How to Prove Complex Trigonometric Identity?

[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.

 

[assyrian_77]

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

 

[clook]

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

no..

hmm let me write it out another way

how can i prove

tanX
______
sinx+cosx

=

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

 

[CarlB]

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

Carl

 

[assyrian_77]

In other words, there were parenthesis missing...

 

[VietDao29]

no..

hmm let me write it out another way

how can i prove

tanX
______
sinx+cosx

=

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

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

 

[clook]

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

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

 

[arildno]

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.

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.

 

[vaishakh]

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

 

[clook]

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.

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.

 

[Hootenanny]

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

Just rewritting what clook posted in a more organised manor:

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

 

[topsquark]

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

Just rewritting what clook posted in a more organised manor:

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

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

-Dan

 

[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.

 

[VietDao29]

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.

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

 

[Hootenanny]

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

Yes solve would make a lot more sense (and would be very easy as well)