Author Topic: Proving trig identities with compound, douple and triple angle formulae (Read 1386 times) Tweet Share

hinakamishiro · « **on:** July 24, 2016, 11:38:35 am »

Hi guys! I was having some trouble proving trig identities using compound, double and triple angle theorems. I've attached the questions (the circled ones are the ones I can't do) below and I would really appreciate if someone could explain how to prove them! Thanks

RuiAce · « **Reply #1 on:** July 24, 2016, 11:49:56 am »

Please refrain from asking too many questions related to the same thing all at once. It's very off-putting and really takes away the will to help.
$\text{For a), note that we want to get }\sin 2\alpha\text{ on the denominator. With a direction in mind, start tackling with the identities.}\\ \begin{align*}LHS&= \cot 2\alpha + \tan \alpha\\ &= \frac{1-\tan^2 \alpha}{2\tan \alpha}+\tan \alpha \\&= \frac{1-\tan^2 \alpha}{2\tan \alpha}+\frac{2\tan^2 \alpha}{2\tan \alpha}\\ &= \frac{1+\tan^2\alpha}{2\tan \alpha}\\ &= \frac{\sec^2 \alpha}{2\tan \alpha}\\ &= \frac{1}{2\sin\alpha\cos\alpha}\tag{*}\\ &= \frac{1}{\sin 2\alpha}\\&= \csc 2\alpha\\ &=RHS \end{align*}\\ \text{Line * multiplied top and bottom by }\cos^2 \alpha\\ \text{Some double angles had to be used on top of Pythagorean identities.}$
$\text{For n), we have to take the hint, tidy things up and hope for the best.}\\ \begin{align*}LHS&=(\tan \theta + \tan 2\theta)(\cot \theta + \cot 3\theta)\\ &= \left(\frac{\tan \theta - \tan^3\theta}{1-\tan^2\theta}+\frac{2\tan\theta}{1-\tan^2\theta} \right)\left(\frac{1}{\tan\theta}+\frac{1-3\tan^2\theta}{3\tan\theta-\tan^3\theta}\right)\\ &= \left(\frac{3\tan\theta-\tan^3\theta}{1-\tan^2\theta}\right)\left(\frac{3-\tan^2\theta}{3\tan\theta-\tan^3\theta}+\frac{1-3\tan^2\theta}{3\tan\theta-\tan^3\theta}\right) \\ &=\left(\frac{3\tan\theta-\tan^3\theta}{1-\tan^2\theta}\right)\left(\frac{4-4\tan^2\theta}{3\tan\theta-\tan^3\theta}\right)\end{align*}\\ \text{This shows that hoping for the best works. Brute force made the expression so much more simplified. Everything cancels down to 4.}$
Edit: Had a look at m). That one is really hard so I'll also post a solution for it.
$\text{For m), at first glance it's not obvious at all what we're trying to do to the numerator.}\\ \text{But the sum of two cubes in the denominator is interesting. Attempt a factorisation process first.}$
$\begin{align*}&\, 2\cos\alpha+\sin\alpha-2(\cos^3\alpha+\sin^3\alpha)\\&=\cos\alpha+(\cos\alpha+\sin\alpha)-2(\cos\alpha+\sin\alpha)(\cos^2\alpha-\cos\alpha\sin\alpha+\sin^2\alpha)\\ &=\cos\alpha+(\cos\alpha+\sin\alpha)-2(\cos\alpha+\sin\alpha)(1-\cos\alpha\sin\alpha)\end{align*}$
$\text{Now we factor out the entire } (\cos\alpha+\sin\alpha)\\ \begin{align*}&=\cos\alpha+(\cos\alpha+\sin\alpha)-(\cos\alpha+\sin\alpha)(2-2\cos\alpha\sin\alpha)\\ &= \cos\alpha+(\cos\alpha+\sin\alpha)(1-2+2\cos\alpha\sin\alpha)\\&=\cos\alpha+ (\cos\alpha+\sin\alpha)(2\cos\alpha\sin\alpha-1)\end{align*}$
$\text{And here I give up. Putting the double angle back in there looks like it's useless.}\\ \text{Time to hope for the best.}\\ \begin{align*}LHS&=\frac{\sin 2\alpha+\cos 2\alpha}{2\cos\alpha+\sin\alpha-2(\cos^3\alpha-\sin^3\alpha)}\\&= \frac{2\sin\alpha\cos\alpha+2\cos^2\alpha-1}{\cos\alpha+ (\cos\alpha+\sin\alpha)(2\cos\alpha\sin\alpha-1)}\\ &= \frac{2\sin\alpha\cos\alpha+2\cos^2\alpha-1}{\cos\alpha+2\cos^2\alpha\sin\alpha-\cos\alpha+2\cos\alpha\sin^2\alpha-\sin\alpha}\\ &= \frac{2\sin\alpha\cos\alpha+2\cos^2\alpha-1}{2\cos^2\alpha\sin\alpha+2\cos\alpha\sin^2\alpha-\sin\alpha}\\ &= \frac{2\sin\alpha\cos\alpha+2\cos^2\alpha-1}{\sin\alpha(2\cos^2\alpha+2\cos\alpha\sin\alpha-1)}\\ &= \frac{1}{\sin \alpha}\\&=\csc \alpha \\ &= RHS\end{align*}$

