If $sin θ = \frac{1}{3}$ $sintheta=1/3$ and $θ$ $theta$ is in quadrant I, how do you evaluate $sin 2 θ$ $sin2theta$ ?

Question

9898 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-02T05:22:57

$\frac{4 \sqrt{2}}{9}$ $(4sqrt 2)/9$ .

The first quadrant $θ = {sin}^{- 1} (\frac{1}{3}) = {19.47}^{o}$ $theta=sin^(-1)(1/3)=19.47^o$ , nearly. So, $2 θ$ $2theta$ is

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

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 $(180^{o} - θ)$ $(180^o-theta)$

for which sin is $sin θ = \frac{1}{3}$ $sin theta= 1/3$ , and $cos θ < 0$ $cos theta < 0$ .

Here, $sin 2 θ = - \frac{4 \sqrt{2}}{9}$ $sin 2 theta=-(4 sqrt2)/9$ .

If sintheta=1/3sinθ=13sintheta=1/3 and thetaθtheta is in quadrant I, how do you evaluate sin2thetasin2θsin2theta?