Trig identities and equations

Jellicorse · Postby **Jellicorse** » Thu Jul 18, 2013 11:02 am

I have been having a lot of problems with trig identities and their use in solving formulae, and wondered if anyone could help me out with this.

The following is an equation which I thought I had solved but my answer is different from that provided by my book, so I must have made a mistake. The problem is, I can't see where.

Solve the equation $cos 2\theta - 5 cos\theta = 2$

One of my big difficulties with this kind of question is knowing which identity to use.

For example, in this question, saying $cos2\theta = 5 cos\theta +2$ means there are three possible starting points:

$cos^2\theta-sin^2\theta = 5 cos\theta +2$

or

$2 cos^2 \theta -1 = 5 cos \theta +2$

or

[ $1-2 sin^2 \theta = 5 cos\theta +2$

After looking at these three possibilities, I decided the second may be the best as everything is in terms of cos theta and it's quadratic...

$2 cos^2 \theta -1 = 5 cos \theta +2$ = $2 cos^2 \theta - 5 cos\theta - 3= 0$

And factorizing this:

$(2cos\theta -1)(cos \theta -3)$

So $2cos\theta = 1$ or [ $cos\theta = 3$ . The latter is outside the domain, so can be discounted.

$cos\theta = \frac{1}{2}$

$\theta = cos^{-1} \frac{1}{2}$

$\theta = \frac{\pi}{3} = 1.047$

But this is not what the books says the answer should be ( $\pm 2.09$ )

I would really appreciate it if anyone could tell me what's going wrong here because these trig identities are really giving me a hard time..!

FWT · Postby **FWT** » Thu Jul 18, 2013 5:41 pm

Jellicorse wrote:One of my big difficulties with this kind of question is knowing which identity to use.

If there's a shortcut or formula for knowing which ident to use, I've never heard of it. You got to try different ones until you find one that works. Sometimes you and a friend will use different ones but still come out to the same answer.

Jellicorse wrote:The following is an equation which I thought I had solved but my answer is different from that provided by my book, so I must have made a mistake. The problem is, I can't see where.

Solve the equation $cos 2\theta - 5 cos\theta = 2$

For example, in this question, saying $cos2\theta = 5 cos\theta +2$ means there are three possible starting points:

$cos^2\theta-sin^2\theta = 5 cos\theta +2$

or

$2 cos^2 \theta -1 = 5 cos \theta +2$

or

[ $1-2 sin^2 \theta = 5 cos\theta +2$

After looking at these three possibilities, I decided the second may be the best as everything is in terms of cos theta and it's quadratic...

Yeh, that's what I'd do, too.

Jellicorse wrote: $2 cos^2 \theta -1 = 5 cos \theta +2$ = $2 cos^2 \theta - 5 cos\theta - 3= 0$

And factorizing this:

$(2cos\theta -1)(cos \theta -3)$

Nah: it's 5 not 7 so you need (2cos(@) + 1)(cos(@) - 3) = 0. Do the sign change & maybe it comes out right.

Jellicorse · Postby **Jellicorse** » Fri Jul 19, 2013 12:46 pm

Thanks a lot FWT, that did lead to the correct solution...

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Trig identities and equations

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