Trignometric identities - Express as a single sine or cosine function

• Sep 9th 2009, 02:59 PM
Kakariki
Trignometric identities - Express as a single sine or cosine function
Okay, I am having trouble understanding how to use trigonometric equations. The question states:
Express as a single sine or cosine function.

i) cos 2a cos a - sin 2a sin a

ii) sin 2m cos m + cos 2m sin m

My attempt:

i) cos 2a cos a - sin 2a cos a
= cos^2 a - sin^2 a cos a - 2sin a cos a sin a
= cos^2 a - 1 - cos^2 a cos a - 2 sin a cos a sin a
I do not know what to do from here. The answer in the book is cos 3a

ii) sin 2m cos m + cos 2m sin m
I do not know where to start with this question.

I was hoping someone could show, as an example, how to answer the first question, and maybe give a few hints as to how to go about answering the second one.

Thank you.
• Sep 9th 2009, 03:59 PM
Jameson
Quote:

Originally Posted by Kakariki
My attempt:

i) cos 2a cos a - sin 2a cos a
= cos^2 a - sin^2 a cos a - 2sin a cos a sin a
= cos^2 a - 1 - cos^2 a cos a - 2 sin a cos a sin a
I do not know what to do from here. The answer in the book is cos 3a

The line in red I have a problem with. When you change $\displaystyle \cos(2a)$ into $\displaystyle \cos^2(a)-\sin^2(a)$, you need to keep this second expression in parentheses because the $\displaystyle \cos(a)$ should distribute to both of them. I think you just multiplied the cosine to the cosine-squared.
• Sep 9th 2009, 07:55 PM
Kakariki
Quote:

Originally Posted by Jameson
The line in red I have a problem with. When you change $\displaystyle \cos(2a)$ into $\displaystyle \cos^2(a)-\sin^2(a)$, you need to keep this second expression in parentheses because the $\displaystyle \cos(a)$ should distribute to both of them. I think you just multiplied the cosine to the cosine-squared.

Okay, then I have no idea how to go about doing the problem. Do you have any examples or can you show me how to do it? I am really stuck right now.
• Sep 10th 2009, 12:46 AM
walleye
these are both addition of angle formulae
there 4 of these that relate to sin and cos

sin(x+y) = sin x cos y + cos x sin y
( i wont write them all)

but look them up in your book or wherever and go to town on it

the questions are obviously written to trick you into thinking you need the doubleangle formulae.
• Sep 10th 2009, 10:53 AM
Kakariki
Quote:

Originally Posted by walleye
these are both addition of angle formulae
there 4 of these that relate to sin and cos

sin(x+y) = sin x cos y + cos x sin y
( i wont write them all)

but look them up in your book or wherever and go to town on it

the questions are obviously written to trick you into thinking you need the doubleangle formulae.

Okay, I think I understand. As both of the variables were "a" I thought that that would not work, but since one is 2a and the other is a, I am to treat them as different variables?

So I would use the formula: Cos (x + y) = cos x cos y - sin x sin y

Which means: Cos 2a cos a - sin 2a sin a
= cos (2a + a)
= cos (3)

Can someone explain why a and 2a are treated as "apples and oranges"? Because, assuming I am understanding correctly, 2a + a = 3a, so why are they being considered as x and y?

Thanks.
• Sep 10th 2009, 11:01 AM
Jameson
Quote:

Originally Posted by Kakariki
Okay, I think I understand. As both of the variables were "a" I thought that that would not work, but since one is 2a and the other is a, I am to treat them as different variables?

So I would use the formula: Cos (x + y) = cos x cos y - sin x sin y

Which means: Cos 2a cos a - sin 2a sin a
= cos (2a + a)
= cos (3)

Can someone explain why a and 2a are treated as "apples and oranges"? Because, assuming I am understanding correctly, 2a + a = 3a, so why are they being considered as x and y?

Thanks.

No, no no... get that idea out of your mind so you don't make this mistake again!

I was trying to show you that $\displaystyle \cos(2a)\cos(a)=\left( \cos^2(a)-\sin^2(a) \right) *\cos(a)$

In your original post, you only multiplied the cos(a) to one of the terms. See what I mean?
• Sep 10th 2009, 12:49 PM
Kakariki
Quote:

Originally Posted by Jameson
No, no no... get that idea out of your mind so you don't make this mistake again!

I was trying to show you that $\displaystyle \cos(2a)\cos(a)=\left( \cos^2(a)-\sin^2(a) \right) *\cos(a)$

In your original post, you only multiplied the cos(a) to one of the terms. See what I mean?

So my answer is incorrect? I think I was going about it the long way in my original post.
• Sep 11th 2009, 02:52 AM
walleye
Quote:

Originally Posted by Kakariki

Can someone explain why a and 2a are treated as "apples and oranges"? Because, assuming I am understanding correctly, 2a + a = 3a, so why are they being considered as x and y?

Thanks.

um
cos(2a) and cos(a)
if you draw the graphs for each of these on the same set of axes (over the top of one another), youll see that they barely relate to one another

for example if you graph
2cos(a) and cos(a) on the same set of axes, youll see that the highpoints and the low points are in line with each other, so then you can easily add them together

but since cos(2a) and cos(a) have completely different highpoints and lowpoints from each other, you cannot add them together.

do you understand this?

another way of thinking about it:

cos(a) + cos(a) = 2cos(a) and definitely not cos(2a)

does that one make sense to you?
• Sep 11th 2009, 02:55 AM
walleye
as for this bit that you wrote

i) cos 2a cos a - sin 2a cos a
= cos^2 a - sin^2 a cos a - 2sin a cos a sin a

the other guy is right, you made a mistake here

cos(2a)cos(a) = cos^2 a cos(a) - sin^2 cos(a)
= (cos^2 a - sin^2)cos(a)

this is similar to saying a x (a + b) = aa + ab