Need some help with trig identities!

[jordan83]

tan(3theta) = 3tan(theta)-tan^3(theta) / 1-3tan^2(theta)

cos(alpha-beta)cos(alpha+beta) = cos^2(alpha) - sin^2(beta)

ln = natural log

ln|1+cos(theta)| + ln|1-cos(theta)| = 2ln|sin(theta)|


I need to prove these to be equal. These are the ones im stuck on on my worksheet thanks

 

[galactus]

tan(3theta) = 3tan(theta)-tan^3(theta) / 1-3tan^2(theta)

cos(alpha-beta)cos(alpha+beta) = cos^2(alpha) - sin^2(beta)

ln = natural log

ln|1+cos(theta)| + ln|1-cos(theta)| = 2ln|sin(theta)|

Here's some good hints:

Use the laws of logs. Remember, \(\displaystyle ln(a)+ln(b)=ln((a)\cdot(b))\)

Also, \(\displaystyle (1+cos(\theta))(1-cos(\theta))=1-cos^{2}(\theta)=sin^{2}(\theta).\)

Another log law, \(\displaystyle ln(a^{b})=(b)\cdot(ln(a))\)


I need to prove these to be equal. These are the ones im stuck on on my worksheet thanks

 

[soroban]

Hello, jordan83!

You're expected to know the Compound Angle Formula: .\(\displaystyle \tan(A\,+\,B)\;=\;\frac{\tan(A)\,+\,\tan(B)}{1\,-\,\tan(A)\tan(B)}\)

And the Double Angle formula: .\(\displaystyle \tan(2A)\;=\;\frac{2\cdot\tan(A)}{1\,-\,\tan^2(A)}\)

\(\displaystyle \tan(3\theta)\:=\:\frac{3\cdot\tan(\theta)\,-\,\tan^3(\theta)}{1\,-\,3\cdot\tan^2(\theta)}\)

We have: .\(\displaystyle \tan(2\theta\,+\,\theta)\;=\;\frac{\tan(2\theta)\,+\,\tan(\theta)}{1\,-\,\tan(2\theta)\cdot\tan(\theta)}\;=\;\L\frac{\frac{2\cdot\tan(\theta)}{1\,-\,\tan^2(\theta)}\,+\,\tan(\theta)}{1\,-\,\frac{2\cdot\tan(\theta)}{1\,-\,\tan^2(\theta)}\cdot\tan(\theta)}\)

Multiply top and bottom by \(\displaystyle [1- \tan^2(\theta)]:\;\;\frac{2\cdot\tan(\theta)\,+\,\tan(\theta)[1\,-\,\tan^2(\theta)]}{[1\,-\,\tan^2(\theta)]\,-\,2\cdot\tan(\theta)\cdot\tan(\theta)}\)


. . . \(\displaystyle =\;\frac{2\cdot\tan(\theta)\,+\,\tan(\theta)\,-\,\tan^3(\theta)}{1\,-\,\tan^2(\theta)\,-\,2\cdot\tan^2(\theta)}\;=\;\frac{3\cdot\tan(\theta)\,-\,\tan^3(\theta)}{1\,-\,3\cdot\tan^2(\theta)}\)