Question

find the exact value of the following, without using a calculator.
\\( \sin \left( \sin ^ { - 1 } \left( \frac { 1 } { 2 } \
ight) + \tan ^ { - 1 } ( - 5 ) \
ight) \\)
\\( \sin \left( \sin ^ { - 1 } \left( \frac { 1 } { 2 } \
ight) + \tan ^ { - 1 } ( - 5 ) \
ight) = \square \\)
(simplify your answer. type an exact answer, using radicals as needed. rationalize all denominators.)

Explanation:

Step1: Calculate \(\sin^{-1}(\frac{1}{2})\)

We know that \(\sin^{-1}(x)\) has a range of \([-\frac{\pi}{2},\frac{\pi}{2}]\). Since \(\sin(\frac{\pi}{6})=\frac{1}{2}\), then \(\sin^{-1}(\frac{1}{2})=\frac{\pi}{6}\).

Step2: Let \(\theta = \tan^{-1}(-5)\)

By the definition of the inverse - tangent function, \(\tan\theta=-5\), and \(\theta\in(-\frac{\pi}{2},\frac{\pi}{2})\). Using the identity \(\sin(A + B)=\sin A\cos B+\cos A\sin B\), where \(A=\frac{\pi}{6}\) and \(B = \theta\).
We know that \(\sin(A + B)=\sin(\frac{\pi}{6}+\theta)=\sin\frac{\pi}{6}\cos\theta+\cos\frac{\pi}{6}\sin\theta\).
Since \(\tan\theta=-5=\frac{\sin\theta}{\cos\theta}\), and \(\sin^{2}\theta+\cos^{2}\theta = 1\). Substituting \(\sin\theta=-5\cos\theta\) into \(\sin^{2}\theta+\cos^{2}\theta = 1\), we get \((-5\cos\theta)^{2}+\cos^{2}\theta=1\).
\(25\cos^{2}\theta+\cos^{2}\theta = 1\), \(26\cos^{2}\theta=1\), \(\cos\theta=\frac{1}{\sqrt{26}}\) (because \(\theta\in(-\frac{\pi}{2},\frac{\pi}{2})\) and \(\tan\theta<0\), so \(\cos\theta>0\)) and \(\sin\theta=-\frac{5}{\sqrt{26}}\).

Step3: Calculate \(\sin(\frac{\pi}{6}+\theta)\)

\(\sin(\frac{\pi}{6}+\theta)=\sin\frac{\pi}{6}\cos\theta+\cos\frac{\pi}{6}\sin\theta\)
\(\sin\frac{\pi}{6}=\frac{1}{2}\), \(\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}\)
\(\sin(\frac{\pi}{6}+\theta)=\frac{1}{2}\times\frac{1}{\sqrt{26}}+\frac{\sqrt{3}}{2}\times(-\frac{5}{\sqrt{26}})\)
\(=\frac{1 - 5\sqrt{3}}{2\sqrt{26}}=\frac{\sqrt{26}(1 - 5\sqrt{3})}{52}=\frac{\sqrt{26}-5\sqrt{78}}{52}\)

Answer:

\(\frac{\sqrt{26}-5\sqrt{78}}{52}\)