^2/AC^2=AC^2/AC^2

(Opposite/hypotenuse)^2+(adjacent/Hypotenuse)^2=AC^2/AC^2

Sin^2x+cos^2x=1


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zeeshankamal007 | Student, Grade 10 | Honors

Posted February 10, 2012 at 4:30 PM (Answer #27)

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Since, SIN X =P/H & COS X = B/H

Therefore,SIN^2X=(P/H)^2 & COS^2X= (B/H)^2

Therefore, SIN^2X + COS^2X=(P/H)^2 + (B/H)^2

Therefore, (P/H)^2 + (B/H)^2 = (H/H)^2     [since, H^2=P^2 + B^2]

Therefore, (H/H)^2= H^2/H^2

Threfore, H^2/H^2 = 1 (ANSWER)

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pitabasa007 | Student, Grade 10 | Honors

Posted July 9, 2011 at 6:15 PM (Answer #28)

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It is the most simple answer.

PROOF- In a right angled triangle ABC

SIN Ф=BC/AC

SIN^2Ф=BC^2/AC^2

COS Ф=AB/AC

COS^2 Ф= AB^2/AC^2

On adding, SIN^2 Ф + COS^2 Ф

= BC^2/AC^2 + AB^2 AC^2 SIN^2 Ф + COS^2 Ф

= (BC^2 + AB^2)/AC^2= AC^2/AC^2=1 (BY PYTHAGORUS           THEOREM,AC^2=BC^2+AC^2)

SIN^2 Ф + COS^2 Ф =1

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gradegrumber | Student, Undergraduate | Salutatorian

Posted August 15, 2011 at 12:52 AM (Answer #24)

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this is probably the most simple identity question there is becausewhen you plug in any value for x, you will get an answer of 1.

1 = sin^2 x + cos^2 x

sin^2 x + cos^2 x  = 1

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shivangivardhan | Student, Grade 11 | Honors

Posted September 5, 2011 at 11:48 AM (Answer #26)

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sin theta=p/h

cos theta=b/h

Therefore,

sin^2 theta=p^2/h^2

cos^2 theta=b^2/h^2

Now,

sin^2 theta + cos^2 theta=(p^2/h^2) + (b^2/h^2)

=(p^2 + b^2)/h^2

By pythagoras' theorem,

p^2 + b^2=h^2

therefore,

(p^2 + b^2)/h^2 = h^2/h^2

=1

Hence, proved

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mksb | Student | Salutatorian

Posted September 10, 2011 at 6:58 PM (Answer #30)

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sin^2X + cos^2X = 1

L.H.S

=sin^2X + cos^2X

=sinXsinX+cosXcosX

=cos(X-X)

=cos0

=1

R.H.S

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iqbalwassan | Student, Undergraduate | eNoter

Posted September 13, 2011 at 8:32 PM (Answer #32)

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sin^2X + cos^2X = 2

L.H.S

=sin^2X + cos^2X

=sinXsinX+cosXcosX

=cos(X-X)

=cos0

=1

R.H.S

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rnk | Student, Grade 10 | Honors

Posted October 5, 2011 at 5:09 PM (Answer #33)

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Suppose in a right angled triangle 'a' is the base, 'b' is the perpendicular and c is the perpendicular thnn by Pythagoras theorom,

a² + b² = c²

Dividing by c² throughout we get,

(a/c)² + (b/c)² = 1

Since base/hypotenus = cosØ and perpendicular/hypotenus = sinØ

Therefore,

cosز + sinز = 1

or sinز + cosز = 1

 

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sushila69 | Student, Undergraduate | eNoter

Posted October 26, 2011 at 5:35 AM (Answer #34)

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I must report that the question formulation is incorrectly written:

Sin^2x is not a correct expression.  It should be written as:

(Sin x)^2

The same goes for   Cos^2x .   It should be expressed as:

(Cos x)^2

Hence, the expression should be written as:

 

(Sin x)^2 + (Cos x)^2 = 1.0

The proof presented by others appears to be appropriate.

