# How do I verify these identities?

• Sep 11th 2008, 04:25 AM
azuresonata
How do I verify these identities?
I was doing an advance reading on trigonometry (identities, specifically) when I encountered these problems on verification:
$\displaystyle (tan^3 x + sin x sec x -sin x cos x)/(sec x - cos x) = tan x sec x + sin x$

$\displaystyle (1 + sin 2B + cos 2B)/(1 + sin 2B - cos 2B) = cot B$

$\displaystyle (sin x + sin 3x + sin 5x + sin 7x)/(cos x + cos 3x + cos 5x + cos 7x) = tan 4x$

I'll try to edit this when I come up with answers. Thanks for those who will help!
• Sep 11th 2008, 05:00 AM
mr fantastic
Quote:

Originally Posted by azuresonata
I was doing an advance reading on trigonometry (identities, specifically) when I encountered these problems on verification:
$\displaystyle (tan^3 x + sin x sec x -sin x cos x)/(sec x - cos x) = tan x sec x + sin x$

Mr F says: At a glance, I'd suggest making the sustitutions $\displaystyle {\color{red}\tan x = \frac{\sin x}{\cos x}}$ and $\displaystyle {\color{red}\sec x = \frac{1}{\cos x}}$ on each side and then simplifying the resulting expressions.

$\displaystyle (1 + sin 2B + cos 2B)/(1 + sin 2B - cos 2B) = cot B$

Mr F says: At a glance I'd suggest substituting from the double angle formulae in the left hand side and simplifying the resulting expression.

$\displaystyle (sin x + sin 3x + sin 5x + sin 7x)/(cos x + cos 3x + cos 5x + cos 7x) = tan 4x$

I'll try to edit this when I come up with answers. Thanks for those who will help!

For the last, you can start by using the following identities (derivable from the compound angle formulae):

1. sin(A + B) + sin(A - B) = 2 sin A cos B.

From which it follows that:

sin(x) + sin(3x) = 2 sin(2x) cos(x)

sin(5x) + sin(7x) = 2 sin(6x) cos(x)

2. cos(A + B) + cos(A - B) = 2 cos A cos B.

From which it follows that

cos(x) + cos(3x) = ........

cos(5x) + cos(7x) = ......
• Sep 11th 2008, 05:03 AM
thelostchild
ok for problems such like this knowing the standard identities is VERY useful, a list of them can be found here
Table of Trigonometric Identities

for example lets look at the first of your problems

looking at the denominator we can write it as
$\displaystyle \sec x - \cos x=\frac{1}{\cos x} - \cos x = \frac{1-cos^2 x}{\cos x} = \frac{sin^2x}{\cos x}=\tan x \sin x$

so the LHS becomes

$\displaystyle \frac{\tan^2x}{\sin x}+\frac{\sec x}{\tan x}-\frac{cos x}{\tan x}$

now writing tan x in terms of sine and cosine we have

$\displaystyle \frac{1}{\sin x}(\tan^2 x +1 -\cos^2 x)$

then using $\displaystyle \cos^2 x +\sin^2 x=1$

$\displaystyle \frac{1}{\sin x}(tan^2 x + sin^2 x)$

once again writing tan in terms of sine and cosine gives us the RHS

$\displaystyle \tan x \sec x + \sin x$

so all this topic is really is learning identities and practice (hence why this is deathly boring!)