1. Trigonometric Formulars

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

Given that $
0<\theta<\frac{1}{2}pi
$

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

using appropriate trigonometric formulas find the exact values of the following
$
(i) cos(2\theta)
$

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

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

$
(ii) cos(\theta)
$

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

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

Positive in this case

$
(iii) sin(3\theta)
$

Rewrite as
$
sin(2\theta+\theta)
$

find $
sin(2\theta)
$
using $
sin(2\theta) = 2 sin\theta cos \theta
$

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

and then use

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

Giving

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

So

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

Again positive

2. Greetings:

Your work is well organized and, in the case at hand, 100% correct! Nice work. Now we need only work on that confidence. You are not a "failed miserably as usual" kind of person. You are learning a complex discipline and, from what I can see, you are learning well!

Regards,

Rich B.

3. Originally Posted by Rich B.
Greetings:

Your work is well organized and, in the case at hand, 100% correct! Nice work. Now we need only work on that confidence. You are not a "failed miserably as usual" kind of person. You are learning a complex discipline and, from what I can see, you are learning well!

Regards,

Rich B.
Thanks that's given me a big confidence boost !!