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abc-ahmed | Student, Undergraduate | eNoter

Posted December 21, 2011 at 2:02 PM (Answer #35)

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It could be solved by polar coordinates as

(sin x)^2 + (cos x)^2  = 1        multiplying by r^2

(r sin x)^2 + (r cos x)^2 = r^2

 

but if we have any inclined line with length r and angle like (x), then we can to resolve it by two components H (horizontal) and V (vertical), so that:

H= r cos x   and V=r sin x

hence: H^2   +   V^2  =  r^2

but, Phythagoras Theorem says: for any inclined line, its sequare  length (r) equal to sum of sequares of the verical and horizontal projection (H and V), so that: r^2= H^2   +  V^2

Lastly,

r^2 = r^2

i. e. 1=1

 

 

 

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shihan | Student, Undergraduate | Salutatorian

Posted January 5, 2012 at 11:17 AM (Answer #36)

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sin^2X + cos^2X = 1 L.H.S =sin^2X + cos^2X =sinXsinX+cosXcosX =cos(X-X) =cos0 =1

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kareemaths | College Teacher | Honors

Posted February 11, 2012 at 11:25 PM (Answer #37)

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Let consider a right angled triangle ABC in which angle C is a right angle and measure of angle A is x

so we have three side for triangle ABC

i.e a (opposite to angle x), b and c (opposite to right angle)

by pythagoras theorem a^2 + b^2 = c^2

dividing both sides by c^2 we get

a^2 / c^2 + b^2/ c^2 = c^2 / c^2

or   (a/c)^2 + (b/c)^2 = 1

or  (sinx)^2 + (cosx)^2 = 1     :since a/c = sinx, b/c = cosx

thus sin^2x + cos^2x = 1

proved...

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ashleyblowes | Student, Grade 9 | Honors

Posted February 27, 2012 at 6:55 AM (Answer #38)

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Keep in mind that sin t = y/r, cos t = x/r, and r^2 = x^2+y^2.  By substitution, sin^2x = y^2/r^2 and cos^2x = x^2/r^2.  So the left hand side of the identitiy becomes (y^2/r^2) + (x^2/r^2).  Since these already have a common denominator, we can write them as one fraction:  (y^2+x^2)/r^2 = r^2/r^2 = 1

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vikas17 | College Teacher | eNoter

Posted February 29, 2012 at 3:21 AM (Answer #39)

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sin x=alt/hyp

cos x=base/hyp

now sin^2 x +cos^2 x =(alt^2/hyp^2)+(base^2/hyp^2)=(alt^2+base^2)/(hyp^2)

now b'z as per pythagorous theoram, alt^2+base^2=hyp^2. replacing the same in the above equation.

 

Sin^2 x+ Cos^2 x = Hyp^2/Hyp^ 2 =1 hence proved.

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twinkysweet30 | Student, Grade 11 | Salutatorian

Posted March 9, 2012 at 12:45 PM (Answer #40)

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sin^2X + cos^2X = 1

L.H.S

=sin^2X + cos^2X

=sinXsinX+cosXcosX

=cos(X-X)

=cos0

=1

R.H.S

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gcroc12321 | Student, Undergraduate | eNoter

Posted March 17, 2012 at 5:21 PM (Answer #41)

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consider a right triangle ABC with right angle at B. AB(a) and BC(b) are the two legs and AC(c) is the hypotenuse. By pythagoras theorem, you get:

a^2+b^2=c^2. Divide the whole thing by c^2.

(a/c)^2+(b/c)^2=1 (for the angle A, cosA=a/c and sinA=b/c)

=>(cos^2)A+(sin^2)A=1

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terellhoward | Honors

Posted September 4, 2011 at 9:22 AM (Answer #29)

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Use the Pythagorean Theorem

r^2=y^2+x^2

Then divide throughout by r^2

y^2/r^2 + x^2/y^2 = 1

then subsituting sin Θ = y/r, cos Θ = x/r

sin^2x + Cos^2x = 1

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sayantani-basu-modern-high-school | Student, Grade 10 | Honors

Posted February 21, 2012 at 3:35 PM (Answer #43)

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Since the radius is also the hypotenuse of the right triangle formed by the angle "x" within the circle, the sine is y and the cosine is x.  By the Pythagorean Theorem, x^2 + y^2 = 1^2 ... or x^2 + y^2 = 1

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utilityfan | Student, College Freshman | Valedictorian

Posted April 19, 2012 at 4:22 AM (Answer #45)

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If you subsitute any two same values as x into the equation: (Sin x)^2+(Cos X)^2; the resultant will always be 1. Try it!

