# Trigonometric Formulars

• Mar 18th 2006, 12:18 PM
macca101
Trigonometric Formulars
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
• Mar 19th 2006, 01:51 AM
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.
• Mar 19th 2006, 02:01 AM
macca101
Quote:

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 !!