Please demonstrate this equation:

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

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.

(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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