hinakamishiro · « **Reply #2 on:** July 24, 2016, 12:11:37 pm »

Quote from: RuiAce on July 24, 2016, 11:49:56 am

Please refrain from asking too many questions related to the same thing all at once. It's very off-putting and really takes away the will to help.
$\text{For a), note that we want to get }\sin 2\alpha\text{ on the denominator. With a direction in mind, start tackling with the identities.}\\ \begin{align*}LHS&= \cot 2\alpha + \tan \alpha\\ &= \frac{1-\tan^2 \alpha}{2\tan \alpha}+\tan \alpha \\&= \frac{1-\tan^2 \alpha}{2\tan \alpha}+\frac{2\tan^2 \alpha}{2\tan \alpha}\\ &= \frac{1+\tan^2\alpha}{2\tan \alpha}\\ &= \frac{\sec^2 \alpha}{2\tan \alpha}\\ &= \frac{1}{2\sin\alpha\cos\alpha}\tag{*}\\ &= \frac{1}{\sin 2\alpha}\\&= \csc 2\alpha\\ &=RHS \end{align*}\\ \text{Line * multiplied top and bottom by }\cos^2 \alpha\\ \text{Some double angles had to be used on top of Pythagorean identities.}$
$\text{For n), we have to take the hint, tidy things up and hope for the best.}\\ \begin{align*}LHS&=(\tan \theta + \tan 2\theta)(\cot \theta + \cot 3\theta)\\ &= \left(\frac{\tan \theta - \tan^3\theta}{1-\tan^2\theta}+\frac{2\tan\theta}{1-\tan^2\theta} \right)\left(\frac{1}{\tan\theta}+\frac{1-3\tan^2\theta}{3\tan\theta-\tan^3\theta}\right)\\ &= \left(\frac{3\tan\theta-\tan^3\theta}{1-\tan^2\theta}\right)\left(\frac{3-\tan^2\theta}{3\tan\theta-\tan^3\theta}+\frac{1-3\tan^2\theta}{3\tan\theta-\tan^3\theta}\right) \\ &=\left(\frac{3\tan\theta-\tan^3\theta}{1-\tan^2\theta}\right)\left(\frac{4-4\tan^2\theta}{3\tan\theta-\tan^3\theta}\right)\end{align*}\\ \text{This shows that hoping for the best works. Brute force made the expression so much more simplified. Everything cancels down to 4.}$

Thank you so much! This really helped. Sorry about all the questions though. XD

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Author Topic: Proving trig identities with compound, douple and triple angle formulae (Read 1386 times) Tweet Share

hinakamishiro

Proving trig identities with compound, douple and triple angle formulae

RuiAce

Re: Proving trig identities with compound, douple and triple angle formulae

hinakamishiro

Re: Proving trig identities with compound, douple and triple angle formulae

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