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pankajgarg9696 | Student, Grade 10 | Honors

Posted September 13, 2011 at 11:40 AM (Answer #31)

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HIT & TRIAL METHOD - JUST PUT THE ANY VALUE OF ANGLE  (say x ),

you will u get the answer 1.

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crazynitin998 | Student, Grade 10 | Salutatorian

Posted May 25, 2012 at 12:32 PM (Answer #48)

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sin^2X + cos^2X = 1

L.H.S

=sin^2X + cos^2X

=sinXsinX+cosXcosX

=cos(X-X)

=cos0

=1

Hence proved

LHS=RHS

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chaobas | College Teacher | Valedictorian

Posted June 30, 2012 at 12:45 AM (Answer #50)

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It can be done through unit circle method. But using pythogorous therorem would be easy.

 

Let us consider a right angled triangle with one of the angle as "x" which is formed by  "l" as opposite and "b" as adjacent.

Now,

Sin x = opposite / hypotenuse 

Cos x = adjacent / hypotenuse

Now,

(sinx)^2 + (cos x)^2 = (opposite / hypotenuse )^2 + (adjacent / hypotenuse)^2

= opposite^2+adjacent^2/hypotenuse^2

=hypotenuse^2/hypotenuse^2                                                   [as opposite^2+adjacent^2 = hypotenuse^2 ---pythogorous therorem]

=1^2


So

(sin x)^2 + (Cos)^2 = 1

Sources:

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thedon75 | eNoter

Posted February 23, 2012 at 8:44 AM (Answer #47)

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You can also reverse this formula backwards to the Pythagorean theorem by thinking in terms of the definitions of sin = o/h
and cos = a/h using opposite, adjacent and hypoteneuse.

Then sin^2+cos^2 = 1 can be redefined as:



Which is the same as:




Rearranging this we get:



Look familiar?

It's the same as the good old Pythagorean theorm which is:

with a and b as sides o and a and c as the hypoteneuse.

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magnetic | Student, Undergraduate | eNoter

Posted May 29, 2012 at 2:32 PM (Answer #52)

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let 2x=A

sin(A)=exp(iA)-exp(-iA))/2i

cos(A)=exp(iA)+exp(-iA))/2

and solve (exp(iA)-exp(-iA))/2i)^2+(exp(iA)+exp(-iA))/2)^2

remembering that i=sqrt(-1) and (X^a)(Y^b)=XY^(a+b)

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lochana2500 | Student, Undergraduate | Valedictorian

Posted June 2, 2012 at 5:22 PM (Answer #53)

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Q: Prove - sin²x + cos²x =1

A :  L:H:S ≡ sinx.sinx + cosx.cosx

               = cos(x-x)

               = cos 0

               = 1

 ∴ L:H:S ≡ R:H:S

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dancer4life15 | Student, Grade 11 | Honors

Posted June 12, 2012 at 7:15 PM (Answer #54)

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y= rsinx

x=rcosx

r^2=x^2+y^2

r^2=(rsinx)^2+(rcosx)^2

r^2=(r^2)(sinx)^2+(r^2)(cosx)^2

r^2=r^2[(sinx)^2+(cosx)^2]

(r^2)/(r^2)=(sinx)^2+(cosx)^2

1=(sinx)^2+(cosx)^2

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odamesuccess | Student, Undergraduate | eNotes Newbie

Posted July 12, 2012 at 5:07 AM (Answer #55)

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sin^2X + cos^2X = 1

L.H.S

=sin^2X + cos^2X

=sinXsinX+cosXcosX

=cos(X-X)

=cos0

=1

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forigami | Student, Undergraduate | eNoter

Posted August 8, 2012 at 4:56 PM (Answer #57)

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we can use the triangular circle and the the pethagorian theorem,we can also use this;

cos(a-b)=cosa*cosb+sina*sinb(also proved in the triangular circle) 

 we put; a=b

      we get;cos0=cos²a+sin²a

  we know that cos0=1

 so;         cos²a+sin²a=1

    

 

 

 

 

 

 

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brendanfern | Student, Grade 10 | eNoter

Posted November 2, 2012 at 5:51 AM (Answer #58)

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SinX=Opp/Hyp

CosX=Adj/Hyp

Therefore: Sin²X+Cos²X=(Opp/Hyp)²+(Adj/Hyp)²

                                   = Opp²+Adj²/Hyp² (Taking LCM and adding)

                                   =  Scince Hyp²=Opp²+Adj²

                                   =Hyp²/Hyp²=1

Hence Proved.

