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.

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

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.

sin^2X + cos^2X = 1

L.H.S

=sin^2X + cos^2X

=sinXsinX+cosXcosX

=cos(X-X)

=cos0

=1

R.H.S

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.

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**

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)

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

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

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

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

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.

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

**L.H.S**

**=sin^2X + cos^2X**

**=sinXsinX+cosXcosX**

**=cos(X-X)**

**=cos0**

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

sin^2X + cos^2X = 1

L.H.S

=sin^2X + cos^2X

=sinXsinX+cosXcosX

=cos(X-X)

=cos0

=1

R.H.S

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

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

- 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. - 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

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

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

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.

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)

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

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*

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

sin^2X + cos^2X = 1

L.H.S

=sin^2X + cos^2X

=sinXsinX+cosXcosX

=cos(X-X)

=cos0

=1

R.H.S

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.

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

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...

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

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.**

sin^2X + cos^2X = 2

L.H.S

=sin^2X + cos^2X

=sinXsinX+cosXcosX

=cos(X-X)

=cos0

=1

R.H.S

sin^2X + cos^2X = 1

L.H.S

=sin^2X + cos^2X

=sinXsinX+cosXcosX

=cos(X-X)

=cos0

=1

R.H.S

*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*

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

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

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!

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

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,

cosØ² + sinØ² = 1

or sinØ² + cosØ² = 1

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

sin^2X + cos^2X = 1

L.H.S

=sin^2X + cos^2X

=sinXsinX+cosXcosX

=cos(X-X)

=cos0

=1

Hence proved

LHS=RHS

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

you will u get the answer 1.

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

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*

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.

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**

sin^2X + cos^2X = 1

L.H.S

=sin^2X + cos^2X

=sinXsinX+cosXcosX

=cos(X-X)

=cos0

=1

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

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

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)

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.

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

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)

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

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

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 **

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

**L.H.S**

**=sin^2X + cos^2X**

**=sinXsinX+cosXcosX**

**=cos(X-X)**

**=cos0**

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

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

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

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

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!

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)

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.

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

## best ANSWER!!

## Sin^2x + Cos^2x = 1

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

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

**c²**

**therefore c²/c² is equal to 1... **

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

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

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

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 :)

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

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

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

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

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.

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!

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

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........