How do you calculate the $cos [2 {sin}^{- 1} (\frac{1}{2})]$ $cos[2 sin^ -1 (1/2)]$ ?

Question

How do you calculate the $cos [2 {sin}^{- 1} (\frac{1}{2})]$ $cos[2 sin^ -1 (1/2)]$ ?

1 Answer

Impact of this question

12217 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-26T02:44:39

$\frac{1}{2}$ $1/2$

Explanation:

First, focus on just ${sin}^{- 1} (\frac{1}{2})$ $sin^-1(1/2)$ . This is another way of asking, for what angle is sine equal to $1/2$ $"1/2"$ ? Recall that

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

Thus, since $sin$ $sin$ and $arcsin$ $arcsin$ are inverse functions,

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

Substitute this into the original expression:

$cos [2 {sin}^{- 1} (\frac{1}{2})] = cos (2 \cdot \frac{π}{6}) = cos (\frac{π}{3}) = \frac{1}{2}$ $cos[2sin^-1(1/2)]=cos(2*pi/6)=cos(pi/3)=1/2$

Science

Math

Humanities

... and beyond

How do you calculate the $cos [2 {sin}^{- 1} (\frac{1}{2})]$ $cos[2 sin^ -1 (1/2)]$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you calculate the cos[2sin−1(12)]cos[2 sin^ -1 (1/2)]?

1 Answer

Explanation:

Related questions

Impact of this question

How do you calculate the $cos [2 {sin}^{- 1} (\frac{1}{2})]$ $cos[2 sin^ -1 (1/2)]$ ?