Question

Find the exact value of the expression.

Answer 1

$\sin(a-b)=\sin a \cos b-\cos a \sin b$. If we let $a=\cos^{-1} \left(\frac{5}{6} \right)$ and $b=\tan^{-1} \left(\frac{1}{2} \right)$, then the given expression is equal to:

$\sin \left(\cos^{-1} \left(\frac{5}{6}} \right) \right) \cos \left(\tan^{-1} \left(\frac{1}{2} \right) \right)-\cos\left(\cos^{-1} \left(\frac{5}{6} \right) \right) \sin \left( \tan^{-1} \left(\frac{1}{2} \right) \right)$

Using the Pythagorean identities $\sin^{2} x+\cos^{2} x=1$ and $\tan^{2} x+1=\sec^{2} x$,

$1) \sin^{2} \left(\cos^{-1} \left(\frac{5}{6} \right) \right)+\cos^{2} \left(\cos^{-1} \left(\frac{5}{6} \right) \right)=1\\\sin^{2} \left(\cos^{-1} \left(\frac{5}{6} \right) \right)+\frac{25}{36}=1\\\sin^{2} \left(\cos^{-1} \left(\frac{5}{6} \right) \right)=\frac{11}{36}\sin \left(\cos^{-1} \left(\frac{5}{6} \right) \right)=\frac{\sqrt{11}}{6}$

$2) \tan^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)+1=\sec^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)\\\frac{1}{4}+1=\sec^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)\\\frac{5}{4}=\sec^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)\\\sec \left(\tan^{-1} \left(\frac{1}{2} \right) \right)=\frac{\sqrt{5}}{2}\\\implies \cos \left(\tan^{-1} \left(\frac{1}{2} \right) \right)=\frac{2}{\sqrt{5}}=\frac{2\sqrt{5}}{5}$

$\cos^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)+\sin^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)=1\\\frac{4}{5}+\sin^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)=1\\\sin^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)=\frac{1}{5}\\\left(\tan^{-1} \left(\frac{1}{2} \right) \right)=\frac{1}{\sqrt{5}}=\frac{\sqrt{5}}{5}$

This means we can write the original expression as:

$\left(\frac{\sqrt{11}}{6} \right) \left(\frac{2\sqrt{5}}{5} \right)-\left(\frac{5}{6} \right) \left(\frac{\sqrt{5}}{5} \right)\\=\frac{2\sqrt{11}\sqrt{5}}{30}-\frac{5\sqrt{5}}{30}\\=\boxed{\frac{\sqrt{5}(2\sqrt{11}-5)}{30}}$

Answer 2

Step-by-step explanation:

let

$a = \cos {}^{ - 1} ( \frac{5}{6} )$

$b = \tan {}^{ - 1} ( \frac{1}{2} )$

$\sin(a - b) = \sin(a) \cos(b) - \cos(a) \sin(b)$

Substitute

$\sin( \cos {}^{ - 1} ( \frac{5}{6} ) ) \cos( \tan {}^{ - 1} ( \frac{1}{2} ) ) - \cos( \cos {}^{ - 1} ( \frac{5}{6} ) ) \sin( \tan {}^{ - 1} ( \frac{1}{2} ) )$

$\frac{ \sqrt{11} }{6} \frac{2}{ \sqrt{5} } - \frac{5}{6} \frac{1}{ \sqrt{5} }$

$\frac{ \sqrt{11} }{6} \frac{ 2\sqrt{5} }{5} - \frac{5 \sqrt{5} }{30}$

$\frac{ 2\sqrt{11} \sqrt{5} - 5 \sqrt{5} }{30}$

$\frac{ \sqrt{5} (2 \sqrt{11} - 5)}{30}$