Can You Solve This Compound Angle Formula Question?

[Firepanda]

It's a previous exam question using compound angle forumlas. I've tried everything to try and get the answer involving rearrangments!

I know the forumlas i just can't get them to work to show this.

I assume its something to do with adding or subtracting the forumla away from each other, but I've tried it. The furthest i got was this:

cos (alpha) + (root 3)sin (alpha)

but i can't get it into the format of what they want

Here's the question :)

http://img251.imageshack.us/my.php?image=mathelpsk7.jpg

Thanks!

 

[VietDao29]

...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? :)

 

[Firepanda]

yes! thankyou :D