2212.12. coSCOS-1-2 sin

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Solution

The inverse of a function f:A→B exists if f is one-one onto i.e.,

y=f( x )⇒ f −1 ( y )=x .

The given inverse trigonometry function is cos −1 ( 1 2 )+2 sin −1 ( 1 2 ) .

Let,

cos −1 ( 1 2 )=x cosx= 1 2 =cos( π 3 )

Therefore, cos −1 ( 1 2 )= π 3

Let,

sin −1 ( 1 2 )=y

siny= 1 2 =sin( π 6 )

Therefore, sin −1 ( 1 2 )= π 6

According to the question, summation of all the functions gives,

cos −1 ( 1 2 )+2 sin −1 ( 1 2 ) = π 3 + 2π 6 = π+π 3 = 2π 3

Thus, the value of cos −1 ( 1 2 )+2 sin −1 ( 1 2 ) is 2π 3 .

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