1. ## simplify trig

A.2sinx+1=0
B.2cos^2x-1=0
C.3tan^3x-tanx=0
D.2sin^2x=2+cosx

I really need help.. I have my answers here but im not sure if they are correct:

A.x=7/6pi, 11/6pi
B.x=pi/4 7/4pi
C.x=pi/6 7/6pi 2pi pi
D.x=pi/2, 3/2pi pi/3 5/3pi

I also need these checked
Using a sum of difference identity find exact values for the following:

A.Cos75(degrees)
B.Sin15(degrees)
C.tan345(degrees)

I got
A.*squareroot*6-2/4
B.*squareroot*6-2/4
C.3-*squareroot*3/3

For problems 2 and 3 find the following.
A.cos(x-y): sin(x-y); tan (x-y); the quadrant (x-y) is in.
B.cos(x+y); sin(x+y) tan(x+y) the quadrant (x+y) is in.

2.If (pi/2)<x<pi and pi<y<(3pi/2) and cosx=-5/13 and tany=3/4

3.If pi<x< or equal too(3pi/2) and (3pi/2)<y<2pi, and secx=-41/9 and tany=-12/5

If you could show me how to do them it would be appreciated.

2. A.x=7/6pi, 11/6pi
right

B.x=pi/4 7/4pi
C.x=pi/6 7/6pi 2pi pi
these solutions work but there are others as well. Remember that if $\displaystyle x^2$ = a then $\displaystyle x = \pm \sqrt {a}$

D.x=pi/2, 3/2pi pi/3 5/3pi
You have lost a minus sign somewhere here, you should have ended up with cos x = 0 or cos x = -1/2, but the good news is that you appear to know what you are doing.

A.Cos75(degrees)
B.Sin15(degrees)
C.tan345(degrees)
$\displaystyle cos (75^o)$= $\displaystyle cos(45^o+30^o)$
=$\displaystyle cos(45^o)cos(30^o)-sin(45^o)sin(30^o)$
=$\displaystyle \frac{1}{\sqrt{2}}\frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}\frac{1}{2}$
=$\displaystyle \frac{\sqrt{3}-1}{2\sqrt{2}}$

$\displaystyle sin (15^o)$ = $\displaystyle cos (75^o)$, as you probably noticed.

$\displaystyle tan (345^o)$ = $\displaystyle -tan (15^o)$
=$\displaystyle -\frac {sin(15^o)}{cos(15^o)}$
=$\displaystyle -\frac{sin(15^o)}{cos (45^o+-30^o)}$
=$\displaystyle -\frac{sin(15^o)}{cos (45^o)cos(-30^o)-sin(45^o)sin(-30^o)}$
=$\displaystyle -\frac{\frac{\sqrt{3}-1}{2\sqrt{2}}}{\frac{1}{\sqrt{2}}\frac{\sqrt{3}}{2 }+\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}}$
=$\displaystyle -\frac{\sqrt{3}-1}{\sqrt{3}+1}$
=$\displaystyle \frac{1-\sqrt{3}}{1+\sqrt{3}}$

3. For problems 2 and 3 find the following.
A.cos(x-y): sin(x-y); tan (x-y); the quadrant (x-y) is in.
B.cos(x+y); sin(x+y) tan(x+y) the quadrant (x+y) is in.

2.If (pi/2)<x<pi and pi<y<(3pi/2) and cosx=-5/13 and tany=3/4
The first thing you need to do is draw up the triangles for x and y, using the definition of cos as adj/hyp and tan as opp/adj and pythagoras theorem.

Now you can read off the absolute values of anything else you may need and combine it with the quadrants of x and y to get the actual values. Then its just a matter of using the formulae for addition of angles in trigonometric functions.

To find the quadrant of x-y, you will need to combine the restrictions imposed by the range for x and y and the signs of the results you get for sin, cos, tan etc of x-y

4. wouldnt cos(-5/13+4/5) be:

(-5/13.-4/5)-(12/14.-3/5)

Because cos in the third quadrant is always negative.
And sin in the third quadrant is always negative.

5. Originally Posted by ninelivesmeow
wouldnt cos(-5/13+4/5) be:

(-5/13.-4/5)-(12/14.-3/5)

Because cos in the third quadrant is always negative.
And sin in the third quadrant is always negative.
I presume that you are trying to do a difference of angles formula for cosine, but I can't identify which (or even what) problem you are working on.

I think you are trying to do $\displaystyle cos(x + y)$ where $\displaystyle cos(x) = -\frac{5}{13}$ and $\displaystyle sin(y) = \frac{4}{5}$?

I need you to confirm this before I go any further.

-Dan