Question

Find the exact value of the expression.

Answer 1

Explanation:

let

`a = \cos {}^( - 1) ( (5)/(6) )`

`b = \tan {}^( - 1) ( (1)/(2) )`

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

Substitute

`\sin( \cos {}^( - 1) ( (5)/(6) ) ) \cos( \tan {}^( - 1) ( (1)/(2) ) ) - \cos( \cos {}^( - 1) ( (5)/(6) ) ) \sin( \tan {}^( - 1) ( (1)/(2) ) )`

`( √(11) )/(6) (2)/( √(5) ) - (5)/(6) (1)/( √(5) )`

`( √(11) )/(6) ( 2√(5) )/(5) - (5 √(5) )/(30)`

`( 2√(11) √(5) - 5 √(5) )/(30)`

`( √(5) (2 √(11) - 5))/(30)`

Answer 2

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

`\sin \left(\cos^(-1) \left((5)/(6)} \right) \right) \cos \left(\tan^(-1) \left((1)/(2) \right) \right)-\cos\left(\cos^(-1) \left((5)/(6) \right) \right) \sin \left( \tan^(-1) \left((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((5)/(6) \right) \right)+\cos^(2) \left(\cos^(-1) \left((5)/(6) \right) \right)=1\\\sin^(2) \left(\cos^(-1) \left((5)/(6) \right) \right)+(25)/(36)=1\\\sin^(2) \left(\cos^(-1) \left((5)/(6) \right) \right)=(11)/(36)\sin \left(\cos^(-1) \left((5)/(6) \right) \right)=(√(11))/(6)`

`2) \tan^(2) \left(\tan^(-1) \left((1)/(2) \right) \right)+1=\sec^(2) \left(\tan^(-1) \left((1)/(2) \right) \right)\\(1)/(4)+1=\sec^(2) \left(\tan^(-1) \left((1)/(2) \right) \right)\\(5)/(4)=\sec^(2) \left(\tan^(-1) \left((1)/(2) \right) \right)\\\sec \left(\tan^(-1) \left((1)/(2) \right) \right)=(√(5))/(2)\\\implies \cos \left(\tan^(-1) \left((1)/(2) \right) \right)=(2)/(√(5))=(2√(5))/(5)`

`\cos^(2) \left(\tan^(-1) \left((1)/(2) \right) \right)+\sin^(2) \left(\tan^(-1) \left((1)/(2) \right) \right)=1\\(4)/(5)+\sin^(2) \left(\tan^(-1) \left((1)/(2) \right) \right)=1\\\sin^(2) \left(\tan^(-1) \left((1)/(2) \right) \right)=(1)/(5)\\\left(\tan^(-1) \left((1)/(2) \right) \right)=(1)/(√(5))=(√(5))/(5)`

This means we can write the original expression as:

`\left((√(11))/(6) \right) \left((2√(5))/(5) \right)-\left((5)/(6) \right) \left((√(5))/(5) \right)\\=(2√(11)√(5))/(30)-(5√(5))/(30)\\=\boxed{(√(5)(2√(11)-5))/(30)}`