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Section 3.5 #13: Find the exact value of $\cos \left( \dfrac{7\pi}{12} \right)$.
Solution: We will rewrite $\dfrac{7\pi}{12}$ as a difference of two angles we can handle using the unit circle. Notice that
$$\dfrac{7\pi}{12} = \dfrac{9\pi}{12} - \dfrac{2\pi}{12} = \dfrac{3\pi}{4} - \dfrac{\pi}{6}.$$
Therefore using the difference identity
$$\cos(\alpha - \beta) = \cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta)$$
to compute
$$\begin{array}{ll}
\cos \left( \dfrac{7\pi}{12} \right) &= \cos \left( \dfrac{3\pi}{4} - \dfrac{\pi}{6} \right) \\
&= \cos \left( \dfrac{3\pi}{4} \right) \cos \left( \dfrac{\pi}{6} \right) + \sin \left( \dfrac{3\pi}{4} \right) \sin \left( \dfrac{\pi}{6} \right) \\
&= \left( - \dfrac{\sqrt{2}}{2} \right) \left( \dfrac{\sqrt{3}}{2} \right) + \left( \dfrac{\sqrt{2}}{2} \right) \left( \dfrac{1}{2} \right) \\
&= - \dfrac{\sqrt{6}}{4} + \dfrac{\sqrt{2}}{4} \\
&= \dfrac{\sqrt{2}-\sqrt{6}}{4}.
\end{array}$$
note: the solution that Wolfram Alpha calculates, $-\dfrac{\sqrt{3}-1}{2\sqrt{2}}$, is equivalent to this -- just multiply by $\dfrac{\sqrt{2}}{\sqrt{2}}$ and you will see it
Section 3.5 #29: Find the exact value of
$$\sin \left( \dfrac{\pi}{12} \right) \cos \left( \dfrac{7\pi}{12} \right) - \cos \left( \dfrac{\pi}{12} \right) \sin\left( \dfrac{7\pi}{12} \right).$$
Solution: Recall the difference identity for sine:
$$\sin(\alpha - \beta) = \sin(\alpha) \cos(\beta) - \cos(\alpha) \sin(\beta).$$
Ths given problem looks like the right hand side of that identity with $\alpha=\dfrac{\pi}{12}$ and $\beta=\dfrac{7\pi}{12}$. So we may compute
$$\begin{array}{ll}
\sin \left( \dfrac{\pi}{12} \right) \cos \left( \dfrac{7\pi}{12} \right) - \cos \left( \dfrac{\pi}{12} \right) \sin\left( \dfrac{7\pi}{12} \right) &=\sin \left( \dfrac{\pi}{12} - \dfrac{7\pi}{12} \right) \\
&=\sin \left(-\dfrac{6\pi}{12} \right) \\
&=\sin \left( -\dfrac{\pi}{2} \right)\\
&=-1
\end{array}$$
Section 3.5 #77: Find the exact value of
$$\cos \left( \tan^{-1} \left( \dfrac{4}{3} \right) + \cos^{-1} \left( \dfrac{5}{13} \right) \right).$$
Solution: Recall the sum identity for cosine:
$$\cos(\alpha + \beta) = \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta).$$
Therefore we see that
$$\cos \left( \tan^{-1} \left( \frac{4}{3} \right) + \cos^{-1} \left( \frac{5}{13} \right) \right) = \cos \left( \tan^{-1} \left( \frac{4}{3} \right) \right) \cos \left( \cos^{-1} \left( \frac{5}{13} \right) \right) - \sin \left( \tan^{-1} \left( \frac{4}{3} \right) \right) \sin \left( \cos^{-1} \left( \frac{5}{13} \right) \right).$$
Now compute each piece.
$\underline{\cos \left( \tan^{-1} \left( \dfrac{4}{3} \right) \right)}$
Let $\theta=\tan^{-1} \left( \dfrac{4}{3} \right)$ so $\theta$ lies in quadrant I or IV. We see that $\tan(\theta)=\dfrac{4}{3}$ and so we can conclude that $\theta$ lies in quadrant I or III. Therefore $\theta$ lies in quadrant I (and so $\cos(\theta)>0$).
