# Proving Trig Identities

1. Jan 17, 2008

### krogsty

[SOLVED] Proving Trig Identities

1. The problem statement, all variables and given/known data

Prove that

tanx - sinx---=---- tanxsinx
________------- _________
tanxsinx -------- tanx + sinx

2. Relevant equations

Is this impossible, so far it has been for me, what about you?

3. The attempt at a solution

I litterlly have 4 pages of written of attempts, I just can't figure it out

2. Jan 17, 2008

### rock.freak667

Actually it isn't that hard.

I used the LHS and just simply converted all the tanx to sinx/cosx and simplified it. Did you do that?

3. Jan 17, 2008

### krogsty

What does LHS stand for?, and I have converted the tanx's to sinx/cosx, but there are so many ways to go from there, I just can't seem to get one, Can you possibly tell me the final simplified version of either side that you got?

4. Jan 17, 2008

### rock.freak667

LHS is Left Hand Side.

$$\frac{tanx-sinx}{tanxsinx}= \frac{\frac{sinx}{cosx}-sinx}{\frac{sin^2x}{cosx}}$$

then multiply the numerator and denominator by cosx then what do you get?

And always remember that $sin^2 x+cos^2 x=1$

5. Jan 17, 2008

### krogsty

Am I aloud to multiply the sinx/cosx before finding common denominators with -sinx?

$$\frac{\{sinx-sinxcosx}{cosx}$$

Is this what you mean?

Last edited: Jan 17, 2008
6. Jan 17, 2008

### rock.freak667

Well if you multiply the Numerator and denominator by cos x then it is the same as multiplying by 1. So that it remains the same.

But you were supposed to get

$$\frac{\{sinx-sinxcosx}{sin^2 x}$$

then recall that $sin^2 x=1-cos^2 x$

7. Jan 17, 2008

### krogsty

So I'm suppose to reduce the left side to

$$\frac{\{sinx-sinxcosx}{sin^2 x}$$

and then do that to the right?

8. Jan 17, 2008

### rock.freak667

No. Work with one side only. After you reduce it to what is above. Just substitute sin^2x and you will get the answer in another 4 lines

9. Jan 17, 2008

### krogsty

I'm sorry, I don't follow any of this, Normally I understand this stuff, but this "puzzle" is just confusing me

$$\frac{\frac{sinx}{cosx}-\frac{sinxcosx}{cosx}}{\frac{sinx^{2}}{cosx}}$$

Last edited: Jan 17, 2008
10. Jan 17, 2008

### rock.freak667

Using the LHS
$$\frac{tanx-sinx}{tanxsinx}$$

sub $tanx=\frac{sinx}{cosx}$

$$\frac{\frac{sinx}{cosx}-sinx}{\frac{sin^2x}{cosx}}$$

multiply both the numerator and denominator by cosx

$$\frac{cosx (\frac{sinx}{cosx}-sinx)}{(\frac{sin^2x}{cosx})cosx}$$

$$\frac{sinx-sinxcosx}{sin^2x}}$$

sub $sin^2x=1-cos^2x$

Following better now?

11. Jan 17, 2008

### krogsty

Then I want to get

$$\frac{sinx(1-cosx)}{(1+cosx)(1-cosx)}$$

correct?

12. Jan 17, 2008

### rock.freak667

yes, what do you have left now when the 1-cosx cancels out?

13. Jan 17, 2008

### krogsty

you get

$$\frac{sinx}{1+cosx}$$ Then I reduce the right side to the same thing, right?

14. Jan 17, 2008

### rock.freak667

Nope. you need to somehow make the Left side the same as the right side. Which is what you are doing

What you want is

$$\frac{sinx}{1+cosx}$$

to somehow become

$$\frac{tanxsinx}{tanx+sinx}$$

Now if you take what you have and multiply the numerator and denominator by tanx

the numerator will become tanxsinx right? Isn't that the numerator that you want?
Deal with the denominator now.

15. Jan 17, 2008

### krogsty

(1+cosx)(tanx)---------------tanx + cosxtanx---------------tanx + cosxsinx/cosx
cancel out the cosx and you get tanx + sinx

16. Jan 17, 2008

### rock.freak667

Yes, the denominator is now tanx+sinx and the numerator is tanxsinx

so you now have what they wanted...

17. Jan 17, 2008

### krogsty

Thank you so much for your help, You were the 3rd person I asked, and the only one who actually knew how to solve this, you are the man, thanks again.