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Math Help - identity 2

  1. #1
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    identity 2

    1. prove that cos (\frac{3\pi}{2}+x)= sin x

    2. is sin 2x sin x- cos x = \frac{\sqrt3}{4} is true for \frac{7\pi}{6}? Explain
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  2. #2
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    Quote Originally Posted by william View Post
    1. prove that cos (\frac{3\pi}{2}+x)= sin x
    \cos{\left(A+B\right)} = \cos{A}\cos{B} - \sin{A}\sin{B}

    So substitute what you have in question 1 into this formula.
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  3. #3
    MHF Contributor red_dog's Avatar
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    Quote Originally Posted by william View Post

    2. is sin 2x sin x- cos x = \frac{\sqrt3}{4} is true for \frac{7\pi}{6}? Explain
    Replace x with \frac{7\pi}{6}

    \sin\left(2\cdot\frac{7\pi}{6}\right)=\sin\frac{7\  pi}{3}=\sin\left(2\pi+\frac{\pi}{3}\right)=\sin\fr  ac{\pi}{3}=\frac{\sqrt{3}}{2}

    \sin\frac{7\pi}{6}=\sin\left(\pi+\frac{\pi}{6}\rig  ht)=-\sin\frac{\pi}{6}=-\frac{1}{2}

    \cos\frac{7\pi}{6}=\cos\left(\pi+\frac{\pi}{6}\rig  ht)=-\cos\frac{\pi}{6}=-\frac{\sqrt{3}}{2}

    I used \sin(2\pi+\alpha)=\sin\alpha, \ \sin(\pi+\alpha)=-\sin\alpha, \ \cos(\pi+\alpha)=-\cos\alpha
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