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Sameera W | eNoter

Posted January 2, 2013 at 4:38 PM (Answer #59)

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From Basics
Sin(x)= BC/AC

Cos(x)=AB/AC

Now jump to the question

Sin^2 (X)= (BC/AC)^2

COS ^2(X)= (AB/AC)^2

Sin^2 (X) +COS ^2(X) =  (BC/AC)^2 + (AB/AC)^2

 Sin^2 (X) +COS ^2(X)= (BC^2+ AC^2)/AC^2 ------(1)

From Pythagoras biography 

BC^2+ AC^2 = AC^2 -------(2)

Substituting (1) with (2)

Sin^2 (X) +COS ^2(X) =AC^2/AC^2

Sin^2 (X) +COS ^2(X)= 1

                                    

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wanasyraf | Student, Grade 9 | eNoter

Posted November 17, 2011 at 8:12 PM (Answer #49)

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best ANSWER!!

Sin^2x + Cos^2x = 1

ARE BASIC IDENTITIES OF TRINONOMETRIC FUNCTION.HENCE,SIN^2X+COS^2X=1

 

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vaaruni | High School Teacher | Salutatorian

Posted June 13, 2012 at 6:22 AM (Answer #64)

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L.H.S =sin^2x+cos^2x=(p/h)^2+(b/h)^2 [ where, p is perpndicular , b  

  is the base and h is the  ypotenuse of the right angled triangle ]            

       = (p^2/h^2) + (b^2/h^2) = (p^2 + b^2)/(h^2)= (h^2)/(h^2)

 [ using pythagoros theroem In right angled triangje h^2=p^2+b^2  ]

        = 1 = R.H.S   Proved 

 

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pwnd | Student, Grade 10 | eNoter

Posted July 15, 2012 at 4:07 PM (Answer #65)

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sin^2X + cos^2X = 1

since sin^a+cos^a=1,

a=2X

therefore sin^2X+cos^2X=1

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eravick | College Teacher | eNoter

Posted July 18, 2012 at 3:05 AM (Answer #56)

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cool

 

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juniorsilvamath | Student, Undergraduate | Honors

Posted August 8, 2012 at 8:54 PM (Answer #68)

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In a right triangle, according to the Pythagorean theorem, we have:
a ² + b ² = c ².

Thus, if x is a measure of the angle B, then:

sin ² x + cos ² x = (b / a) ² + (c / a) ² =
b ² / a ² + c ² / a ² =
(b ² + c ²) / a ² =

Indeed,
a ² / a ² = 1.

---------------------

You can draw a picture to show. It is even more intuitive.

Sources:

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dlccanada | High School Teacher | eNoter

Posted August 12, 2012 at 9:59 PM (Answer #69)

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sin^2x + cos^2x=1

 

Let we consider a point p(X, Y). Using the unit circle means radious is 1.Connect a line between the origin(0,0) and the point P(X,Y) and form the right angle triangle which base is "X", height is "Y" and hypotaneous is 1. The hypotaneous forms the angle ""x" with base X.Using SOH CAH TOA formula

sinx=opposite/hypotenouse

sinx = Y/1

Y=1*sinx

Y=sinx  ...(1)

again,

cosx =adjacent/hypotenouse

cosx=X/1

X=cosx .......(2)

Now, squaring eqn (1) and (2) and add, we get

X^2 +Y^2 =cos^2 +sin^2 ....(3)

Now using the distance formula to find the distance between the origin(0,0) and  the point (X,Y)/ or pythagoreous theorem, we get

d^2 = (X-0)^2 +(Y-0)^2

d^2=X^2+Y^2, .....(4)

as we drow the unit circle and it's radius is 1 i'e d=1, there fotr, d^2 1,

Now ,from equation (3) = eqn (4)

X^2+Y^2 =cos^2x+sin^2x =d^2 =1

Therefore, sin^2x+cos^2x=1 (proved)

Sources:

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ayushkaka | Student, Grade 9 | eNoter

Posted August 20, 2012 at 1:14 PM (Answer #71)

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sin t = y/r, cos t = x/r, and r^2 = x^2+y^2.  By substitution, sin^2x = y^2/r^2 and cos^2x = x^2/r^2.  So the left hand side of the identitiy becomes (y^2/r^2) + (x^2/r^2).  Since these already have a common denominator, we can write them as one fraction:  (y^2+x^2)/r^2 = r^2/r^2 = 1. 

