# Trig Review

• Apr 3rd 2006, 03:17 PM
drewms64
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
• Apr 3rd 2006, 06:31 PM
ThePerfectHacker
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.
• Apr 3rd 2006, 06:44 PM
drewms64
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
• Apr 4th 2006, 03:56 PM
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

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
• Apr 4th 2006, 06:28 PM
ThePerfectHacker
Quote:

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, $=2$

$\mathbb{Q}.\mathbb{E}.\mathbb{D}$
• Apr 4th 2006, 06:38 PM
drewms64
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 :)