Question

sin-1 (sin 5) is equal to​

Answer 1

Let  sin−1(sin5)=x∀ −π/2≤x≤π/2  

sinx=sin5  

x=2nπ+5  or  2nπ+π−5  

Where,  n  is any integer i.e.  n=0,±1,±2,±3,…  

But the only value of  x  satisfying the condition:  x∈[−π/2,π/2]  is obtained by setting  n=−1  in above general solution. Hence we get the solution

x=5−2π≈−73.52110243

Answer 2

Answer:

= sin 1 (sin 5)

- range of sin x is - π/2 to π/2

- -3.14/2

= -1.57 = + 1.57

5-2π

= 5-2×3.14 = 5 - 6.28 = 1.28

Ans = sin - 1 ( sin 5 ) = 5-2π

Hope it help u