Since $\tan(\theta)=\dfrac{4}{3}$ and $\cos(\theta)>0$ we take $y=4$ and $x=3$. This means that $r^2=3^2+4^2=9+16=25$ so we can conclude that $r=\pm \sqrt{25}=\pm 5$, but a radius is never negative so $r=5$. Now using the idea that $\cos(\theta) = \dfrac{x}{r}$, we may compute
$$\begin{array}{ll}
\cos \left( \tan^{-1} \left( \dfrac{4}{3} \right) \right) &= \cos (\theta) \\
&= \dfrac{3}{5}.
\end{array}$$
$\underline{\cos \left( \cos^{-1} \left( \dfrac{5}{13} \right) \right)}$
Using the inverse property, we may compute
$$\cos \left( \cos^{-1} \left( \dfrac{5}{13} \right) \right) = \dfrac{5}{13}.$$
$\underline{\sin \left( \tan^{-1} \left( \dfrac{4}{3} \right) \right)}$
Let $\theta = \tan^{-1} \left(\dfrac{4}{3} \right)$. The analysis we did above for compute $\cos \left( \tan^{-1} \left( \dfrac{4}{3} \right) \right)$ all works the same and we have the same values $x=3$, $y=4$, and $r=5$. Therefore using the idea that $\sin(\theta) = \dfrac{y}{r}$ we compute
$$\sin \left( \tan^{-1} \left( \dfrac{4}{3} \right) \right) = \sin (\theta) = \dfrac{4}{5}.$$
$\underline{\sin \left( \cos^{-1} \left( \dfrac{5}{13} \right) \right)}$
Let $\theta = \cos^{-1} \left( \dfrac{5}{13} \right)$ so we may conclude that $\theta$ lies in quadrant I or II. From this we have $\cos(\theta) =\dfrac{5}{13}$ from which we can conclude that $\theta$ lies in quadrant I or IV. Therefore $\theta$ lies in quadrant I and so we can conclude that $\sin(\theta)>0$.
Using $\cos(\theta) = \dfrac{5}{13}$ and the idea that $\cos(\theta) = \dfrac{x}{r},$ we can take $x=5$ and $r=13$. We can find the value of $y$ using the equation $x^2+y^2=r^2$. Plugging in our values yields $5^2+y^2=13^2$, or equivalently
$$y^2 = 169 - 25 = 144,$$
so
$$y = \pm \sqrt{144}=\pm 12$$
Since $\theta$ lies in quadrant I, we must take the positive value for $y$: $y=12$. Now we may use the idea that $\sin(\theta) = \dfrac{y}{r}$ to compute
$$\begin{array}{ll}
\sin \left( \cos^{-1} \left( \dfrac{5}{13} \right) \right) &= \sin(\theta)= \dfrac{12}{13}
\end{array}$$
Now we put all this information together (continuing our original calculation) to compute
$$\begin{array}{ll}
\cos \left( \tan^{-1} \left( \dfrac{4}{3} \right) + \cos^{-1} \left( \dfrac{5}{13} \right) \right) &= \left(\dfrac{3}{5} \right) \left( \dfrac{5}{13} \right) - \left( \dfrac{4}{5} \right)\left( \dfrac{12}{13} \right) \\
&=\dfrac{15}{65} - \dfrac{48}{65} \\
&= -\dfrac{33}{65} \\
\end{array}.$$
Section 3.5 #85: Write as an algebraic expression containing $u$ and $v$:
$$\cos(\cos^{-1}(u)+\sin^{-1}(v)).$$
Solution: Recall the sum identity for cosine:
$$\cos(\alpha+\beta)=\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta).$$
Applying this to the given problem gives us
$$\begin{array}{ll}
\cos(\cos^{-1}(u)+\sin^{-1}(v)) &=\cos(\cos^{-1}(u))\cos(\sin^{-1}(v))-\sin(\cos^{-1}(u))\sin(\sin^{-1}(v)) \\
&=u \cos(\sin^{-1}(v)) - v \sin(\cos^{-1}(u)).
