Question

If `sintheta=1/3` and `theta` is in quadrant I, how do you evaluate `sin2theta`?

Answer 1

`(4sqrt 2)/9`.

Explanation:

The first quadrant `theta=sin^(-1)(1/3)=19.47^o`, nearly. So, `2theta` is

also in the first quadrant, and so, `sin 2theta >0`.

Now, `sin 2theta=2 sin theta cos theta.=2(1/3)(sqrt(1-(1/3)^2))=(4sqrt 2)/9`.

If theta is in the 2nd quadrant as `(180^o-theta)`

for which sin is `sin theta= 1/3`, and `cos theta < 0`.

Here, `sin 2 theta=-(4 sqrt2)/9`.