 

Be Carefull!

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anmes | Student, Grade 11 | eNotes Newbie

Posted August 25, 2012 at 11:11 AM (Answer #73)

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the best way to do so is by using pythagoras theorm.

we know sinx = p/h

             cosx = b/h

so using this we can find that,

   (p/h)^2 + (b/h)^2

= (p^2 + b^2)/h^2

= h^2/h^2

= 1

         hence proved

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moukthika | Student, Grade 10 | eNoter

Posted August 29, 2012 at 10:22 AM (Answer #74)

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to prove sin^2x + cos^2x = 1

proof:from the identity sin^2theeta +cos^2 theeta=1

therefore, sin^2 theeta=1- cos^2 theeta

L.H.S    1-cos^2x + cos^2x =1 

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gavchristian | Student, Undergraduate | Honors

Posted September 8, 2012 at 12:14 PM (Answer #75)

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You can prove it by using pythagoras' theorem using the following equation:   r^2=y^2+x^2

-Then divide throughout by r^2

-y^2/r^2 + x^2/y^2 = 1

then subsituting sin Θ = y/r, cos Θ = x/r

sin^2x + Cos^2x = 1

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twinkysweet30 | Student, Grade 11 | Salutatorian

Posted September 16, 2012 at 4:39 AM (Answer #76)

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sin^2X + cos^2X = 1

L.H.S

=sin^2X + cos^2X

=sinXsinX+cosXcosX

=cos(X-X)

=cos0

=1= R.H.S hence proved!!

 

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krishnasis123 | Student, Grade 9 | Honors

Posted September 17, 2012 at 9:28 AM (Answer #77)

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We know, sin x = p/h

and,         cos x = b/h

sin²x+cos²x

=(p/h)²+(b/h)²

=p²/h²+b²/h²

=(p²+b²)/h²

But we know, p²+b²=h²    (By pythagoras theorem) so,

=h²/h²

=1

=1 

 

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johnnychi | Student, Grade 11 | eNoter

Posted October 11, 2012 at 9:37 PM (Answer #78)

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the identity is sin^2x + cos^2x=1

so, 1=1

or you can do the long way

Sin^2x=1-cos^2x

1-cos^2x+cos^2x     (the -cos^2x and the +cos^2x  cancle out )

1=1

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chili32 | Student | eNoter

Posted November 11, 2012 at 3:15 AM (Answer #81)

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Q: (SinX)^2 + (Cos X)^2 =1

A:Substitute X = 45 degrees or pi/ 4

         Sin(45 deg) =1/(square root of 2) and Sin(pi/4) = 1/(square root of 2)

Cos(45 deg) =1/(square root of 2) and Cos(pi/4) = 1/(square root of 2)

So, (SinX)^2 = 1/2 -------- (a). Similarly (CosX)^2 =1/2-----(b)

When (a) and (b) equations are added then we get

(SinX)^2 + (Cos X)^2 =1/2 + 1/2 =1.

therefore (SinX)^2 + (Cos X)^2 =1

             OR

Similarly Sin(0 deg) =0, Cos(0 deg)=1

when substituiting these in the LHS of ques we get

(SinX)^2 + (Cos X)^2 = 0 +1 =1

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eravick | College Teacher | eNoter

Posted December 20, 2012 at 11:25 PM (Answer #83)

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sin^2x + cos^2x=1

 

Let we consider a point p(X, Y). Using the unit circle means radious is 1.Connect a line between the origin(0,0) and the point P(X,Y) and form the right angle triangle which base is "X", height is "Y" and hypotaneous is 1. The hypotaneous forms the angle ""x" with base X.Using SOH CAH TOA formula

sinx=opposite/hypotenouse

sinx = Y/1

Y=1*sinx

Y=sinx  ...(1)

again,

cosx =adjacent/hypotenouse

cosx=X/1

X=cosx .......(2)

Now, squaring eqn (1) and (2) and add, we get

X^2 +Y^2 =cos^2 +sin^2 ....(3)

