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.
Hello, ZKG9744!
If you "just started trig", this is a very fancy problem.
$\displaystyle \text{Find }R\text{ and }a\text{ such that: }\:9\sin\theta + 4\cos\theta \:=\: R\sin(\theta + a)$
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: .$\displaystyle \sqrt{97}(\cos a\sin\theta + \sin a\cos\theta) \;=\;\sqrt{97}\sin(\theta + a)$
$\displaystyle \text{Therefore: }\:R \,=\,\sqrt{97},\;\; a \,=\,\sin^{\text{-}1}\!\left(\tfrac{4}{\sqrt{97}}\right)$
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 $\displaystyle sin(\theta+a)$