1. There's a task in my math book that says:
Write the expression

\/2*(cos(3x+ π/4))

using cos3x and sin3x. (\/2 means the square root of 2, I couldn't locate the square root sign on my keyboard).

I ended up with \/2*(cos3x-sin3x), but the answer is supposed to be cos3x-sin3x. What did I do wrong?

Oh, and, sorry if my translation makes no sense. I'm Norwegian and not very used to talk about mathematics in any other language than Norwegian   2. ### Related Discussions:

3. Originally Posted by daniel
There's a task in my math book that says:
Write the expression

\/2*(cos(3x+ ?/4))

using cos3x and sin3x. (\/2 means the square root of 2, I couldn't locate the suare root sign on my keyboard).

I ended up with \/2*(cos3x-sin3x), but the answer is supposed to be cos3x-sin3x. What did I do wrong?

Oh, and, sorry if my translation makes no sense. I'm Norwegian and not very used to talk about mathematics in any other language than Norwegian Am I to assume your "n" is supposed to be pi?

If so, then that's the problem.

cheers
william  4. Yeah, I just copied the pi sign from a website and it showed up mysteriously similar to an "n" here. How is that the problem?  5. Okay, not really the problem so much as the "solution".

cos(pi/4) = sqrt(2)/2 = sin(pi/4)

sqrt(2)*sqrt(2)/2 = 1

Hope that helps,
william  6. I assume you're using
cos(a+b) = cos(a)cos(b) - sin(a)sin(b)
by the way....

cheers  7. Well, it says cos(3x+pi/4), not cos(pi/4)
Here's what I did to reach the answer I got:
sqrt2*(cos(3x+ pi/4))
=
sqrt2*(cos3x*sin(pi/4) - cos(pi/4)*sin3x)
=
sqrt2*(cos3x - sin3x)  8. Oh, snap, I just got understood why the sqrt2 goes away. Thanks for your excellent assistance to my enlightenment, dear William.  9. Originally Posted by daniel
Well, it says cos(3x+pi/4), not cos(pi/4)
Here's what I did to reach the answer I got:
sqrt2*(cos(3x+ pi/4))
=
sqrt2*(cos3x*sin(pi/4) - cos(pi/4)*sin3x)
=
sqrt2*(cos3x - sin3x)
You're confusing cos(a+b) with sin(a+b)...
and what you wrote indeed says "cos(pi/4)" (and sin(pi/4)) no?
(But you still need to use the correct trig identity....)

cheers  10. Originally Posted by daniel
Oh, snap, I just got understood why the sqrt2 goes away. Thanks for your excellent assistance to my enlightenment, dear William.
No problem.
But make sure you use the correct trig identity. It looks like you are using the one for sin(a-b)....

cheers  11. Originally Posted by william
You're confusing cos(a+b) with sin(a+b)...
Ah, yes, I didn't see that until now ^_^  Bookmarks
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