How do you verify $sin^2 x + sin^2(pi/2 -x) = 1$ ?

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How do you verify $sin^2 x + sin^2(pi/2 -x) = 1$ ?

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Answer 1 · 2015-05-15T19:20:19

On the trig unit circle:

$sin ( pi/2 - x) = cos x$ , then

$sin^2 x + sin^2 (pi/2 - x) = sin^2 x + cos^2 x = 1$

Answer 2 · 2015-05-17T19:37:44

If $0 < x < pi/2$ then another way to picture this is as follows:

Construct a right angled triangle with angles $x$ , $(pi/2 - x)$ and $pi/2$ , with hypotenuse of length $1$ . It is possible to do this since these angles add up to $pi$ .

Then the side opposite the angle $x$ will have length $sin x$ and the side opposite the angle $(pi/2 - x)$ will have length $sin (pi/2 - x)$ .

By Pythagoras theorem, the sum of the squares of the lengths of these sides is equal to the square of the length of the hypotenuse.

So $sin^2 x + sin^2 (pi/2 - x) = 1^2 = 1$

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How do you verify $sin^2 x + sin^2(pi/2 -x) = 1$ ?

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