Question

Lets say for instance your verifying identities and you have sin^3(x)+cos^3(x)/1-2cos^2(x)=sec(x)-sin(x)/tan(x)-1

Answer 1

\[\frac{\sin^3x+\cos^3x}{1-2\cos^2x}=\frac{\sec x-\sin x}{\tan x-1}?\]

Answer 2

sorry um which side would you start to work on first.

Answer 3

I'm asking if that's the right identity you have to show.

Answer 4

yup

Answer 5

My first instinct is to start with the left side. The sum of cubes in the numerator can be factored, so you have
\[\frac{(\sin x+\cos x) (\sin^2x-\sin x\cos x + \cos^2x)}{1-2\cos^2x}=\cdots\]

Answer 6

would I simplify from that point because i already did that or would I began to work on the right.

Answer 7

There's nothing you can simplify here. When you're proving identities, you're supposed to work with only one side and derive the other side.

Then, rewrite the denominator, \[\frac{(\sin x+\cos x)(\sin^2x-\sin x\cos x+\cos^2x)}{1-\cos^2x-\cos^2x}=\cdots\]
From the Pythagorean identity, you have\[\frac{(\sin x+\cos x)(\sin^2x-\sin x\cos x+\cos^2x)}{\sin^2x-\cos^2x}=\cdots\]
You can factor the denominator due to the difference of squares:\[\frac{(\sin x+\cos x)(\sin^2x-\sin x\cos x+\cos^2x)}{(\sin x+\cos x)(\sin x-\cos x)}=\cdots\]
Then you can cancel out the common (sinx + cosx) term on top and bottom, leaving you with \[\frac{\sin^2x-\sin x\cos x+\cos^2x}{\sin x-\cos x}=\cdots\]

Answer 8

From the Pythagorean identity, you then have\[\frac{1-\sin x\cos x}{\sin x-\cos x}=\cdots\]
Multiply both numerator and denominator by secx:
\[\frac{1-\sin x\cos x}{\sin x-\cos x}\cdot\frac{\sec x}{\sec x}=\cdots\\
\frac{\sec x-\sin x}{\tan x-1}=\cdots\]

Answer 9

That last step is due to
\[\sec x=\frac{1}{\cos x}\\
\tan x = \frac{\sin x}{\cos x}\]

Answer 10

thank you so much you didn't have to type all of that but I'm grateful nonetheless.