Simplifying a trig identity

Printable View

November 28th 2012, 05:20 PM
Lethargic

Simplifying a trig identity

Simplify the expression:

cos(θ + π/6) - sin (θ + π/6)

Write your answer as an exact value.

I'm taking grade 12 math through an independent learning course and I'm really stuck on this question, I have no teacher to ask for help, could someone give me some pointers? I tried expanding the brackets but could only simplify to:

√3/2 cos θ - 1/2 sin θ

Which doesn't seem right to me...
November 28th 2012, 05:53 PM
zhandele

Re: Simplifying a trig identity

When you expanded the brackets, what did you come up with?

I you must've misinterpreted the meaning of the brackets. For instance, cos(theta + pi/6) means the cosine of the quantity "theta + pi/6', not "cosine theta plus cosine pi/6."

If you need a table of trig identities, try

Table of Trigonometric Identities

I did come up with a different answer than you did.

I just checked out my result on a CAS, and I'm pretty sure I have the right formulation. Now, I'm assuming that by simplifying you mean a form where you have terms containing only sin and cos theta with coefficients.

If you have a calculator or CAS, one way to check your result is to substitute a value for theta, in both the original and simplified form, and see if they're equal.
November 28th 2012, 06:09 PM
MarkFL

Another approach would be to essentially use a linear combination identity as follows:

$\sqrt{2}\left(\sin\left(\frac{\pi}{4} \right)\cos\left(\theta+\frac{\pi}{6} \right)-\cos\left(\frac{\pi}{4} \right)\sin\left(\theta+\frac{\pi}{6} \right) \right)$

Now, within the brackets, use the angle-difference identity for sine to write:

$\sqrt{2}\sin\left(\frac{\pi}{4}-\left(\theta+\frac{\pi}{6} \right) \right)=\sqrt{2}\sin\left(\frac{\pi}{12}-\theta \right)$

Now, if we prefer, we may use the identity $\sin(\pi-x)=\sin(x)$ to write this as:

$\sqrt{2}\sin\left(\pi-\left(\frac{\pi}{12}-\theta \right \right)=\sqrt{2}\sin\left(\theta+\frac{11\pi}{12} \right)$
November 28th 2012, 06:10 PM
Lethargic

Re: Simplifying a trig identity

I expanded and got:

cos θ cos π/6 - sin θ sin π/6

I know from special triangles that cos π/6 = √3/2 and sin π/6 = 1/2, which is where I got the solution that I posted at first
November 28th 2012, 06:13 PM
Lethargic

Re: Simplifying a trig identity

Quote:

Originally Posted by MarkFL2

Another approach would be to essentially use a linear combination identity as follows:

$\sqrt{2}\left(\sin\left(\frac{\pi}{4} \right)\cos\left(\theta+\frac{\pi}{6} \right)-\cos\left(\frac{\pi}{4} \right)\sin\left(\theta+\frac{\pi}{6} \right) \right)$

Now, within the brackets, use the angle-difference identity for sine to write:

$\sqrt{2}\sin\left(\frac{\pi}{4}-\left(\theta+\frac{\pi}{6} \right) \right)=\sqrt{2}\sin\left(\frac{\pi}{12}-\theta \right)$

Now, if we prefer, we may use the identity $\sin(\pi-x)=\sin(x)$ to write this as:

$\sqrt{2}\sin\left(\pi-\left(\frac{\pi}{12}-\theta \right \right)=\sqrt{2}\sin\left(\theta+\frac{11\pi}{12} \right)$

I believe this is at a level beyond the lesson that I'm on, I have yet to hear the term "linear combination identity"
November 28th 2012, 06:15 PM
MarkFL

Re: Simplifying a trig identity

Quote:

Originally Posted by Lethargic

I expanded and got:

cos θ cos π/6 - sin θ sin π/6

I know from special triangles that cos π/6 = √3/2 and sin π/6 = 1/2, which is where I got the solution that I posted at first

That is the expansion for the first term only. You need to also apply the angle-sum identity for sine on the second term.
November 28th 2012, 06:16 PM
zhandele

Re: Simplifying a trig identity

OK, it looks like you applied the sum of cosines identity. After that, you still have to subtract sin(th + pi/6). That requires using the other identity, for the sine of a sum.
November 28th 2012, 06:16 PM
topsquark

Re: Simplifying a trig identity

Not needed!

-Dan
November 28th 2012, 06:19 PM
Lethargic

Re: Simplifying a trig identity

Oh wow that was such a simple mistake, thank you so much everyone, I've been tearing my hair out.
November 28th 2012, 06:23 PM
MarkFL

Re: Simplifying a trig identity

Quote:

Originally Posted by Lethargic

I believe this is at a level beyond the lesson that I'm on, I have yet to hear the term "linear combination identity"

You may not have heard the term yet, but you can see that I multiplied the first by $1=\sqrt{2}\sin\left(\frac{\pi}{4} \right)$ and the second term by $1=\sqrt{2}\cos\left(\frac{\pi}{4} \right)$ and then the rest should make sense if you know standard the identities used.