# Factor Formulae

• Jun 23rd 2011, 11:11 AM
Googl
Factor Formulae
Hi all,

I have an exam tomorrow but the questions which will come up I have never come across before. Some of the questions involve the Factor Formulae which I have never seen before so I was wondering whether you might give some examples and perhaps links to other pages where I can take a long revision.

1. Using the factor formulae evaluate the integrals of the type (sinx cos3x)dx

2. Use the factor formulae to prove trig identities.

I have a general understanding of Trigonometry and I have just completed my Calculus.

We've just been told today about this exam and I have no idea how the factor formulae looks like and how to use it.

Thanks.
• Jun 23rd 2011, 03:53 PM
TheCoffeeMachine
Re: Factor Formulae
These are the factor formulas:

\begin{aligned}& \cos \theta \cos \varphi = {\cos(\theta - \varphi) + \cos(\theta + \varphi) \over 2}\\& \sin \theta \sin \varphi = {\cos(\theta - \varphi) - \cos(\theta + \varphi) \over 2}\\&\sin \theta \cos \varphi = {\sin(\theta + \varphi) + \sin(\theta - \varphi) \over 2}\\& \cos \theta \sin \varphi = {\sin(\theta + \varphi) - \sin(\theta - \varphi) \over 2}.\\& \end{aligned}
• Jun 23rd 2011, 03:56 PM
emakarov
Re: Factor Formulae
I am not sure, but maybe they mean product-to-sum trigonometric identities.
• Jun 23rd 2011, 04:19 PM
Googl
Re: Factor Formulae
Quote:

Originally Posted by TheCoffeeMachine
These are the factor formulas:

\begin{aligned}& \cos \theta \cos \varphi = {\cos(\theta - \varphi) + \cos(\theta + \varphi) \over 2}\\& \sin \theta \sin \varphi = {\cos(\theta - \varphi) - \cos(\theta + \varphi) \over 2}\\&\sin \theta \cos \varphi = {\sin(\theta + \varphi) + \sin(\theta - \varphi) \over 2}\\& \cos \theta \sin \varphi = {\sin(\theta + \varphi) - \sin(\theta - \varphi) \over 2}.\\& \end{aligned}

Hi,

Could you show me some examples where you prove trig identities. I have been revising the factor formulae but they look very different to yours.
• Jun 23rd 2011, 04:27 PM
Googl
Re: Factor Formulae
Here is an example which I have been able to finish using the factor formulae that I found.

Prove that: 4(cosx + cos2x)sin3xsinx = cos4x - cos8x

My take:

=2(cosx + cos2x) 2sin3xsinx
Reverse using factor formulae
=2(cosx + cos2x)(cos4x - cos2x)

From here I know I am wrong because I have tried it all the way and the outcome is wrong.
• Jun 23rd 2011, 04:44 PM
emakarov
Re: Factor Formulae
Quote:

Prove that: 4(cosx + cos2x)sin3xsinx = cos4x - cos8x
This does not seem to be right: see the graph.
• Jun 24th 2011, 01:38 AM
Googl
Re: Factor Formulae
Hi,

So no one understands the factor formulae? If I could learn how to integrate this: (sinx cos3x)dx using the factor formulae that would be great, because I have done some research on proving trig using the factor formulae and revise enough.

Thanks.
• Jun 24th 2011, 02:14 AM
emakarov
Re: Factor Formulae
Have you tried rewriting sin(x)cos(3x) as a sum of two sines according to the formulas in post #2 and integrating them separately?
• Jun 24th 2011, 04:19 AM
Googl
Re: Factor Formulae
Quote:

Originally Posted by emakarov
Have you tried rewriting sin(x)cos(3x) as a sum of two sines according to the formulas in post #2 and integrating them separately?

Thanks for that. I can't believe I did not realise that. So is this correct?

We change the integral into sum of two sines as above using the factor formulas. Then we use the chain rule to integral each one separately...

The outcome will be:

3/2cos(x+3x) + -3/2cos(x-3x) +c

Is that right?
• Jun 24th 2011, 08:02 AM
emakarov
Re: Factor Formulae
Quote:

Originally Posted by Googl
The outcome will be:

3/2cos(x+3x) + -3/2cos(x-3x) +c

Is that right?

No, there is a mistake somewhere.

$\int \sin x\cos 3x\,dx={}$
$\frac{1}{2}\int (\sin(x+3x) + \sin(x-2x))\,dx={}$
$\frac{1}{2}\int (\sin 4x-\sin2x)\,dx={}$
$\frac{1}{2}(-\frac{1}{4}\cos4x+\frac{1}{2}\cos2x)+C={}$
$\frac{1}{4}\cos2x-\frac{1}{8}\cos4x+C$