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Math Help - 2 Trig Questions

  1. #1
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    2 Trig Questions

    A) I have to find the exact value of cos8*cos38* + sin8*sin38*(the *'s are degrees)

    Now, I'm 90% sure I have to use cos(A+/-B)=cosAcosB +/- sinAsinB, but I've never understood what to do from that point.

    B) I also have to prove that sin^2(x)/1-cos(x)=sec(x)+1/sec(x), yet I've tried several times and reach nothing but deadends.

    Mucho Gracias for the help in advance!
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  2. #2
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    Quote Originally Posted by Blahdkm View Post
    A) I have to find the exact value of cos8*cos38* + sin8*sin38*(the *'s are degrees)

    Now, I'm 90% sure I have to use cos(A+/-B)=cosAcosB +/- sinAsinB, but I've never understood what to do from that point.
    \cos (A - B) = \cos A \cos B + \sin A \sin B, so

    \cos 8^{\circ} \cos 38^{\circ} + \sin 8^{\circ} \sin 38^{\circ} = \cos (8^{\circ} - 38^{\circ}) = \cos (-30^{\circ}) = \frac{\sqrt{3}}{2}


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  3. #3
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    Quote Originally Posted by Blahdkm View Post
    B) I also have to prove that sin^2(x)/1-cos(x)=sec(x)+1/sec(x), yet I've tried several times and reach nothing but deadends.
    You mean this?
    \frac{\sin^2 x}{1 - \cos x} = \frac{\sec x + 1}{\sec x}?


    \frac{\sin^2 x}{1 - \cos x}
    \begin{aligned}<br />
&= \frac{1 - \cos^2 x}{1 - \cos x} \\<br />
&= \frac{(1 + \cos x)(1 - \cos x)}{1 - \cos x} \\<br />
&= 1 + \cos x<br />
\end{aligned}
    \begin{aligned}<br />
&= 1 + \frac{1}{\sec x} \\<br />
&= \frac{\sec x}{\sec x} + \frac{1}{\sec x} \\<br />
&= \frac{\sec x + 1}{\sec x}<br />
\end{aligned}


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