How do you evaluate ${csc}^{- 1} (\sqrt{2})$ $csc^-1(sqrt2)$ ?

Question

How do you evaluate ${csc}^{- 1} (\sqrt{2})$ $csc^-1(sqrt2)$ ?

1 Answer

Impact of this question

27067 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-24T03:25:39

$\frac{π}{4}$ $pi/4$

Explanation:

Let $θ = {csc}^{- 1} (\sqrt{2})$ $theta=csc^-1(sqrt2)$ .

Since $csc (x)$ $csc(x)$ and ${csc}^{- 1} (x)$ $csc^-1(x)$ are inverse functions, this means that $csc (θ) = \sqrt{2}$ $csc(theta)=sqrt2$ .

Another way of reaching that fact is to take the cosecant of both sides: $csc (θ) = csc ({csc}^{- 1} (\sqrt{2}))$ $csc(theta)=csc(csc^-1(sqrt2))$ , and since $csc ({csc}^{- 1} (x)) = x$ $csc(csc^-1(x))=x$ , this becomes $csc (θ) = \sqrt{2}$ $csc(theta)=sqrt2$ .

Taking the reciprocal of both sides, the equation becomes $sin (θ) = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$ $sin(theta)=1/sqrt2=sqrt2/2$ .

So, we want to find $θ$ $theta$ , or the angle where the value of sine is $\frac{\sqrt{2}}{2}$ $sqrt2/2$ .

This is a well known value of sine. It occurs when $θ = \frac{π}{4}$ $theta=pi/4$ , which means that ${csc}^{- 1} (\sqrt{2}) = \frac{π}{4}$ $csc^-1(sqrt2)=pi/4$ .

Science

Math

Humanities

... and beyond

How do you evaluate ${csc}^{- 1} (\sqrt{2})$ $csc^-1(sqrt2)$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you evaluate csc−1(√2)csc^-1(sqrt2)?

1 Answer

Explanation:

Related questions

Impact of this question

How do you evaluate ${csc}^{- 1} (\sqrt{2})$ $csc^-1(sqrt2)$ ?