# Math Help - Trigonometry

1. ## Trigonometry

If $sin\;\beta = \frac{2}{7}$ find $cos\;2\beta$ without using a calculator.

First off I know that $cos\;2\beta = 1 - 2\;sin^2\;\beta$

But I'm having trouble plugging in the $sin\;\beta = \frac{2}{7}$

Anyone have any thoughts on this?

2. Originally Posted by OzzMan
If $sin\;\beta = \frac{2}{7}$ find $cos\;2\beta$ without using a calculator.

First off I know that $cos\;2\beta = 1 - 2\;sin^2\;\beta$

But I'm having trouble plugging in the $sin\;\beta = \frac{2}{7}$

Anyone have any thoughts on this?
Why not $cos\;2\beta = 1 - 2\;sin^2\;\beta = 1 - 2 \left( \frac{2}{7}\right)^2$ .....?

Hmmmmmm ...... you do realise that $\sin^2 \beta$ means $(\sin \beta)^2$, right ....?

3. $
cos\;2\beta = 1 - 2\;sin^2\;\beta
$

Remember that:

$sin^2\;\beta = (sin \beta)^2$

So simply plug in $\frac {2}{7}$ :

$cos\;2\beta = 1 - 2 \left( \frac{2}{7}\right)^2$

4. I thought you had to leave a sin in there because it was $sin = \frac{2}{7}$ not $sin^2 = \frac{2}{7}$

Alright so the answers $cos\;2\beta = \frac{41}{49}$

5. Originally Posted by OzzMan
I thought you had to leave a sin in there because it was $sin = \frac{2}{7}$ not $sin^2 = \frac{2}{7}$
You are right, it was $sin = \frac{2}{7}$ not $sin^2 = \frac{2}{7}$ .

That is why you have to square the $\frac {2}{7}$ .

Originally Posted by OzzMan
Alright so the answers $cos\;2\beta = \frac{41}{49}$
Yes but it should be negative.

6. $1 - 2\left(\frac{2}{7}\right)^2$

$1 - 2\left(\frac{4}{49}\right)$

$1 - \frac{8}{49}$

$\frac{49}{49} - \frac{8}{49}$

$\frac{41}{49}$

7. Oops, sorry you are right.