MORE trig help please?

[jordan83]

I need help. I need to establish the following identity.

sin(3theta)/sin(theta) - cos(3theta)/cos(theta) + cos(alpha + beta)/ cos(alpha)cos(beta) = 3 - tan(alpha)tan(beta)

Any help would be great I know there are alot of steps but I get lost everytime.

I usually get the first part down to sin(theta)cos2(theta) + sin2(theta)cos(theta) / sin(theta) - cos(theta)cos2(theta) - sin(theta)sin2(theta) / cos(theta) the third part of the problem is pretty easy and i can get it down to 1-tan(alpha)tan(beta).

 

[soroban]

Hello, jordan83!

You'll need all your identities for this one . . .

\(\displaystyle \frac{\sin(3\theta)}{\sin(\theta)}\,-\,\frac{\cos(3\theta)}{\cos(\theta)}\,+\,\frac{\cos(A + B)}{\cos(A)\cos(B)}\;=\;3\,-\,\tan(A)\tan(B)\)

Let's work with the first two fractions:

\(\displaystyle \frac{\sin(3\theta)}{\sin(\theta)}\,-\,\frac{\cos(3\theta)}{\cos(\theta)}\;=\;\frac{\sin(3\theta)\cos(\theta)\,-\,\sin(\theta)\cos(3\theta)}{\sin(\theta)\cos(\theta)}\;=\;\frac{\sin(3\theta-\theta)}{\sin(\theta)\cos(\theta)}\)

. . . \(\displaystyle =\;\frac{\sin(2\theta)}{\sin(\theta)\cos(\theta)}\;=\;\frac{2\cdot\sin(\theta)\cos(\theta)}{\sin(\theta)\cos(\theta)} \;= \;2\) . . . . . wow!


The second fraction is:
. . \(\displaystyle \frac{\cos(A)\cos(B)\,-\,\sin(A)\sin(B)}{\cos(A)\cos(B)} \;=\;\frac{\cos(A)\cos(B)}{\cos(A)\cos(B)}\,-\,\frac{\sin(A)\sin(B)}{\cos(A)\cos(B)}\;=\;1\,-\,\tan(A)\tan(B)\)


Therefore, the left side is: .\(\displaystyle 2\,+\,1\,-\,\tan(A)\tan(B)\;=\;3\,-\,\tan(A)\tan(B)\) . . . . There!

 

[jordan83]

Thank you so much!! You are a genius.