Question

The value of `sin(2cos^(-1)(1/2))` is what?

when I plug it into the calculator (in radians), the answer is `sqrt(3)/2` but the answer is supposed to be `1/3` and i don't understand why

Answer 1

`sin 2 arccos(1/2) = pm sqrt{3}/2`

Explanation:

It doesn't matter if it's done in degrees or radians.

We'll treat the inverse cosine as multivalued. Of course a cosine of `1/2` is one of the two tired triangles of trig.

`arccos(1/2) = pm 60^circ + 360^circ k quad` integer `k`

Double that, `2 arccos(1/2) = pm 120^circ`

So `sin 2 arccos(1/2) = pm sqrt{3}/2`

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Even when the question writers don't have to use 30/60/90 they do. But let's do

`sin 2 arccos (a/b)`

We have `sin(2a) = 2 sin a cos a` so

`sin 2 arccos(a/b) = 2 sin arccos (a/b) cos arccos(a/b)`

`sin 2 arccos(a/b) = {2a}/b sin arccos(a/b)`

If the cosine is `a/b` that's a right triangle with adjacent `a` and hypotenuse `b`, so opposite `pm sqrt{b^2-a^2}.`

`sin 2 arccos(a/b) = {2a}/b cdot (pm sqrt{b^2-a^2})/b`

`sin 2 arccos(a/b) = pm {2a}/b^2 sqrt{b^2-a^2}`

In this problem we have `a=1 and b=2` so

`sin 2 arccos(1/2) = pm 1/2 sqrt{3} quad sqrt`

The principal value is positive.