Have I got this right or have I failed miserably as usual

Given that $\displaystyle

0<\theta<\frac{1}{2}pi

$

and $\displaystyle

sin (\theta)= \frac{1}{4}

$

using appropriate trigonometric formulas find the exact values of the following

$\displaystyle

(i) cos(2\theta)

$

Use

$\displaystyle

cos(2\theta) = 1-2 sin^2\theta

$

gives

$\displaystyle

cos(2\theta) = 1-2 * \frac{1}{4}^2 = \frac{7}{8}

$

$\displaystyle

(ii) cos(\theta)

$

use

$\displaystyle

cos(\theta) = \pm \sqrt{1-sin^2\theta}

$

gives

$\displaystyle

cos(\theta) = \pm \sqrt{1-\frac{1}{4}^2}= \pm\sqrt{\frac{15}{16}} = \frac{1}{4}\sqrt{15}

$

Positive in this case

$\displaystyle

(iii) sin(3\theta)

$

Rewrite as

$\displaystyle

sin(2\theta+\theta)

$

find $\displaystyle

sin(2\theta)

$ using $\displaystyle

sin(2\theta) = 2 sin\theta cos \theta

$

$\displaystyle

sin(2\theta) = 2 * \frac {1}{4}*\frac{1}{4}\sqrt{15} = \frac{1}{8}\sqrt{15}

$

and then use

$\displaystyle

sin(\phi+\theta) = sin\phi cos\theta + cos\phi sin\theta

$

Giving

$\displaystyle

sin(2\theta+\theta) = \frac{1}{8}\sqrt{15}*\frac{1}{4}\sqrt{15} + \frac{7}{8}*\frac{1}{4} = \frac{11}{16}

$

So

$\displaystyle

sin(3\theta) = \frac{11}{16}

$

Again positive