1. ## Trigonometry

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

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

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

Anyone have any thoughts on this?

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

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

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

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

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

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

Remember that:

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

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle \frac{41}{49}$

7. Oops, sorry you are right.