Question

Question #f51a4

Answer 1

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

`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.`

Explanation:

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

`sin^-1x=theta, |x|le1 iff sintheta=x, theta in [-pi/2,pi/2].`

`Ans. (1)`

We have, `sin((5pi)/4)=sin(pi+pi/4)=-sin(pi/4)=-1/sqrt2,`

so, `sin^-1(sin((5pi)/4))=sin^-1(-1/sqrt2).`

Now, knowing that `sin(-pi/4)=-1/sqrt2, and, -pi/4 in [-pi/2,pi/2],` we

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

`sin^-1(sin((5pi)/4))=sin^-1(-1/sqrt2)=-pi/4,` 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.`