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