Now using the distance formula to find the distance between the origin(0,0) and  the point (X,Y)/ or pythagoreous theorem, we get

d^2 = (X-0)^2 +(Y-0)^2

d^2=X^2+Y^2, .....(4)

as we drow the unit circle and it's radius is 1 i'e d=1, there fotr, d^2 1,

Now ,from equation (3) = eqn (4)

X^2+Y^2 =cos^2x+sin^2x =d^2 =1

Therefore, sin^2x+cos^2x=1 (proved)

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shihan | Student, Undergraduate | Salutatorian

Posted January 21, 2013 at 2:08 PM (Answer #85)

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sin^2X + cos^2X = 1

L.H.S

=sin^2X + cos^2X

=sinXsinX+cosXcosX

=cos(X-X)

=cos0

=1

R.H.S

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sat2010 | Student, Grade 10 | eNoter

Posted February 27, 2013 at 2:59 PM (Answer #99)

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sin^2X + cos^2X = 1

L.H.S

=sin^2X + cos^2X

=sinXsinX+cosXcosX

=cos(X-X)

=cos0

=1

L.H.S=R.H.S

hence proved

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leedygood7 | High School Teacher | eNoter

Posted May 12, 2012 at 1:48 AM (Answer #90)

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In a right triangle with a as hypotenuse and b and c as other legs,
sinA=c/a and cosA=b/a

(sinA)^2+(cosA)^2
= (c/a)^2+(b/a)^2
=(b^2+c^2)/a^2
=a^2/a^2       (by the pythagorean theorem in which a^2=b^2+c^2)
=1

Therefore, (sinA)^2+(cosA)^2 =1

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aniljha127 | High School Teacher | eNoter

Posted June 11, 2012 at 4:39 PM (Answer #92)

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sinx=p/h

cosx=b/h

(sinx)^2+(cosx)^2 = (p^2+b^2)/h^2=h^2/h^2=1

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wimudu | Student, Undergraduate | eNoter

Posted July 16, 2012 at 6:30 AM (Answer #93)

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we can write cos4x = 1- 2 sin^2 x

and cos4x = 2Cos^2x -1

so sin^2 2x+ cos^2 2x = 1/2- cos4x/2+ cos4x/2 +1/2

                                    1/2+1/2=1

 

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jeanclaudedusse | eNoter

Posted October 16, 2012 at 12:01 PM (Answer #79)

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The problem when you ask to prove something is that it must be clear what you already know.

Many people propose to solve it by calling upon pythagoras' theorem, which is basically what you want to prove. And if you already have defined the sinus and cosinus of an angle as the ratio of the length of the small sides of a right-angled triangle to its hypotenuse, then it is straightforward to see your answer in a circle of radius 1.

BUT. If all you are supposed to know about sinus and cosinus is that they are the solutions of the differential equation y" + y = 0 such that y(0) = 0 and y'(0) = 1 for the former and  y(0) = 1 and y'(0) = 0  for the latter, then you have to do otherwise:

1) prove that sin' = cos and cos' = -sin (using the fact that there is only one function such as y" + y = 0 and y(0) = a and y'(0) = b)

2) differentiate cos² + sin²

3) conclude

 

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ds6 | Student, Undergraduate | Honors

Posted December 11, 2012 at 1:05 AM (Answer #94)

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sin^2X + cos^2X = 1

left side

=sin^2X + cos^2X                 use the compound angle fomula

=sinXsinX+cosXcosX              of cos(x-y) = sinxsiny + cosxcosy  

=cos(X-X)

=cos0

=1

 

 

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wescarlyon337 | Student, Grade 10 | Salutatorian

Posted February 3, 2013 at 4:27 PM (Answer #86)

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sin^2x=1-cos2a/2,cos^2x=1+cos2a/2

=(1-cos2a/2)+(1+cos2a/2)

=2/2

=1

If you haven't got the answer already :)

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pramodpandey | College Teacher | Valedictorian

Posted February 23, 2013 at 7:42 AM (Answer #98)

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Consider right angle triangle with one its angle measure degree x. let its sides other than hypotenuse are m and n.so hypotenuse r=squar( squa m +squa n)

so if sinx= m/r so cos x=n/r

squa sinx+ squar cosx= squar(m/r)+squar(n/r)= (squarm+squarn)/squar r= 1

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vivek2708 | Student, Undergraduate | eNotes Newbie

