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Math Help - Trigonometry

  1. #1
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    Trigonometry

    What does \sin \left(x+\frac14k\pi\right)+\cos\left(x+\frac14k\pi  \right) equal when convering it using the form:

    a\sin\theta+b\cos\theta = R\sin(\theta+\alpha)?
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  2. #2
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    Quote Originally Posted by Air View Post
    What does \sin \left(x+\frac14k\pi\right)+\cos\left(x+\frac14k\pi  \right) equal when convering it using the form:

    a\sin\theta+b\cos\theta = R\sin(\theta+\alpha)?
    \sin \left(x+\frac14k\pi\right)+\cos\left(x+\frac14k\pi  \right) = \sqrt{2}\sin \left(x+\frac14k\pi + \frac{\pi}4\right) = \sqrt{2}\sin \left(x+\frac{(k+1)\pi}4\right)
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  3. #3
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    Quote Originally Posted by Air View Post
    What does \sin \left(x+\frac14k\pi\right)+\cos\left(x+\frac14k\pi  \right) equal when convering it using the form:

    a\sin\theta+b\cos\theta = R\sin(\theta+\alpha)?
    Expand the right hand side of a\sin\theta+b\cos\theta = R\sin(\theta+\alpha) using the compound angle formula and compare each side:

    a = R \cos \alpha .... (1)

    b = R \sin \alpha .... (2)

    (1)^2 + (2)^2 \Rightarrow R = \sqrt{a^2 + b^2}.

    \frac{(2)}{(1)} \Rightarrow \tan \alpha = \frac{b}{a}.

    In your case, a = b = 1.

    Note: \theta = x + \frac{1}{4} k \pi.
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