more trig

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From the beginning, the question drops out quite nicely if you use the sum/product formulae for sines and cosines.

Sorry, are the sum/product formulae the same thing as double angle formulae?

$\cos \theta + \cos 3\theta = \sin \theta + \sin 3\theta$range is from 0 to 2 pi and we have to find the solutions to the equation.

So from about 2 pages of workings I think I have managed to get it down to $1-2 \sin 2\theta = \tan \theta$ but I'm starting to think this is wrong as I can't see where to go next.

Basically I used the idea that 3= 2 + 1 and use compound angle formula on the sin3x and cos3x and moved on from there

Of course, excellent. It's just that I prefer to convert everything to s and c first and then use the usual identities. So sin(3x) becomes 3s-4s^3, etc., because this method always works with A level trig.

For this one I would use two different notations.

$s_1=\sin a$, $c_1=\cos a$ (where a=3x), and

$s=\sin x$, $c=\cos x$.

Left hand side = $\sin x+\sin (2(3x)+x)$
$= \sin x+\sin(2(3x))\cos x+\sin x\cos(2(3x))$
$=s+2s_1c_1c+s(c_1^2-s_1^2)$.

Right hand side = $\cos x+\cos(2(3x)+x)$
$=\cos x+\cos(2(3x))\cos x-\sin(2(3x))\sin x$
$=c+(c_1^2-s_1^2)c-2s_1c_1s$.

So the equation is

$s+2s_1c_1c+s(c_1^2-s_1^2)=c+(c_1^2-s_1^2)c-2s_1c_1s$

$(c-s)(1+c_1^2-s_1^2)=2s_1c_1(c+s)$ (*)

$\frac{s}{c-s}=\frac{c_1-s_1}{2s_1}$

$s_1(s+c)=c_1(c-s)$

$s_1^2(1+2sc)=c_1^2(1-2sc)$

$s_1^2(1+2sc)=(1-s_1^2)(1-2sc)$

$s_1^2+2s_1^2sc=1-2sc-s_1^2+2s_1^2sc$

$s_1^2=1-2sc-s_1^2$

$1-2s_1^2=2sc$

$\cos(2(3x))=\sin(2x)$

And then check for extraneous solutions at the end, since we squared.

The only caveat is (*) when we divided by $2s_1c_1(c-s)$, so we need to find when this is 0. This happens when either $c=s$ or $s_1=0$ or $c_1=0$, which are all easy to dispose of.

arkanm, that method is so clumsy and unfruitful that I am contemplating suicide right now.

Credits to Zen for LaTeX'ing my solution because I'm a recumbent bastard.

I think you're on the verge of "trolling" here again.

Your method is what's known in professional circles as using a sledgehammer to crack a walnut

Too mainstream for me.

qtsie tootsie xxx

Look in your formula book!

You have formulae for cos A + cos B and sin A + sin B which will help you enormously

As already suggested, a veteran like *me* would get to this point in one line using the sum/product formulae.

From the beginning, the question drops out quite nicely if you use the sum/product formulae for sines and cosines.

This may seem like a stupid question but do we use the sum formulae when we have sinx +/- siny where x and y are different? i.e. x= $2\theta$ y=$57\theta$ (so does it work for all amounts of theta or does it break down anywhere).

Finally, we had $\cos \theta + \cos 3\theta = \sin \theta + \sin 3\theta$, would this method work if we had different amounts of theta on each side?

Thanks I did a bit of research on the sum formulae and managed to do the question correctly!

This may seem like a stupid question but do we use the sum formulae when we have sinx +/- siny where x and y are different? i.e. x= $2\theta$ y=$57\theta$ (so does it work for all amounts of theta or does it break down anywhere).

Finally, we had $\cos \theta + \cos 3\theta = \sin \theta + \sin 3\theta$, would this method work if we had different amounts of theta on each side?

Yes to the first one.

Depends on the values for the second one.

However, if you haven't covered the sum/product formulae, you probably aren't meant to use them. Your original method would work, but it looks as if you made an algebraic slip somewhere.

Not quite sure what you're asking!

The formulae you have for sin A + sin B etc are identities - they are true for any values of A and B (and therefore for any multiples of theta).

If you mean could you get an equation like sin A + sin B = cos C + cos D that you couldn't solve because A, B, C, D didn't combine to give nice numbers, then that's possible - but it won't happen in an exam - they'll choose "nice" angles that let you solve the problem!