Question

19.The value of sin^(-1)sin(16)+cos^(-1)cos(10) is (1) 26 (2) -26 (3) 6+pi (4) 9 pi-26​

Answer 1

The correct answer is option (3) 6+π.

Given:

The expression $sin^{-1}$(sin(x))+$cos^{-1}$(cos(x))

To Find:

The value of the given expression.

Solution:

$sin^{-1}$(sin(x))=x  when x ∈ [−π/2 ,π/2]

Now, 16 ∉ [−π/2 ,π/2]

But 16−5π ∈ [−π/2, π/2]

16= 16+ 5π-5π = 5π +(16-5π)

∴ $sin^{-1}$(sin(16)) = $sin^{-1}$(sin(5π +(16-5π)))

Now, sin(5π+θ) = -sin(θ)

Hence the above equation becomes,

$sin^{-1}$(sin(16)) =  $sin^{-1}$(-sin(16-5π)) = -(16-5π) = 5π-16  ......................(1)

Also,

$cos^{-1}$(cos(x)) = x when x ∈ [0, π]

10 ∉ [0, π]

But 10-4π ∈ [0, π]

10 = 10+4π-4π= 4π+(10-4π)

∴$cos^{-1}$(cos(10)) = $cos^{-1}$(cos(4π+(10-4π)))

Now cos(4π+θ) = cos(θ)

Hence the above equation becomes

$cos^{-1}$(cos(10)) = $cos^{-1}$(cos(10-4π))) = 10-4π   ......................(2)

∴ From equations (1) and (2),

$sin^{-1}$(sin(16)) + $cos^{-1}$(cos(10)) = 5π-16 + 10-4π = 6+π, which is option (3)

Therefore, the correct answer is option (3) 6+π.

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