The following problem is from the $8$th edition of the book Calculus, by James Stewart. It is problem number $9$ in section $6.6$.
Problem:
Find an exact value for the expression:$$ \cos{\left( 2 \sin^{-1}\left( \frac{5}{13} \right) \right) } $$
Answer:
\begin{align*} \cos{\left( 2 \sin^{-1}\left( \frac{5}{13} \right) \right) } &= \sqrt{ 1 - \sin^2{\left( 2 \sin^{-1}\left( \frac{5}{13} \right) \right) } } \\ % \cos{\left( 2 \sin^{-1}\left( \frac{5}{13} \right) \right) } &= \sqrt{ 1 - 2 \sin^2{\left( \sin^{-1}\left( \frac{5}{13} \right) \right) } \cos^2{\left( \sin^{-1}\left( \frac{5}{13} \right) \right) } } \\ % \cos{\left( 2 \sin^{-1}\left( \frac{5}{13} \right) \right) } &= \sqrt{ 1 - 2 \left( \frac{25}{13^2} \right) \cos^2{\left( \sin^{-1}\left( \frac{5}{13} \right) \right) } } \\ % \cos{\left( 2 \sin^{-1}\left( \frac{5}{13} \right) \right) } &= \sqrt{ 1 - \left( \frac{50}{13^2} \right) \cos^2{\left( \sin^{-1}\left( \frac{5}{13} \right) \right) } } \\ \end{align*}\begin{align*} \cos^2{\left( \sin^{-1}\left( \frac{5}{13} \right) \right) } &= 1 - \sin^2{\left( \sin^{-1}\left( \frac{5}{13} \right) \right) } \\ \cos^2{\left( \sin^{-1}\left( \frac{5}{13} \right) \right) } &= 1 - \frac{25}{169} = \frac{169 - 25}{169} \\ \cos^2{\left( \sin^{-1}\left( \frac{5}{13} \right) \right) } &= \frac{144}{169} \\ \cos{\left( 2 \sin^{-1}\left( \frac{5}{13} \right) \right) } &= \sqrt{ 1 - \left( \frac{50}{13^2} \right) \left( \frac{144}{169} \right) } \\ \cos{\left( 2 \sin^{-1}\left( \frac{5}{13} \right) \right) } &= \sqrt{ \frac{13^4 - 50(144)}{13^4} } \\ \cos{\left( 2 \sin^{-1}\left( \frac{5}{13} \right) \right) } &= \sqrt{ \frac{21361}{13^4} } \\ \cos{\left( 2 \sin^{-1}\left( \frac{5}{13} \right) \right) } &= \frac{ \sqrt{ 21361 } } { 169 } \end{align*}
The book's answer is $\frac{119}{169}$ and SciLab matches the book. Where did I go wrong?
$\endgroup$ 13 Answers
$\begingroup$Your mistake is when you squared $\sin2\theta$, it should be $4\sin^{2}\theta\cos^{2}\theta$ not $2\sin^{2}\theta\cos^{2}\theta$, but in case you want an easier method than going for a long method, here is one:
Just let $\displaystyle \theta=\sin^{-1}\left(\frac{5}{13}\right)$ so that we have $\sin \theta=\dfrac{5}{13}$ and now we know that $\displaystyle \cos 2\theta=1-2\sin^{2}\theta=1-2\left(\frac{5}{13}\right)^2=\frac{119}{169}.$
$\endgroup$ 0 $\begingroup$In the second line there is a mistake!$$\sin^22\alpha=4\sin^2\alpha\cos^2\alpha.$$
Now, $$\cos2\arcsin\frac{5}{13}=\sqrt{1-4\left(\frac{5}{13}\right)^2\left(\frac{12}{13}\right)^2}=\frac{119}{169}.$$
$\endgroup$ $\begingroup$Alternative solution: Let $sin^{-1}(\frac{5}{13})=x$, then $\sin x=\frac{5}{13},\space \cos x=\frac{12}{13}$
$\cos 2x=1-2\sin^2x=1-2\times(\frac{5}{13})^2=\frac{119}{169}$
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