Question

if `sin(sin^(-1)  1/5 + cos^(-1) x)  = 1` then find the value of x

Answer 1

`sin(sin^(-1)  1/5  + cos^(-1) x ) = 1`

`=> sin (sin^(-1)  1/5) cos(cos^(-1)x) + cos(sin^(-1)  1/5) sin(cos^(-1) x) = 1`

`[sin(A+B) = sin A cosB + cosA sin B]`

`=> 1/5 xx x + cos(sin^-1  1/5) sin(cos^(-1) x)  = 1 `

`=> x/5 + cos(sin^(-1)  1/5) sin (cos^(-1) x) = 1`     (1)

Now let `sin^(-1)  1/5 = y`

Then, `siny = 1/5  => cos y = sqrt(1 - (1/5)^2) = (2sqrt6)/5 => y = cos^(-1) ((2sqrt6)/5)`

`:. sin^(-1)  1/5 = cos^(-1)  ((2sqrt6)/5) `  ...(2)

Let `cos^(-1) x = z`

Then `cos z = x => sin z = sqrt(1-x^2) => z = sin^(-1) (sqrt(1-x^2))`

`:. cos^(-1) x = sin^(-1) (sqrt(1-x^2))`

From 1, 2 and 3 we have

`x/5 + cos(cos^(-1) (2sqrt6)/5). sin(sin^(-1)sqrt(1- x^2)) = 1`

`=>  x/5  + (2sqrt6)/5. sqrt(1 - x^2) = 1`

`=> x + 2sqrt6sqrt(1-x^2) = 5`

`= 2sqrt6sqrt(1-x^2) = 5 - x`

On squaring both sides, we get:

`(4)(6)(1-x^2) = 25 + x^2 - 10x`

`=> 24 - 24x^2 = 25 + x^2 - 10x`

`=> 25x^2 - 10x + 1 = 0`

`=> (5x - 1)^2 = 0`

=> (5x -1) = 0

`=> x = 1/5`

Hence, the value of x is `1/5`