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Trig equation help

Heres the question http://gyazo.com/f100c8ee789906d686dd5f841a2220bc
From there rearrange the equation:3= sin(3x)/sin(x)
or
0=sin(3x)/sin(x) - 3

But dont really understand what to do after that
Original post by T-GiuR
Heres the question http://gyazo.com/f100c8ee789906d686dd5f841a2220bc
From there rearrange the equation:3= sin(3x)/sin(x)
or
0=sin(3x)/sin(x) - 3

But dont really understand what to do after that

Remember that if you divide at any point, then you have to take separate account of the case that the dividend is zero.

The first way that springs to mind is a bit brute-force. Do you know the triple-angle formula for sin (or can you deduce it from the addition formulae)?
Original post by Smaug123
Remember that if you divide at any point, then you have to take separate account of the case that the dividend is zero.

The first way that springs to mind is a bit brute-force. Do you know the triple-angle formula for sin (or can you deduce it from the addition formulae)?

We have only covered the double angle formula, and not triple and I assume we wont either, as we have moved onto a different topic.

With that being said is the "triple angle formula" the only way to solve this ?
Original post by T-GiuR
We have only covered the double angle formula, and not triple and I assume we wont either, as we have moved onto a different topic.

With that being said is the "triple angle formula" the only way to solve this ?

You should be capable of deriving the triple angle formula, from the fact that sin⁡(3x)=sin⁡(x+2x)\sin(3x) = \sin(x+2x).

I'll have a think and see if I come up with any other ways to solve it. None immediately spring to my attention.
Original post by T-GiuR
We have only covered the double angle formula, and not triple and I assume we wont either, as we have moved onto a different topic.

With that being said is the "triple angle formula" the only way to solve this ?


Do sin(2x+x) by using the double angle formula and the sin addition formula
Reply 5
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TeeEm
19
Original post by T-GiuR
We have only covered the double angle formula, and not triple and I assume we wont either, as we have moved onto a different topic.

With that being said is the "triple angle formula" the only way to solve this ?


follow smaug'a advice.

The only sensible way is to derive the triple angle formula.

The triple angle formula is not a topic! It is just an "exercise/application" involving the compound and double angle identities.

There is no sensible alternative to this question
Reply 6
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davros
16
Original post by Smaug123
You should be capable of deriving the triple angle formula, from the fact that sin⁡(3x)=sin⁡(x+2x)\sin(3x) = \sin(x+2x).

I'll have a think and see if I come up with any other ways to solve it. None immediately spring to my attention.



Original post by TeeEm
follow smaug'a advice.

The only sensible way is to derive the triple angle formula.

The triple angle formula is not a topic! It is just an "exercise/application" involving the compound and double angle identities.

There is no sensible alternative to this question


Here's a sneaky suggestion: subtract sin x from both sides :smile:
Reply 7
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TeeEm
19
Original post by davros
Here's a sneaky suggestion: subtract sin x from both sides :smile:


I have been teaching non stop for 12 hours today and perhaps I cannot do 3 x 4 at the moment ...
but...
...am I missing the obvious or is it a joke that I am too slow to get?
Reply 8
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davros
16
Original post by TeeEm
I have been teaching non stop for 12 hours today and perhaps I cannot do 3 x 4 at the moment ...
but...
...am I missing the obvious or is it a joke that I am too slow to get?


There is a standard formula for sin A - sin B. I wondered if it would be helpful in this case, so I rewrote the equation as sin(3x) - sin x = 2sinx, then applied the formula to the LHS(actually I rederived it because I haven't memorized it but it's easy to deduce from the sin(X+Y) formula which I can remember!).

This gives a new trig equation which can be factorized and solved :smile:
Reply 9
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TeeEm
19
Original post by davros
There is a standard formula for sin A - sin B. I wondered if it would be helpful in this case, so I rewrote the equation as sin(3x) - sin x = 2sinx, then applied the formula to the LHS(actually I rederived it because I haven't memorized it but it's easy to deduce from the sin(X+Y) formula which I can remember!).

This gives a new trig equation which can be factorized and solved :smile:


Valid method it will definitely work ...

However for most students on a first course of trigonometry it will be a hard act to perform.

usually a question such as this has a prompt such as

By using .... or otherwise

In which case I could easily see your suggestion as a promp.

(lucky you can still think at the moment, unlike me)
Reply 10
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davros
16
Original post by TeeEm
Valid method it will definitely work ...

However for most students on a first course of trigonometry it will be a hard act to perform.

usually a question such as this has a prompt such as

By using .... or otherwise

In which case I could easily see your suggestion as a promp.

(lucky you can still think at the moment, unlike me)


It was a flash of inspiration really...if it hadn't come out with a factor common to both sides I'd just have kept the idea to myself :biggrin:

It's always rewarding when these things come off!
Reply 11
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TeeEm
19
Original post by davros
It was a flash of inspiration really...if it hadn't come out with a factor common to both sides I'd just have kept the idea to myself :biggrin:

It's always rewarding when these things come off!


I know the feeling ...

Keep up the good work and keep away from Dr "listen mate" Einstein ...
Original post by davros
Here's a sneaky suggestion: subtract sin x from both sides :smile:

Nice - I never learnt those formulae, and they never occur to me for that reason.
Reply 13
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davros
16
Original post by Smaug123
Nice - I never learnt those formulae, and they never occur to me for that reason.


I've never troubled to memorize them - too much danger of writing one down the wrong way round or getting a sign wrong - but they're easy enough to derive.

I've seen a couple of AEA questions of the form "Solve sin A - cos B = sin C + cos D" where you need a sneaky rearrangement first then apply the formula, so it just occurred to me to try something vaguely similar here!

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