Firepanda said:
...cos (alpha) + (root 3)sin (alpha)
but i can't get it into the format of what they want
Yup, so far so good. :)
It goes like this. If you want to combine the expression:
a \sin \alpha + b \cos \alpha to get some expression with only one sine, or one cos function, you should pull out the factor: \sqrt{a ^ 2 + b ^ 2}
a \sin \alpha + b \cos \alpha = \sqrt{a ^ 2 + b ^ 2} \left( \frac{a}{\sqrt{a ^ 2 + b ^ 2}} \sin \alpha + \frac{b}{\sqrt{a ^ 2 + b ^ 2}} \sin \alpha \right)
Now, let \beta be some angle such that:
\left\{ \begin{array}{l} \sin \beta = \frac{a}{\sqrt{a ^ 2 + b ^ 2}} \\ \cos \beta = \frac{b}{\sqrt{a ^ 2 + b ^ 2}} \end{array} \right. \quad \mbox{or} \quad \left\{ \begin{array}{l} \sin \beta = \frac{b}{\sqrt{a ^ 2 + b ^ 2}} \\ \cos \beta = \frac{a}{\sqrt{a ^ 2 + b ^ 2}} \end{array} \right.
There will definitely be an angle \beta like that, since, we have:
\left| \sin \beta \right| = \left| \frac{a}{\sqrt{a ^ 2 + b ^ 2}} \right| \leq 1
\left| \cos \beta \right| = \left| \frac{b}{\sqrt{a ^ 2 + b ^ 2}} \right| \leq 1
and
\sin ^ 2 \beta + \cos ^ 2 \beta = \frac{a ^ 2}{a ^ 2 + b ^ 2} + \frac{b ^ 2}{a ^ 2 + b ^ 2} = \frac{a ^ 2 + b ^ 2}{a ^ 2 + b ^ 2} = 1
So, we have:
a \sin \alpha + b \cos \alpha = \sqrt{a ^ 2 + b ^ 2} \left( \frac{a}{\sqrt{a ^ 2 + b ^ 2}} \sin \alpha + \frac{b}{\sqrt{a ^ 2 + b ^ 2}} \cos \alpha \right) = \left[ \begin{array}{l} \sqrt{a ^ 2 + b ^ 2} (\sin \beta \sin \alpha + \cos \beta \sin \alpha) \\ \sqrt{a ^ 2 + b ^ 2} (\cos \beta \sin \alpha + \sin \beta \cos \alpha) \end{array} \right.
= \left[ \begin{array}{l} \sqrt{a ^ 2 + b ^ 2} \cos (\alpha - \beta) \\ \sqrt{a ^ 2 + b ^ 2} \sin (\alpha + \beta) \end{array} \right.-------------------------
Applying this to your problem, we have:
\cos \alpha + \sqrt{3} \sin \alpha
Pull out \sqrt{1 ^ 2 + (\sqrt{3}) ^ 2} = \sqrt{4} = 2, we have:
... = 2 \left( \frac{1}{2} \cos \alpha + \frac{\sqrt{3}}{2} \sin \alpha \right)
Now, we will try to find such angle \beta, we have:
\sin \left( \frac{\pi}{6} \right) = \frac{1}{2}, and
\cos \left( \frac{\pi}{6} \right) = \frac{\sqrt{3}}{2}, so: \beta = \frac{\pi}{6}. Substitute \beta into the expression, yielding:
...= 2 \left( \sin \left( \frac{\pi}{6} \right) \cos \alpha + \cos \left( \frac{\pi}{6} \right) \sin \alpha \right) = 2 \sin \left( \alpha + \frac{\pi}{6} \right) (Q.E.D)
Yay, it's done.
Is it clear?
Can you get it? :)
