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Math Help - Trigo value

  1. #1
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    Trigo value

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

    Q:Trigo value-q5.jpg

    A:Trigo value-a5.jpg
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  2. #2
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    e^(i*pi)'s Avatar
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    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
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  3. #3
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    Quote Originally Posted by e^(i*pi) View Post
    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)?
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  4. #4
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    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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