Trigonoemtric Identity

[basty]

Why $\sin^2 2x + \cos^2 2x = 1$?

Will

$\sin^2 3x + \cos^2 3x$

or

$\sin^2 4x + \cos^2 4x$

and so on, be = 1?

How to proof this?

 

[Bystander]

You know the definitions of the functions sin(x) and cos(x); you also are familiar with the Pythagorean theorem.

 

[basty]

You know the definitions of the functions sin(x) and cos(x); you also are familiar with the Pythagorean theorem.

I am familiar with the Pythagorean theorem but I don't know the definitions of the functions sin(x) and cos(x).

Explain please.

 

[Bystander]

You have got to be kidding me. Given a right triangle, we'll plot its vertices on Cartesian coordinates (0,0), (1,0), and (1,y) where y = 0 to +∞, and the interior angle at (0,0), α = 0 to 90°, or 0 to π/2 radians, sin(α) ≡ y/(x² + y²)^1/2, or the ratio of the length of the side of the triangle opposite ∠α to the length of the hypotenuse of the right triangle. cos(α) ≡ the ratio of the length of the base of the triangle, x, to the length of the hypotenuse.

Now, give it a try.

Edited to include sqrt --- my bad.

 

[basty]

You have got to be kidding me. Given a right triangle, we'll plot its vertices on Cartesian coordinates (0,0), (1,0), and (1,y) where y = 0 to +∞, and the interior angle at (0,0), α = 0 to 90°, or 0 to π/2 radians, sin(α) ≡ y/(x² + y²), or the ratio of the length of the side of the triangle opposite ∠α to the length of the hypotenuse of the right triangle. cos(α) ≡ the ratio of the length of the base of the triangle, x, to the length of the hypotenuse.

Now, give it a try.

Why $\sin (a) = \frac{y}{x^2 + y^2}$?

 

[Mark44]

The identity is $sin^2(\text{whatever}) + cos^2(\text{whatever}) = 1$. Do you see how this fits with the questions you asked in post #1?

Why $\sin (a) = \frac{y}{x^2 + y^2}$?

It's not - there's a radical that's missing. It should be
$\sin (a) = \frac{y}{\sqrt{x^2 + y^2}}$

If you don't understand where that comes from, you need to review right triangle trigonometry.