# Let theta be an angle where: "1)" theta in "Quadrant III" and "2)" sin( theta ) = - 15/17. What Quadrant does 12theta belong to ? No Calculators !!

Feb 16, 2018

$12 \theta$ is in $Q 1$.

#### Explanation:

As $\theta \in \text{Quadrant} I I I$ and $\sin \theta = - \frac{15}{17} = - 0.882353$

$\cos \theta = - \sqrt{1 - {\left(\frac{15}{17}\right)}^{2}} = - \sqrt{1 - \frac{225}{289}} = - \sqrt{\frac{64}{289}} = - \frac{8}{17} = - 0.470588$

Then $\sin 3 \theta = 3 \sin x - 4 {\sin}^{3} x = 3 \left(- 0.882353\right) - 4 {\left(- 0.882353\right)}^{3}$

= $- 2.647059 + 2.747813 = 0.100753$

and $\cos 3 \theta = 4 {\cos}^{3} x - 3 \cos x = 4 {\left(- 0.470588\right)}^{3} - 3 \left(- 0.470588\right)$

= $- 0.416853 + 1.411764 = 0.994911$

Note that as $\sin 3 \theta$ and $\cos 3 \theta$ are positive, $3 \theta$ is in $Q 1$,

Now $\sin 6 \theta = 2 \sin 3 \theta \cos 3 \theta = 2 \times 0.100753 \times 0.994911$

= $0.200481$

and $\cos 6 \theta = {\left(0.994911\right)}^{2} - {\left(0.200481\right)}^{2} = 0.949655$

and $6 \theta$ is in $Q 2$

Continuing this way $\sin a 2 \theta = 2 \sin 6 \theta \cos 6 \theta = 2 \times 0.200481 \times 0.949655$

= $0.380776$

and $\cos 12 \theta = {\left(0.949655\right)}^{2} - {\left(0.200481\right)}^{2} = 0.861652$

Hence $12 \theta$ is in $Q 1$.