# Math Help - Trig problem

1. ## Trig problem

Hi just started trig for university and stuck on this problem highly appreciate if someone can point me in the right direction thanks

Find "R" and "a" such that:

9sin% + 4cos% = Rsin(% + a)

% = angle theta in radian.

2. ## Re: Trig problem

Hello, ZKG9744!

If you "just started trig", this is a very fancy problem.

$\text{Find }R\text{ and }a\text{ such that: }\:9\sin\theta + 4\cos\theta \:=\: R\sin(\theta + a)$

We have: . $9\sin\theta + 4\cos\theta$

Multiply by $\tfrac{\sqrt{97}}{\sqrt{97}}\!:\;\;\tfrac{\sqrt{97 }}{\sqrt{97}}(9\sin\theta + 4\cos\theta) \;=\;\sqrt{97}\left(\tfrac{9}{\sqrt{97}}\sin\theta + \tfrac{4}{\sqrt{97}}\cos\theta\right)$

$\text{Let }\,\sin a \,=\,\tfrac{4}{\sqrt{97}},\:\cos a \,=\,\tfrac{9}{\sqrt{97}}$

We have: . $\sqrt{97}(\cos a\sin\theta + \sin a\cos\theta) \;=\;\sqrt{97}\sin(\theta + a)$

$\text{Therefore: }\:R \,=\,\sqrt{97},\;\; a \,=\,\sin^{\text{-}1}\!\left(\tfrac{4}{\sqrt{97}}\right)$

3. ## Re: Trig problem

Thanks understanding is better less fuzzy now. Sqr(97/97) still confused where that came from sorry

4. ## Re: Trig problem

If you think of a right angled triangle with the side opposite a having length equal to 4 and the side adjacent to a having the length 9. The hypotenuse, by pythagoras' theorem is $\sqrt{4^2+9^2}=\sqrt{97}$
Then with the basic definition of sin: $\sin{a}=\frac{opposite}{hypotenuse}=\frac{4}{\sqrt {97}}$. Similar thing with cos.

You might find it easier to do the problem by first expanding $sin(\theta+a)$

5. ## Re: Trig problem

@ Shakarri
Thank you very very much