# Math Help - Precalculus Trig Problem

1. ## Precalculus Trig Problem

Hello,

Here is a pre calculus problem I am having trouble with. The problem is from a review sheet for pre calculus, but I do not recall doing anything like this in my pre calculus class.

(a) Express the function $f(x) = - \sin 2x + \sqrt{3} \cos 2x$ in terms of sine only.

(b) Find the amplitude A, period P, and phase shift B of the function, and graph one complete period.

Any help is greatly appreciated.

Thanks

2. for part (a) use the identity $\cos 2x = 1 - 2 \sin^2 x$
and then you'll have it in terms of sin

3. Originally Posted by khodaja
Hello,

Here is a pre calculus problem I am having trouble with. The problem is from a review sheet for pre calculus, but I do not recall doing anything like this in my pre calculus class.

(a) Express the function $f(x) = - \sin 2x + \sqrt{3} \cos 2x$ in terms of sine only.

(b) Find the amplitude A, period P, and phase shift B of the function, and graph one complete period.

Any help is greatly appreciated.

Thanks
What razorfever said for part (a) is fine, and you will get an answer in terms of sine that is correct, however, the form that you will get likely will not be one that helps you in part (b) a lot.

alternatively, note that there is a formula that goes like this (i do not know what it's called, or if it even has a name), but it is based on the addition formula for sine.

$\boxed{A \sin x + B \cos x = C \sin (x + \theta)}$

here, $A, ~B,~C, \text{ and } \theta$ are constants.

basically, here (if we write 2 instead of x), $A = -1, \text{ and } B = \sqrt{3}$ and we want to find $C$ and $\theta$ to have a nice form in sine that we can graph easily and find all the things we are required to find. first, we will modify the formula, to write 2x instead of just x.

so here goes:

$C \sin (2x + \theta ) = C(\sin 2x \cos \theta + \sin \theta \cos 2x)$ .............by the addition formula for sine

$= \underbrace{C \cos \theta}_{\text{this is our }A} \sin 2x + \underbrace{C \sin \theta}_{\text{this is our }B} \cos 2x$

note that here we have

$C \cos \theta = -1$ ...........(1)
$C \sin \theta = \sqrt{3}$ .........(2)

thus we have two simultaneous equations, with two unknowns, we can solve for our unknowns. (try doing that yourself, it can come in handy to recall $\sin^2 \theta + \cos^2 \theta = 1$)

anyway, there are infinitely many solutions here, any one will suffice. the nicest one i think is: $C = 2$, $\theta = \frac {2 \pi}3$

thus, we have:

$f(x) = - \sin 2x + \sqrt{3} \cos 2x = 2 \sin \left( 2x + \frac {2 \pi}3\right)$

or in other words,

$f(x) = 2 \sin 2 \left( x + \frac {\pi}3 \right)$

can you take it from here?

EDIT: in retrospect, it would perhaps have been easier to note the following:

$f(x) = - \sin 2x + \sqrt{3} \cos 2x$

$= 2 \left( \underbrace{- \frac 12}_{\cos \frac {2 \pi}3} \sin 2x + \underbrace{\frac {\sqrt{3}}2}_{\sin \frac {2 \pi}3} \cos 2x \right)$

$= 2 \left( \sin 2x \cos \frac {2 \pi}3 + \sin \frac {2 \pi}3 \cos 2x \right)$

$= 2 \sin \left( 2x + \frac {2 \pi }3\right)$

$= 2 \sin 2 \left( x + \frac {\pi}3 \right)$

4. so, according to this solution, the amplitude is 2. right?

5. Originally Posted by khodaja
so, according to this solution, the amplitude is 2. right?
yes

what about the period and phase shift?

6. phase shift is : pi/3

period is : pi

7. Originally Posted by khodaja
phase shift is : pi/3
not exactly.

look back at the general form to see what the phase shift is

period is : pi
correct