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Math Help - trigo problem again

  1. #1
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    trigo problem again

    trigo problem again
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  2. #2
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    Hello, afeasfaerw23231233!

    You showed very few steps.
    I assume you used the product-to-sum identities.

    . . \sin A\sin B \:=\:\frac{1}{2}\left[\cos(A-B) - \cos(A+B)\right]
    . . . \cos A\cos B \:=\:\frac{1}{2}\left[\cos(A-B) + \cos(A + B)\right]



    29 (a) Prove that if \tan\theta\tan(\theta + \alpha) \:=\:k

    . . then: . (k+1)\cos(2\theta+\alpha) \;=\;(1-k)\cos\alpha

    We have: . \frac{\sin\theta}{\cos\theta}\cdot\frac{\sin(\thet  a+\alpha)}{\cos(\theta+\alpha)} \;=\;k . . \Rightarrow\quad\sin\theta\!\cdot\!\sin(\theta +\alpha) \;=\;k\cdot\cos\theta\!\cdot\!\cos(\theta + \alpha)

    Then: . \frac{1}{2}\left[\cos\alpha - \cos(2\theta+\alpha)\right] \;=\;k\cdot\frac{1}{2}\left[\cos\alpha + \cos(2\theta + \alpha)\right]

    . . . . . \cos\alpha - \cos(2\theta +\alpha) \;=\;k\!\cdot\!\cos\alpha + k\!\cdot\!\cos(2\theta+\alpha)

    . . . . . k\!\cdot\!\cos(2\theta+\alpha) + \cos(2\theta+\alpha) \;=\;\cos\alpha - k\!\cdot\!\cos\alpha


    Factor: . (k+1)\cos(2\theta + \alpha) \;=\;(1-k)\cos\alpha

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