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) } } \\
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\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) } } \\
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\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?
Best Answer
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 $\displaystyle \cos 2\theta=1-2\sin^{2}\theta=1-2\left(\frac{5}{13}\right)^2=\frac{119}{169}.$