How do you prove ${sin}^{- 1} (- \frac{1}{2}) = - \frac{π}{6}$ $Sin^-1(-1/2)= -pi/6$ ?

Question

How do you prove ${sin}^{- 1} (- \frac{1}{2}) = - \frac{π}{6}$ $Sin^-1(-1/2)= -pi/6$ ?

1 Answer

Impact of this question

5908 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-02T10:23:07

See the Explanation.

Explanation:

The Defn. of ${sin}^{- 1}$ $sin^-1$ fun. is :

${sin}^{- 1} x = θ, | x | \leq 1 \Leftrightarrow sin θ = x, θ \in [- \frac{π}{2}, \frac{π}{2}] .$ $sin^-1 x=theta, |x| le 1 iff sintheta=x, theta in [-pi/2,pi/2].$

So, we have to find a $θ,$ $theta,$ which satisfies the conds., namely,

$(1) : - \frac{π}{2} \leq θ \leq \frac{π}{2}, & (2) : sin θ = - \frac{1}{2} .$ $(1) : -pi/2 le theta le pi/2," & "(2) : sintheta=-1/2.$

Knowing that, $sin (- \frac{π}{6}) = - sin (\frac{π}{6}) = - \frac{1}{2},$ $sin (-pi/6)=-sin(pi/6)=-1/2,$ and, noting

that, $- \frac{π}{6} \in [- \frac{π}{2}, \frac{π}{2}]$ $-pi/6 in [-pi/2,pi/2]$ , we have,

${sin}^{- 1} (- \frac{1}{2}) = - \frac{π}{6} .$ $sin^-1 (-1/2)=-pi/6.$

Science

Math

Humanities

... and beyond

How do you prove ${sin}^{- 1} (- \frac{1}{2}) = - \frac{π}{6}$ $Sin^-1(-1/2)= -pi/6$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you prove sin−1(−12)=−π6Sin^-1(-1/2)= -pi/6?

1 Answer

Explanation:

Related questions

Impact of this question

How do you prove ${sin}^{- 1} (- \frac{1}{2}) = - \frac{π}{6}$ $Sin^-1(-1/2)= -pi/6$ ?