Question

find the exact value of the expression.

$\cos\left(\sin^{-1}\frac{12}{13}+\tan^{-1}\frac{3}{4}\
ight)$

$\cos\left(\sin^{-1}\frac{12}{13}+\tan^{-1}\frac{3}{4}\
ight)=\square$
(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression.)

Explanation:

Step1: Let \(\alpha=\sin^{-1}\frac{12}{13}\) and \(\beta = \tan^{-1}\frac{3}{4}\)

By the definition of inverse - sine function, if \(\alpha=\sin^{-1}\frac{12}{13}\), then \(\sin\alpha=\frac{12}{13}\) and \(\cos\alpha=\sqrt{1 - (\frac{12}{13})^2}=\frac{5}{13}\) (using the identity \(\cos\alpha=\sqrt{1-\sin^{2}\alpha}\) and since \(\alpha\in[-\frac{\pi}{2},\frac{\pi}{2}]\) and \(\sin\alpha>0\), \(\cos\alpha>0\)).

By the definition of inverse - tangent function, if \(\beta=\tan^{-1}\frac{3}{4}\), then \(\tan\beta=\frac{3}{4}\). Using the identity \(\tan\beta=\frac{\sin\beta}{\cos\beta}\) and \(\sin^{2}\beta+\cos^{2}\beta = 1\), we get \(\sin\beta=\frac{3}{5}\) and \(\cos\beta=\frac{4}{5}\) (since \(\beta\in(-\frac{\pi}{2},\frac{\pi}{2})\) and \(\tan\beta>0\), \(\beta\in(0,\frac{\pi}{2})\)).

Step2: Use the cosine addition formula \(\cos(A + B)=\cos A\cos B-\sin A\sin B\)

Here \(A=\alpha\) and \(B = \beta\).

Substitute \(\cos\alpha=\frac{5}{13}\), \(\sin\alpha=\frac{12}{13}\), \(\cos\beta=\frac{4}{5}\), and \(\sin\beta=\frac{3}{5}\) into the formula:

$$
LATEXBLOCK0
$$

Answer:

\(-\frac{16}{65}\)