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Thread: 2 trig problems pls help...

  1. #1
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    Talking 2 trig problems pls help...

    1) show that cosec$\displaystyle \theta$/(cosec$\displaystyle \theta$-sin$\displaystyle \theta$)=sec^2$\displaystyle \theta$

    2) solve: sin3$\displaystyle \theta$cos2$\displaystyle \theta$=sin2$\displaystyle \theta$cos3$\displaystyle \theta$ for 0<$\displaystyle \theta$<2$\displaystyle \pi$

    thanks in advance
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  2. #2
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    Hello, snowball!

    1) Show that: .$\displaystyle \frac{\csc\theta}{\csc\theta -\sin\theta} \:=\:\sec^2\!\theta$

    On the left side, multiply top and bottom by $\displaystyle \sin\theta$

    . . $\displaystyle \frac{\sin\theta(\csc\theta)}{\sin\theta(\csc\thet a-\sin\theta)} \:=\:\frac{1}{1-\sin^2\!\theta} \;=\;\frac{1}{\cos^2\!\theta} \;=\;\sec^2\!\theta$




    2) Solve: .$\displaystyle \sin3\theta\cos2\theta \:=\:\sin2\theta \cos3\theta,\;\;\text{ for }0 \leq \theta < 2\pi$

    $\displaystyle \text{We have: }\;\underbrace{\sin3\theta\cos2\theta - \sin2\theta\cos3\theta} \;=\;0$

    . . . . . . . . . . .$\displaystyle \overbrace{\sin(3\theta-2\theta)} \;=\;0$

    . . . . . . . . . . . . $\displaystyle \sin\theta \;=\;0$

    Therefore:. $\displaystyle \theta \;=\;0.\:\pi$

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