# Thread: Proving Trigonometric Identities II

1. ## Proving Trigonometric Identities II

Given that $\alpha$ is acute and that $tan \alpha = \frac {3}{4}$, prove that:

$3 sin (\theta + \alpha) + 4 cos (\theta + \alpha) = 5 cos \theta$

I start by saying that $R cos \alpha = 4$ and $R sin \alpha = 3$ but then I get stuck.

2. Originally Posted by Quacky
Given that $\alpha$ is acute and that $tan \alpha = \frac {3}{4}$, prove that:

$3 sin (\theta + \alpha) + 4 cos (\theta + \alpha) = 5 cos \theta$

I start by saying that $R cos \alpha = 4$ and $R sin \alpha = 3$ but then I get stuck.

3. Do you mean as in Sin(A+B)=SinACosB+CosASinB? I've tried but I can't work out how to get an answer in only $\theta$

4. Originally Posted by Quacky
Given that $\alpha$ is acute and that $tan \alpha = \frac {3}{4}$, prove that:

$3 sin (\theta + \alpha) + 4 cos (\theta + \alpha) = 5 cos \theta$

I start by saying that $R cos \alpha = 4$ and $R sin \alpha = 3$ but then I get stuck.
Hi Quacky,

Since $\alpha$ is acute, draw a right-angled triangle.

$Tan\alpha=\frac{3}{4}=\frac{opp}{adj}$

Hence, the hypotenuse is $\sqrt{3^2+4^2}=\sqrt{25}=5$

Hence, $Sin\alpha=\frac{3}{5},\ Cos\alpha=\frac{4}{5}$

Then...

$3Sin(\theta+\alpha)+4Cos(\theta+\alpha)=3(Sin\alph a\ Cos\theta+Cos\alpha\ Sin\theta)+4(Cos\theta\ Cos\alpha-Sin\theta\ Sin\alpha)$

$=3(\frac{3}{5}Cos\theta+\frac{4}{5}Sin\theta)+4(\f rac{4}{5}Cos\theta-\frac{3}{5}Sin\theta)$

$=\frac{9}{5}Cos\theta+\frac{12}{5}Sin\theta+\frac{ 16}{5}Cos\theta-\frac{12}{5}Sin\theta=\frac{25}{5}Cos\theta=5Cos\t heta$

5. Ah. I would never have got that. Thanks again.