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.

Printable View

- Aug 27th 2013, 02:57 PMZKG9744Trig 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 PMSorobanRe: 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 PMZKG9744Re: Trig problem
Thanks understanding is better less fuzzy now. Sqr(97/97) still confused where that came from sorry

- Aug 27th 2013, 05:18 PMShakarriRe: 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 PMZKG9744Re: Trig problem
@ Shakarri

Thank you very very much