# Can anyone help me with this Trig Identity double angle question?

• Dec 26th 2010, 10:26 PM
solarflare
Can anyone help me with this Trig Identity double angle question?
Hi guys I am having real difficulty cracking this tri identity.

square root 2 sin (theta- pi/4) = sin theta - cos theta

I have attempted it ever so slightly but am struggling at transposing the LHS to match the RHS

pi/4 I figured was square root 2/2 as a trigonometric value, I thought that this would have some influenece in cancelling out the initial square root of 2.

Another thought was to calculate the pi/4 = 0.785 and also square root of 2 sin = 1.414.

Again I am not entirely sure which path or direction to take with this one, any help would be massively appreciated.

(Wink)
• Dec 26th 2010, 10:32 PM
TheCoffeeMachine
$\sin\left(\alpha-\beta\right) = \sin{\alpha}\cos{\beta}-\cos{\alpha}\sin{\beta}$
and $\sin\frac{\pi}{4} = \cos{\frac{\pi}{4} = \frac{1}{\sqrt{2}}.$ That should do it.
• Dec 26th 2010, 11:28 PM
solarflare
Thanks Coffee machine,

Very kind, please can you explain how you transposed to get to the result that you did, I was confused about how I can get rid of the initial square root of 2 (space) sin?,...... crikey I have been banging my head on this one for over a week.

Much appreciated
• Dec 26th 2010, 11:58 PM
TheCoffeeMachine
My suggestion was to use the addition identity:

$\sin\left(\alpha-\beta\right) = \sin{\alpha}\cos{\beta}-\cos{\alpha}\sin{\beta}.$

You have:

$\sin\left(\theta-\frac{\pi}{4}\right) = \sin{\theta}\cos{\frac{\pi}{4}}-\cos{\theta}\sin{\frac{\pi}{4}}.$

The value of $\cos\frac{\pi}{4}$ is $\frac{1}{\sqrt{2}}$ and that of $\sin\frac{\pi}{4}$ is $\frac{1}{\sqrt{2}}$, so we have:

$\sin\left(\theta-\frac{\pi}{4}\right) = \sin{\theta}\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\sin{\theta} = \frac{1}{\sqrt{2}}\left(\sin{\theta}-\cos{\theta}\right).$

Thus $\sqrt{2}\sin\left(\theta-\frac{\pi}{4}\right) = \frac{\sqrt{2}}{\sqrt{2}}\left(\sin{\theta}-\cos{\theta}\right) = \sin{\theta}-\cos{\theta}.$
• Dec 27th 2010, 01:41 AM
solarflare
Very kind, thanks ever so much, I can see how this progresses now.