Question #f51a4

1 Answer
Ratnaker Mehta
Sep 30, 2016

Ans.(1) : sin^-1(sin((5pi)/4))=-pi/4.Ans.(1):sin−1(sin(5π4))=−π4.

Ans.(2) : sin^-1(sin((5pi)/3))=-pi/3.Ans.(2):sin−1(sin(5π3))=−π3.

Ans.(3) : cos^-1(cos((-7pi)/4))=pi/4.Ans.(3):cos−1(cos(−7π4))=π4.

Ans.(4) : cos^-1(cos((2pi)/3))=(2pi)/3.Ans.(4):cos−1(cos(2π3))=2π3.

Explanation:

Let us first recall the Defn. of sin^-1sin−1 fun. :

sin^-1x=theta, |x|le1 iff sintheta=x, theta in [-pi/2,pi/2].sin−1x=θ,|x|≤1⇔sinθ=x,θ∈[−π2,π2].

Ans. (1)Ans.(1)

We have, sin((5pi)/4)=sin(pi+pi/4)=-sin(pi/4)=-1/sqrt2,sin(5π4)=sin(π+π4)=−sin(π4)=−1√2,

so, sin^-1(sin((5pi)/4))=sin^-1(-1/sqrt2).sin−1(sin(5π4))=sin−1(−1√2).

Now, knowing that sin(-pi/4)=-1/sqrt2, and, -pi/4 in [-pi/2,pi/2],sin(−π4)=−1√2,and,−π4∈[−π2,π2], we

conclude, from the Defn. of sin^-1, sin^-1(-1/sqrt2)=-pi/4,sin−1,sin−1(−1√2)=−π4, i.e.,

sin^-1(sin((5pi)/4))=sin^-1(-1/sqrt2)=-pi/4,sin−1(sin(5π4))=sin−1(−1√2)=−π4, which is in the

desired range, i.e., −π/2≤θ≤π.

Proceeding as above,

Ans.(2) : sin^-1(sin((5pi)/3))=-pi/3.

Ans.(3) : cos^-1(cos((-7pi)/4))=pi/4.

Ans.(4) : cos^-1(cos((2pi)/3))=(2pi)/3.

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