The value of sin- (sin5) is equal to
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q no. 22
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your answer is here buddy....
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∘
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