\end{array}.$$
$\underline{\cos(\sin^{-1}(v)}$
Let $\theta=\sin^{-1}(v)$, so $\theta$ lies in quadrant I or IV. From this we may conclude that $\cos(\theta)>0$. Also $\sin(\theta)=v$. Therefore using the fact that $\sin^2(\theta)+\cos^2(\theta)=1$ rewritten as $\cos(\theta)= \pm \sqrt{1-\sin^2(\theta)}$ and using the fact that $\cos(\theta)>0$ we discovered above, we may compute
$$\cos(\sin^{-1}(v))=\cos(\theta) = \sqrt{1-\sin^2(\theta)}=\sqrt{1-v^2}.$$
$\underline{\sin(\cos^{-1}(u))}$
Let $\theta=\cos^{-1}(u)$ so $\theta$ lies in quadrant I or II, implying that $\sin(\theta)>0$. Also $\cos(\theta)=u$. Therefore using the fact that $\sin(\theta)=\sqrt{1-\cos^2(\theta)}$ and the formula $\cos(\theta)=u$ we may compute
$$\sin(\cos^{-1}(u))=\sin(\theta)=\sqrt{1-\cos^2(\theta)}=\sqrt{1-u^2}.$$
Putting this all together, we can compute
$$\cos(\cos^{-1}(u)+\sin^{-1}(v))=u\sqrt{1-v^2}-v\sqrt{1-u^2}.$$
Section 3.6 #27: Find an exact value of $\sin \left( - \dfrac{\pi}{8} \right)$.
Solution: Notice that this is the same as $\sin \left( -\dfrac{1}{2} \cdot \dfrac{\pi}{4} \right)$. So we use the half-angle identity, taking the negative value because $\dfrac{\pi}{8}$ lies in QIV to get
$$\sin \left( -\dfrac{\pi}{8} \right) = -\sqrt{ \dfrac{1-\cos(\frac{\pi}{4})}{2}} = -\sqrt{\dfrac{1-\frac{\sqrt{2}}{2}}{2}}=\dfrac{\sqrt{2-\sqrt{2}}}{2}.$$
Section 3.6 #49: Establish the identity
$$\cot(2\theta) = \dfrac{\cot^2(\theta)-1}{2\cot(\theta)}.$$
Solution: Compute
$$\begin{array}{ll}
\cot(2\theta) &= \dfrac{\cos(2\theta)}{\sin(2\theta)} \\
&= \dfrac{\cos^2(\theta)-\sin^2(\theta)}{2\sin(\theta)\cos(\theta)} \\
&= \dfrac{\cos^2(\theta)-\sin^2(\theta)}{2\sin(\theta)\cos(\theta)} \left( \dfrac{\frac{1}{\sin^2(\theta)}}{\frac{1}{\sin^2(\theta)}} \right) \\
&=\dfrac{\cot^2(\theta)-1}{2\cot(\theta)},
\end{array}$$
as was to be shown.
Section 3.6 #79: Find the exact value of
$$\sin \left( 2 \sin^{-1} \left( \dfrac{1}{2} \right) \right).$$
Solution: By the double angle identity,
$$\sin \left(2\sin^{-1} \left( \dfrac{1}{2} \right) \right)=2 \sin \left( \sin^{-1} \left( \dfrac{1}{2} \right) \right) \cos \left( \sin^{-1} \left( \dfrac{1}{2} \right) \right) = \cos \left( \sin^{-1} \left( \dfrac{1}{2} \right) \right).$$
Let $\theta=\sin^{-1}\left( \dfrac{1}{2} \right)$ then $\theta$ lives in QI or QIV. Also $\sin(\theta) = \dfrac{1}{2}$ so $\theta$ lives in QI or QII. Therefore $\theta$ lives in QI.
Since $\sin(\theta) = \dfrac{y}{r}$ we will take $y=1$ and $r=2$. Using $x^2+y^2=r^2$ we plug in our values and solve for $x$ to get $x= \pm \sqrt{4-1}=\pm\sqrt{3}$. Since $\theta$ is in quadrant I we take $x=\sqrt{3}$. Therefore we compute
$$\sin \left( 2 \sin^{-1} \left( \dfrac{1}{2} \right) \right)=\cos\left( \sin^{-1} \left( \dfrac{1}{2} \right) \right)= \dfrac{\sqrt{3}}{2}.$$
