Thread: How do I verify these identities?

1. How do I verify these identities?

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

$

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

$

$

(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!

2. Originally Posted by azuresonata
I was doing an advance reading on trigonometry (identities, specifically) when I encountered these problems on verification:
$
(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 ${\color{red}\tan x = \frac{\sin x}{\cos x}}$ and ${\color{red}\sec x = \frac{1}{\cos x}}$ on each side and then simplifying the resulting expressions.

$

(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.

$

(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) = ......

3. 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
$\sec x - \cos x=\frac{1}{\cos x} - \cos x = \frac{1-cos^2 x}{\cos x} = \frac{sin^2x}{\cos x}=\tan x \sin x$

so the LHS becomes

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

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

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

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

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

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

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

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