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What is this method of solving called

Lets say, I am trying to get −tanθ1−tan2θ -\frac{tan\theta}{1-tan^2\theta} into its most simple form.

So I answser −tanθ1−tan2θ=−12(2tanθ1−tan2θ)=−12tan2θ - \frac{tan \theta}{1-tan^2 \theta}=- \frac{1}{2}(\frac{2tan \theta}{1-tan^2 \theta})=-\frac{1}{2}tan2 \theta What would be the name of what i did when i made
−tanθ1−tan2θ=−12(2tanθ1−tan2θ) - \frac{tan \theta}{1-tan^2 \theta}=- \frac{1}{2}(\frac{2tan \theta}{1-tan^2 \theta})?
(edited 13 years ago)
Reply 1
Avatar for nuodai
nuodai
17
Does it need a name? You multiplied the top and bottom of a fraction by the same number, so I suppose you could call it "multiplication by one".
Reply 2
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Core
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Original post by nuodai
Does it need a name? You multiplied the top and bottom of a fraction by the same number, so I suppose you could call it "multiplication by one".


Im not very good at using this, but ive seen it done before on many trig identities, how do know when to use it?
Reply 3
Avatar for Ollie53
Ollie53
Original post by Core
Im not very good at using this, but ive seen it done before on many trig identities, how do know when to use it?


I think it's more of a case of knowing what you are looking for rather than knowing when to use it. Obviously there will be examples where this "technique" appears to be the only thing to do, but in others it will be down more to trial and error, and experimenting with different identities.
Reply 4
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Core
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Also in the maths book it says that 2cos2θ=2cosθ \sqrt{2cos^2\theta} = \sqrt{2cos\theta} when i checked on my calc it did not.. trg is starting to irritate me. Not because of trig itself but the way the questions are bing solved.
(edited 13 years ago)
Reply 5
Avatar for Pheylan
Pheylan
Original post by Core
Also in the maths book it says that 2cos2θ=2cosθ \sqrt{2cos^2\theta} = \sqrt{2cos\theta} when i checked on my calc it did not.. trg is starting to irritate me. Not because of trig itself but the way the questions are bing solved.


It's a mistake (either in the book or on your part). 2cos2θ≡2cosθ\sqrt{2cos^2\theta}\equiv\sqrt2cos\theta
Reply 6
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Core
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Original post by Pheylan
It's a mistake (either in the book or on your part). 2cos2θ≡2cosθ\sqrt{2cos^2\theta}\equiv\sqrt2cos\theta


are you saying it shouldbe ≡ \equiv and not = = ? Ah 2⋅cosθ \sqrt{2}\cdot cos\theta i though it was everything rooted.
(edited 13 years ago)
Reply 7
Avatar for nuodai
nuodai
17
Original post by Core
are you saying it shouldbe ≡ \equiv and not = = ? Ah 2⋅cosθ \sqrt{2}\cdot cos\theta i though it was everything rooted.


It shouldn't be ≡\equiv because that implies it's true for all values of θ\theta. In fact, 2cos⁡2θ≡2∣cos⁡θ∣\sqrt{2\cos^2 \theta} \equiv \sqrt{2} |\cos \theta| because the LHS always has to be positive... but don't worry too much about the difference between = and ≡\equiv (and if in doubt, just use =).

As for your question about when this method is used, it's used when you can use an identity but what you have is out by some constant factor. So for example, say you saw 3sin⁡θcos⁡θ3\sin \theta \cos \theta. We know that sin⁡2θ=2sin⁡θcos⁡θ\sin 2 \theta = 2\sin \theta \cos \theta, but what we have has a 3 instead of a 2. But, if we multiply both sides by 32\dfrac{3}{2} then we get 32sin⁡2θ=32×2sin⁡θcos⁡θ=3sin⁡θcos⁡θ\dfrac{3}{2} \sin 2 \theta = \dfrac{3}{2} \times 2 \sin \theta \cos \theta = 3 \sin \theta \cos \theta.

This is exactly the same as what you did in your question; you had −tan⁡θ1−tan⁡2θ-\dfrac{\tan \theta}{1-\tan^2 \theta} and you knew the identity tan⁡2θ=2tan⁡θ1−tan⁡2θ\tan 2\theta = \dfrac{2\tan \theta}{1-\tan^2 \theta}, but in order to get what you need you have to divide both sides by -2, and so:
−12tan⁡2θ=−12×2tan⁡θ1−tan⁡2θ=−tan⁡θ1−tan⁡2θ-\dfrac{1}{2}\tan 2\theta = -\dfrac{1}{2} \times \dfrac{2\tan \theta}{1-\tan^2 \theta} = -\dfrac{\tan \theta}{1-\tan^2 \theta}

Hence your answer.
(edited 13 years ago)
Reply 8
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Core
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my mistake thanks very helpful thread.
Reply 9
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Core
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Original post by nuodai
It shouldn't be ≡\equiv because that implies it's true for all values of θ\theta. In fact, 2cos⁡2θ≡2∣cos⁡θ∣\sqrt{2\cos^2 \theta} \equiv \sqrt{2} |\cos \theta| because the LHS always has to be positive... but don't worry too much about the difference between = and ≡\equiv (and if in doubt, just use =).

As for your question about when this method is used, it's used when you can use an identity but what you have is out by some constant factor. So for example, say you saw 3sin⁡θcos⁡θ3\sin \theta \cos \theta. We know that sin⁡2θ=2sin⁡θcos⁡θ\sin 2 \theta = 2\sin \theta \cos \theta, but what we have has a 3 instead of a 2. But, if we multiply both sides by 32\dfrac{3}{2} then we get 32sin⁡2θ=32×2sin⁡θcos⁡θ=3sin⁡θcos⁡θ\dfrac{3}{2} \sin 2 \theta = \dfrac{3}{2} \times 2 \sin \theta \cos \theta = 3 \sin \theta \cos \theta.

This is exactly the same as what you did in your question; you had −tan⁡θ1−tan⁡2θ-\dfrac{\tan \theta}{1-\tan^2 \theta} and you knew the identity tan⁡2θ=2tan⁡θ1−tan⁡2θ\tan 2\theta = \dfrac{2\tan \theta}{1-\tan^2 \theta}, but in order to get what you need you have to divide both sides by -2, and so:
−12tan⁡2θ=−12×2tan⁡θ1−tan⁡2θ=−tan⁡θ1−tan⁡2θ-\dfrac{1}{2}\tan 2\theta = -\dfrac{1}{2} \times \dfrac{2\tan \theta}{1-\tan^2 \theta} = -\dfrac{\tan \theta}{1-\tan^2 \theta}
Hence your answer.


Awsome thanks, ive been trying to use the multply by 1 rule for almost a week without prevail.
Reply 10
Avatar for nuodai
nuodai
17
Original post by Core
Awsome thanks, ive been trying to use the multply by 1 rule for almost a week without prevail.


Pleasure. I just realised my explanation kind of went about it the other way round. Taking the 3sin⁡θcos⁡θ3 \sin \theta \cos \theta example again, you want to see a 2 instead of a 3 so you multiply by 22\dfrac{2}{2} to get 3×2sin⁡θcos⁡θ2\dfrac{3 \times 2 \sin \theta \cos \theta}{2}, and then using the identity gives 3sin⁡2θ2\dfrac{3 \sin 2\theta}{2}. You can use whichever method is easiest for you (i.e. starting with the identity and multiplying by things until you get your function, or starting with your function and "multiplying by one" to get something you can apply the identity to).

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