Math Help - Trig Review

1. Trig Review

we got a bunch of review problems for the test thats coming up in a few days, I got most of them figured out but am stuck on these 3 if anyone could give me a few pointers of where to head

1. Solved
Code:

TANx + TAN^3x + SINx
---------------------- =   COSx + SEC^2x
TANx

TANx(1 + TAN^2x) + SINx
-------------------------- =
TANx

TANx(SEC^2x) + SINx
--------------------- =
TANx

(SEC^2x) + SINx =
2. Solved
Code:

COS(x+y)
--------- = 1 - TANxTANy
COSxCOSy

COSxCOSy - SINxSINy
--------------------- =
COSxCOSy

COSxCOSy - TANxTANy =

not sure how to turn COSxCOSy into 1
3.
Code:
SIN3x    COS3x
------ - ------ = 2
SINx      COSx

SIN(x+2x)   COS(x+2x)
--------- - ---------- =
SINx           COSx

SINxCOS2x + COSxSIN2x      COSxCOS2x - SINxSIN2x
-----------------------  -  ----------------------- =
SINx                               COSx

COSxSINxCOS2x + COS^2xSIN^2x - SINxCOSxCOS2x - SIN^2xSIN^2x
--------------------------------------------------------------- =
SINxCOSx

COS^2xSIN2x -SIN^2xSIN2x
---------------------------- =
SINxCOSx
COS^2x(2SINxCOSx) - SIN^2x(2SINxCOSx)
------------------------------------ =
SINxCOSx

2SINxCOSx(COS^2x - SIN^2x)
----------------------------- =
SINxCOSx

2(COS^2x - SIN^2x) =
if anyone could give me suggestions of where to go for the next steps in these, that would help me out alot, I'll post back if I figure anything out in the mean time

2. In
Code:
=C++

Tanx(Sec^2 x)+Sin x
-----------------------
Tan x
When you divide by tangent you get,
$
\sec^2x+\frac{\sin x}{\tan x}$

But,
$\frac{\sin x}{\tan x}=\cos x$
Because, $\tan x=\frac{\sin x}{\cos x}$
Thus, you are left with,
$\sec^2x+\cos x$ and the proof is complete.

3. ahh, thanks for the help, didnt notice I had to divide by both sides there and that I could use COS = SIN/TAN

for 2. , I figured out what I was doing wrong, just gotta finish the 3rd one now

Code:
COSxCOSy - SINxSINy
--------------------- =
COSxCOSy

COSxCOSy - TANxTANy =

should of been

COSxCOSy     SINxSINy
---------- - -------- =
COSxCOSy      COSxCOSy

1 - TANxTANy = 1 - TANxTANy

4. well I got the 3rd one down to
2(COS^2x - SIN^2x) =
but not sure how to make (COS^2x - SIN^2x) into 1, I know that COS^2x + SIN^2x = 1 though

edit: if I did
Code:
2(1-SIN^2x - 1-COS^2x)
2(-SIN^2x - COS^2x)
does -SIN^2x - COS^2x = 1? or could I just multiply the part inside parenthesis by -1 to become (SIN^2x + COS^2x) so it = 1

5. Originally Posted by drewms64
well I got the 3rd one down to
2(COS^2x - SIN^2x) =
but not sure how to make (COS^2x - SIN^2x) into 1, I know that COS^2x + SIN^2x = 1 though
For #3
Code:
=C++

SIN3x    COS3x
------ - ------ = 2
SINx      COSx
You are doing this the primitive way. Rather add the fractions immediately to get,
$\frac{\sin 3x\cos x-\cos 3x\sin x}{\sin x\cos x}$
Recognizing the sine for angle difference we have,
$\frac{\sin 2x}{\sin x\cos x}$
Using the double angle formula we get,
$\frac{2\sin x\cos x}{\sin x\cos x}$
Thus, $=2$

$\mathbb{Q}.\mathbb{E}.\mathbb{D}$

6. wow, that was alot easier then what I was trying, thanks for the help
I'll post back with my results on the test tomorrow