1. ## more trigonometry

$\displaystyle sin^2(\frac{\pi}{8}+\frac{A}{2})-sin^2(\frac{\pi}{8}-\frac{A}{2})=\frac{1}{\sqrt{2}}sinA$

I was having trouble proving this and was wondering if anyone could help me out... thanks in advance.

2. Originally Posted by slevvio
$\displaystyle sin^2(\frac{\pi}{2}+\frac{A}{2})-sin^2(\frac{\pi}{2}-\frac{A}{2})=\frac{1}{\sqrt{2}}sinA$

I was having trouble proving this and was wondering if anyone could help me out... thanks in advance.
Not true:
$\displaystyle sin \left ( \frac{\pi}{2} + \frac{A}{2} \right ) = sin \left ( \frac{\pi}{2} \right ) cos \left ( \frac{A}{2} \right ) + cos \left ( \frac{\pi}{2} \right ) sin \left ( \frac{A}{2} \right ) = cos \left ( \frac{A}{2} \right )$

and
$\displaystyle sin \left ( \frac{\pi}{2} - \frac{A}{2} \right ) = sin \left ( \frac{\pi}{2} \right ) cos \left ( \frac{A}{2} \right ) - cos \left ( \frac{\pi}{2} \right ) sin \left ( \frac{A}{2} \right ) = cos \left ( \frac{A}{2} \right )$

So
$\displaystyle sin^2 \left ( \frac{\pi}{2} + \frac{A}{2} \right ) - sin^2 \left ( \frac{\pi}{2} + \frac{A}{2} \right ) = cos^2 \left ( \frac{A}{2} \right ) - cos^2 \left ( \frac{A}{2} \right ) = 0$

Unless you are to solve for A? In that case then we have
$\displaystyle 0 = \frac{sin(A)}{\sqrt{2}}$
means that
$\displaystyle A = n \pi$
where n is an integer.

-Dan

3. Thanks a lot, no I only needed to prove the statement I didnt realise that $\displaystyle \cos(\frac{\pi}{2})=0$ hence difficulties... Maybe i shouldn't try maths problems at 10:30 at night lol...

4. Originally Posted by slevvio
Thanks a lot, no I only needed to prove the statement I didnt realise that $\displaystyle \cos(\frac{\pi}{2})=0$ hence difficulties... Maybe i shouldn't try maths problems at 10:30 at night lol...
Anytime you're taking a trigonometric function of a constant with a fraction of pi in it, it is a good idea to check your unit circle. There are a few angles which are extremely common, pi/2 being one of them, you will see these angles all the time, because they come out cleanly (well, i guess if you consider square root of 3, divided by 2 clean lol). Anyway, many equations use these angles, and much of the time it is necessary to be able to convert them.

5. Woops! In my exhaustion I copied down the fraction wrong into the post so I will edit it now and hopefully someone will prove that which is actually true!

6. Originally Posted by slevvio
$\displaystyle sin^2(\frac{\pi}{8}+\frac{A}{2})-sin^2(\frac{\pi}{8}-\frac{A}{2})=\frac{1}{\sqrt{2}}sinA$

I was having trouble proving this and was wondering if anyone could help me out... thanks in advance.
$\displaystyle sin^2\frac{1}{2} \left( {\frac{\pi}{4}+A} \right) -sin^2\frac{1}{2} \left( {\frac{\pi}{4}-A} \right) = \frac{1 - cos (\frac{\pi}{4}+A) - (1 - cos (\frac{\pi}{4}-A))}{2}$
$\displaystyle = \frac{cos (\frac{\pi}{4}-A) - cos (\frac{\pi}{4}+A)}{2}$

$\displaystyle = \frac{cos \frac{\pi}{4}cosA + sin \frac{\pi}{4} sin A - cos \frac{\pi}{4}cosA + sin \frac{\pi}{4} sin A}{2}$

$\displaystyle = sin \frac{\pi}{4} sin A = \frac{1}{\sqrt 2}sin A$

7. Thanks for the help! I was wondering if you could tell me which rule you used in the first line of working i.e. $\displaystyle sin^2\frac{1}{2}(\frac{\pi}{4}+A)-sin^2\frac{1}{2}(\frac{\pi}{4}-A)=\frac{1-cos(\frac{\pi}{4}+A)-(1-cos(\frac{\pi}{4}-A)}{2}$?

I eventually got it to work using the addition formula:

$\displaystyle cos\theta - cos\phi = -2sin(\frac{\theta - \phi}{2})sin(\frac{\theta + \phi}{2})$ which gives the same result as yours but after about 2 pages of working - do you know a quicker way? Or did you just omit your working? Thanks again.

8. Originally Posted by slevvio
Thanks for the help! I was wondering if you could tell me which rule you used in the first line of working i.e. $\displaystyle sin^2\frac{1}{2}(\frac{\pi}{4}+A)-sin^2\frac{1}{2}(\frac{\pi}{4}-A)=\frac{1-cos(\frac{\pi}{4}+A)-(1-cos(\frac{\pi}{4}-A)}{2}$?

I eventually got it to work using the addition formula:

$\displaystyle cos\theta - cos\phi = -2sin(\frac{\theta - \phi}{2})sin(\frac{\theta + \phi}{2})$ which gives the same result as yours but after about 2 pages of working - do you know a quicker way? Or did you just omit your working? Thanks again.
ahh okay..

note that $\displaystyle sin \frac{\theta}{2} = \pm \sqrt{\frac{1 - cos \theta}{2}}$

so, $\displaystyle sin^2 \frac{\theta}{2} = \frac{1 - cos \theta}{2}$