Trig identity

1. Mar 6, 2006

clook

how can i prove

tan X/sinx+cosx=sin^2 X + sinXcosX/cos X - 2cos^3X

so far i've tried using basic identities to figure it out, but i just end up getting confused.

Last edited: Mar 6, 2006
2. Mar 6, 2006

assyrian_77

The way you have written it is confusing. Are there any paranthesis missing?

3. Mar 6, 2006

clook

no..

hmm lemme write it out another way

how can i prove

tanX
______
sinx+cosx

=

sin^2 X + sinXcosX
________________
cos X - 2cos^3X

4. Mar 6, 2006

CarlB

Try putting x = pi/4 and see what you get.

Carl

5. Mar 6, 2006

assyrian_77

In other words, there were parenthesis missing...

6. Mar 7, 2006

VietDao29

The problem is wrong. Are you sure you copied this correctly?
I think it should be:
$$\frac{\tan x}{\sin x + \cos x} = \frac{\sin ^ 2 x + \sin x \cos x}{\cos x + 2 \sin x \cos ^ 2 x}$$

7. Mar 7, 2006

clook

i copied it as it was exactly written in the book..

8. Mar 7, 2006

arildno

No, you haven't copied this correctly: Put X=0
Left hand side yields 2, whereas right-hand side yields -2.

Read your book again; this time with your eyes open.

9. Mar 7, 2006

vaishakh

Try cancelling the common factors and try compreesing the question to its simplest form. You would then probably then get a breakthrough

10. Mar 7, 2006

clook

i am positive i copied it correctly. i would take a picture from the book, but i don't have a camera or scanner right now.

11. Mar 7, 2006

Hootenanny

Staff Emeritus
When you put x = 0 in both sides are equal to zero.

Just rewritting what clook posted in a more organised manor:

$$\frac{\tan x}{\sin x + \cos x} = \frac{\sin^2 x + \sin x \cos x}{\cos x - 2\cos^3 x}$$

12. Mar 7, 2006

topsquark

This statement is equivalent to saying $$sinx=-cosx$$, so something is VERY fishy here.

-Dan

13. Mar 7, 2006

assyrian_77

Ok clook, assuming that

$$\frac{\tan x}{\sin x + \cos x} = \frac{\sin^2 x + \sin x \cos x}{\cos x - 2\cos^3 x}$$

is what you have in your book. I worked with the RHS and got this:

$$\frac{\tan x}{\sin x-\cos x}$$

As you can see, this is different from the LHS in your equation. Something is indeed fishy here.

14. Mar 8, 2006

VietDao29

Okay, look at your problem again, does it say something like:
Prove this identity, or does it tell you to solve the equation?

15. Mar 8, 2006

Hootenanny

Staff Emeritus
Yes solve would make alot more sense (and would be very easy as well)