Posted April 3, 2013 at 8:34 PM (Answer #100)

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we know that ,
sin x=p/h,
cos x=b/h,
L.H.S=sin^2 x + cos^2 x
     =(p/h)^2+ (b/h)^2
     =((p^2)/(h^2)) + ((b^2)+(h^2))

      =((p^2)+ (b^2))/(h^2)

       WE KNOW PYTHAGORAS THEOREM(H^2=P^2+B^2)
    THEN,
     =(H^2)/(H^2)=1 = R.H.S



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user2350623 | eNotes Newbie

Posted July 9, 2013 at 8:08 AM (Answer #102)

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We can answer this by using pythagorean theorem:

soh, cah, toa, but soh cah is enough to prove this;

let: sine=b/c; then cosine=a/c :.take note of this a²+b²=c²

substitute in the equation 

(b/c)²+(a/c)²=b²+a²/c² :.you notice that b²+a² is equal to

therefore c²/c² is equal to 1... 

I hope you accept my Proving...thanks <3

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pattystephens | High School Teacher | (Level 1) Adjunct Educator

Posted July 22, 2013 at 9:45 PM (Answer #103)

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The equation of the unit circle is x^2+y^2=1.

All points on this circle have coordinates that make this equation true. 

For any random point (x, y) on the unit circle, the coordinates can be represented by (cos `theta` , sin `theta` ) where `theta` is the degrees of rotation from the positive x-axis (see attached image). 

By substituting cos `theta` = x and sin `theta` = y into the equation of the unit circle, we can see that (cos `theta` )^2 + (sin `theta` )^2 = 1. 

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babifat16 | Student, Grade 11 | eNoter

Posted June 10, 2012 at 3:37 PM (Answer #91)

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think of a circle. if you have x^2 + y^2 = 1 and you know that cos represents x and sin represents y then you know that sin^2x + cos^2x = 1. 

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tanvipuri | Student | eNotes Newbie

Posted August 16, 2012 at 2:20 PM (Answer #70)

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sin^a + cos^a =1

taking 2x=y

sin^y +cos^y=1

therefore, sin^2x + cos^2x=1

 

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saramsha | Student, Grade 11 | eNotes Newbie

Posted November 1, 2012 at 11:15 AM (Answer #80)

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What a shitty question,............

Do you think im a fool who cant answer it........

answer is..........

                 sin^2x+cos^2x

                 =sin^2x+(1-sin^x)

                  =sin^2x+1-sin^2x

                   =1     [since,sin^2x-sin^2x=0]

          Therefore,1=1          proved........

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eravick | College Teacher | eNoter

Posted December 20, 2012 at 11:26 PM (Answer #84)

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I don't think that's possible!!!

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abhi4307 | Student, Grade 11 | Valedictorian

Posted May 26, 2013 at 6:12 AM (Answer #101)

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L.H.S Sin^2x+Cos^2x =P^2/h^2+b^2/h^2 =(p^2+b^2)/h^2 =h^2/h^2=1=r.h.s. proved.
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vinish | Student, Undergraduate | Valedictorian

Posted May 9, 2012 at 1:13 PM (Answer #89)

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Because the answer should be 1 and no other answer.

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gelsaied | TA , College Freshman | Honors

Posted September 22, 2013 at 3:26 AM (Answer #104)

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sin (x) ^2 + cos (x) ^2=1

cos(x)cos(x)+sin(x)sin(x)=1

cos(x-x)=1

Using the sum difference identity:

cos(A-B)=cos(A)cos(B)+sin(A)Sin(B)

cos(0)=1

1=1

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rimmery | TA , Undergraduate | Honors

Posted May 14, 2014 at 12:15 AM (Answer #105)

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The easiest way to prove this requires knowing the following identity: 

`cos(A-B)=cosAcosB+sinAsinB`

So starting from the equation we have:

`sin^2x+cos^2x=1`

But `sin^2x` is just ` sinx * sinx`

So `sin^2x+cos^x=sinx*sinx+cosxcosx`

Notice that the above equation is in the exact format as the right-hand side of the first identity where `A=B=x`

Therefore we get `cos(x-x)`

Then we are left with `cos0`

Since `cos0=1=R.H.S` 

We are done!

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