Trig equation help Watch

username1393226
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#1
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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
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Smaug123
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(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)?
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username1393226
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(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 ?
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Smaug123
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(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).

I'll have a think and see if I come up with any other ways to solve it. None immediately spring to my attention.
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Sonnyjimisgod
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(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
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TeeEm
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(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
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davros
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(Original post by Smaug123)
You should be capable of deriving the triple angle formula, from the fact that \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
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TeeEm
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(Original post by davros)
Here's a sneaky suggestion: subtract sin x from both sides
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?
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davros
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(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
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TeeEm
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(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
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)
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davros
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(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

It's always rewarding when these things come off!
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TeeEm
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(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

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 ...
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Smaug123
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(Original post by davros)
Here's a sneaky suggestion: subtract sin x from both sides
Nice - I never learnt those formulae, and they never occur to me for that reason.
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davros
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(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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