Trig Help

mmgolf · #1 $Old$ 10-07-2007, 02:55 PM

I have a collection of Trig problems that are giving me some difficulty...

1. Finding the Period, Phase Shift, and Asymptotes for the following expressions
3+4cot(5x-2)
y=-3tan2(pi X-pi/4)
2. Find the exact value for
tan105
sin195
cos51cos9-sin51sin9
3.Simplify the expression
1+cosx + sin x
sin x 1+cos x

1+secx
sinx + tanx

1 +siny
1 +cscy

thanks in advance for all help...these are some HW problems that i am confused on and i have an exam in the coming week, so if explanations could be given for all problems that would be nice as well

mmgolf · #2 $Old$ 10-07-2007, 03:00 PM

The first simplify expression was posted weirdly

it is supposed to read

1 + cosx divided by sinx + sin x divided by 1 + cosx

Soroban · #3 $Old$ 10-07-2007, 03:38 PM

Hello, mmgolf!

There are some "tricks" you should know for #3 . . .

Quote:

3. Simplify:

. . $(a)\;\;\frac{1+\cos x}{\sin x} + \frac{\sin x}{1+\cos x}$

Multiply the second fraction by: $\frac{1-\cos x}{1-\cos x}$

. . We have: . $\frac{\sin x}{1 + \cos x}\cdot\frac{1-\cos x}{1-\cos x} \;=\;\frac{\sin x(1 - \cos x)}{1 - \cos^2\!x} \;=\;\frac{\sin x(1 - \cos x)}{\sin^2\!x} \;=\;\frac{1-\cos x}{\sin x}$

The problem becomes: . $\frac{1+\cos x}{\sin x} + \frac{1-\cos x}{\sin x}\;=\; \frac{2}{\sin x} \;=\;2\csc x$

Quote:

$(b)\;\;\frac{1+\sec x }{\sin x + \tan x}$

Multiply by: $\frac{\cos x}{\cos x}$

. . $\frac{\cos x}{\cos x}\,\cdot\,\frac{1 + \sec x}{\sin x + \tan x} \;=\;\frac{\cos x + 1}{\sin x\cos x + \sin x} \;=\;\frac{\cos x + 1}{\sin x(\cos x + 1)} \;=\;\frac{1}{\sin x} \;=\;\csc x$

Quote:

$(c)\;\frac{1 + \sin y}{1 + \csc y}$

Multiply by: $\frac{\sin y}{\sin y}$

. . $\frac{\sin y}{\sin y}\cdot\frac{1 + \sin y}{1 + \csc y} \;=\;\frac{\sin y(1 + \sin y)}{\sin y + 1} \;=\;\sin y$

mmgolf · #4 $Old$ 10-07-2007, 04:49 PM

thanks...anyone else....??

Soroban · #5 $Old$ 10-07-2007, 07:52 PM

Hello again, mmgolf!

For #2, you need some Sum/Difference formulas.

Quote:

2. Find the exact value for:
$(a)\;\;\tan105$
$(b)\;\;\sin195$
$(c)\;\;\cos51\cos9 - \sin51\sin9$

$(a)\;\;\tan105^o$

We note that: . $105 \:=\:60 + 45$
. . and we use: . $\tan(A + B)\:=\:\frac{\tan A + \tan B}{1 - \tan A\tan B}$

Hence we have: . $\tan105 \:=\:\tan(60 + 45) \;=\;\frac{\tan60 + \tan45}{1 - \tan60\tan45}$

Can you finish it now?

Quote:

$(b)\;\;\sin195^o$

Note that: . $195 \:=\:150 + 45$
. ,and we use: . $\sin(A + B) \:=\:\sin A\cos B + \sin B\cos A$

So we have: . $\sin(150 + 45) \:=\:\sin150\cos45 + \sin45\cos150$ . . . etc.

Quote:

$(c)\;\;\cos51\cos9 - \sin51\sin9$

You're expected to recognize that this is the right side of the formula:
. . $\cos(A + B) \:=\:\cos A\cos B - \sin A\sin B$

Got it?

mmgolf · #6 $Old$ 10-07-2007, 08:10 PM

thanks soroban

any hints on the tangent problems on how to find the asymptotes, period, and phase shift??