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Please demonstrate this equation: (sin^2)x + (cos^2)x=1

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drole | Student, Grade 10 | (Level 1) Honors

Posted March 12, 2009 at 4:07 AM via web

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Please demonstrate this equation:

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

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

Posted March 12, 2009 at 9:41 AM (Answer #1)

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For our purposes, I will also create a right triangle with the sides a = 3, b = 4, and c = 5 to use as my example.  Remember that these sides are opposite of angles using the same letter.  Side a is opposite angle A, side b is opposite angle B, etc.  The hypotenuse is always the largest number in our pythagorean triple.

 If I were finding sine of this triangle using angle A, sine would be (opposite/hypot.) 3/5.  Next cosine of angle A would be (adjacent/hypot.) 4/5. 

The formula sine^2 + cosine^2 = 1 works because (3/5)^2 +(4/5)^2 = 1.  (3/5)^2 is 9/25 and (4/5)^2 is 16/25.  9/25 +16/25 = 25/25 which is one. 

We could also use angle B to prove our formula as sine of angle B is 4/5 and cosine of angle B is 3/5.  (4/5)^2 +(3/5)^2 = 1 since 16/25 + 9/25 = 25/25 which is one. 

Please feel free to ask other questions if you need!

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neela | High School Teacher | (Level 3) Valedictorian

Posted June 4, 2009 at 4:55 PM (Answer #2)

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Take a right angled triagle,ABC with right angle at B. And AC is the hypotenuse.Then sine of angle A and cosine of angle A are  defined as follows:

sineA = opposite side /hypotenuse = BC/AC.

cosineA =adjascent side /hypotenuse= AB/AC.

 

But since ABC being right angled at B ,AB^2+BC^2 =AC^2,by Pythagorus theorem.

If we divide both sides by AC^2 , the result of Pythagrus above becomes:

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

(AB/AC)^2 + (BC/AC)^2 = 1.

(CosA)^2+(sinA)^2=1. QED.

 

 

 

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jeyaram | Student, Undergraduate | (Level 1) Valedictorian

Posted September 13, 2009 at 10:11 PM (Answer #3)

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

L.H.S

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

=sinXsinX+cosXcosX

=cos(X-X)

=cos0

=1

=R.H.S

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