# Trig problem

• Aug 27th 2013, 02:57 PM
ZKG9744
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.
• Aug 27th 2013, 04:10 PM
Soroban
Re: Trig problem
Hello, ZKG9744!

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

Quote:

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

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

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

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

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

$\displaystyle \text{Therefore: }\:R \,=\,\sqrt{97},\;\; a \,=\,\sin^{\text{-}1}\!\left(\tfrac{4}{\sqrt{97}}\right)$
• Aug 27th 2013, 04:20 PM
ZKG9744
Re: Trig problem
Thanks understanding is better less fuzzy now. Sqr(97/97) still confused where that came from sorry
• Aug 27th 2013, 05:18 PM
Shakarri
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 $\displaystyle \sqrt{4^2+9^2}=\sqrt{97}$
Then with the basic definition of sin: $\displaystyle \sin{a}=\frac{opposite}{hypotenuse}=\frac{4}{\sqrt {97}}$. Similar thing with cos.

You might find it easier to do the problem by first expanding $\displaystyle sin(\theta+a)$
• Aug 27th 2013, 05:37 PM
ZKG9744
Re: Trig problem
@ Shakarri
Thank you very very much