Trigo value

**cloud5** · December 11th 2010, 08:25 AM

Can anyone explain how the step(red boxed) formed?

Q:

A:

**e^(i*pi)** · December 11th 2010, 08:31 AM

It's the phase difference between cos(x) and sin(x). It can be shown using the addition formula for cos.

$\cos(A-B) = \cos A\cos B + \sin A \sinB$ . When $A = \frac{\pi}{2}$ and $b = \theta$ you get:

$\cos \left(\dfrac{\pi}{2} - \theta \right) = \cos \left(\dfrac{\pi}{2}\right) \cos \theta + \sin \left(\dfrac{\pi}{2}\right) \sin \theta$

Since $\cos \left(\dfrac{\pi}{2}\right) = 0 \text{ and } \sin \left(\dfrac{\pi}{2}\right) = 1$ the equation above simplifies to give $\cos \left(\dfrac{\pi}{2} - \theta \right) = \sin \theta$

**cloud5** · December 11th 2010, 06:43 PM

Originally Posted by e^(i*pi)

It's the phase difference between cos(x) and sin(x). It can be shown using the addition formula for cos.

$\cos(A-B) = \cos A\cos B + \sin A \sinB$ . When $A = \frac{\pi}{2}$ and $b = \theta$ you get:

$\cos \left(\dfrac{\pi}{2} - \theta \right) = \cos \left(\dfrac{\pi}{2}\right) \cos \theta + \sin \left(\dfrac{\pi}{2}\right) \sin \theta$

Since $\cos \left(\dfrac{\pi}{2}\right) = 0 \text{ and } \sin \left(\dfrac{\pi}{2}\right) = 1$ the equation above simplifies to give $\cos \left(\dfrac{\pi}{2} - \theta \right) = \sin \theta$

What about the $cos 3\theta$ from the beginning? Assume it's equal to $cos (\dfrac{\pi}{2} - \theta)$ ?

**snowtea** · December 11th 2010, 07:10 PM

The problem states that cos(3\theta) = sin(\theta)
and sin(\theta) = cos(pi/2 - \theta) as shown above, so cos(3\theta) = cos(pi/2 - \theta).

Also, another way to see that sin(\theta) = cos(pi/2 - \theta):
Draw a right triangle. There are two acute angles, call it angles A and B.
Isn't sin(angle A) = cos(angle B)?
If angle A = \theta, then angle B = 90 degrees - angle A = pi/2 - \theta

This is not a rigorous proof (only works for 0 < \theta < pi/2), but may help with intuition.

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