Stuck on some precalc/trig questions

gexamb · #1 $Old$ November 1st, 2009, 05:43 PM

I just cant for the life of me figure out these 3 problems below. I will show any work that I have already done, but I just get stuck on some steps and my logic does not let me go any further:

Problem #1:

Prove the following by working with the LEFT side:

$(cot(x)/[1+tan(-x)]) + (tan(x)/[1+cot(-x)]) = cot(x) + tan(x) + 1$

so i have to solve (cot(x)/[1+tan(-x)]) + (tan(x)/[1+cot(-x)]) so that it is equal to cot(x) + tan(x) + 1

here is what I have done so far (I will only show work for the equation on the LEFT side):

$= [cot(x)(1-cot(x)) + tan(x)(1-tan(x))]/(1-tan(x))(1-cot(x))]$
$=[cot(x) - cot^2(x) - tan(x) - tan^2(x)]*[2 - cot(x) - tan(x)]$
$=cot(x) - 3cot^2(x) - tan(x) - tan^2(x) + cot^3(x) + tan^3(x)$
$=cot(x)*[1 - 3cot(x) + cot^2(x)] + tan(x)*[-1 - tan(x) + tan^2(x)]$

After that, I'm stuck. Have i followed the right steps? If not, what's wrong here/how do I solve it?

Problem #2:

Find all solutions in the interval [0,2pi): give exact solutions

$sin(x) = 3cos(x)$

this is what I have done, but I am not sure if is right:

$= 1/cos(x) * sin(x) = 3cos(x) * 1/cos(x)$
$= sin(x)/cos(x) = 3$
$= tan(x) = 3$

3 is out of interval range, therefore there is no solution.

is this right?

Problem #3:

If cos(X) = 2/sqrt(5) (3pi/2 < X < 2pi) and sin(B) = 4/5 (pi/2 < B < pi), compute sin(B-X)

here is what I have so far:

since (3pi/2 < X < 2pi) is in the third quadrant, cos(X) = 2/sqrt(5) in QIII.

i have to solve sin of X but I am only given cos of X, do i have to find out the value of sin of angle X in the 3rd quad?

well, using the Pythagorean theorem a^2 + b^2 = c^2, i concluded that making a graph of a triangle in QIII and assuming c = sqrt(5), a = 2, we have to find b to get the sin of angle X. so b = +-(1), but the opposite side in QIII would be negative, therefore b = -1.

so sin(X) would be = -1/sqrt(5)

then solved for sin(B-X):

$= sin(B-X) = [sin(4/5) * cos(-1/sqrt(5))] - [cos(4/5) * sin(-1/sqrt(5))]$

that is what i have so far. is my logic correct? have i solved the problem correctly?

I need to study these problems for a test I have tomorrow, so I would extremely appreciate any help anyone would be willing to give me. Thank you all for your time in advance.

Note: Prirority would be Problem #1. TY!

Prove It · #2 $Old$ November 1st, 2009, 06:56 PM

Quote:

Originally Posted by gexamb $View Post$

I just cant for the life of me figure out these 3 problems below. I will show any work that I have already done, but I just get stuck on some steps and my logic does not let me go any further:

Problem #1:

Prove the following by working with the LEFT side:

$(cot(x)/[1+tan(-x)]) + (tan(x)/[1+cot(-x)]) = cot(x) + tan(x) + 1$

$\frac{\cot{x}}{1 + \tan{(-x)}} + \frac{\tan{x}}{1 + \cot{(-x)}} = \frac{\frac{\cos{x}}{\sin{x}}}{1 + \frac{\sin{(-x)}}{\cos{(-x)}}} + \frac{\frac{\sin{x}}{\cos{x}}}{1 + \frac{\cos{(-x)}}{\sin{(-x)}}}$

$= \frac{\frac{\cos{x}}{\sin{x}}}{1 - \frac{\sin{x}}{\cos{x}}} + \frac{\frac{\sin{x}}{\cos{x}}}{1 - \frac{\cos{x}}{\sin{x}}}$

$= \frac{\frac{\cos{x}}{\sin{x}}}{\frac{\cos{x} - \sin{x}}{\cos{x}}} + \frac{\frac{\sin{x}}{\cos{x}}}{\frac{\sin{x} - \cos{x}}{\sin{x}}}$

$= \frac{\cos^2{x}}{\sin{x}(\cos{x} - \sin{x})} + \frac{\sin^2{x}}{\cos{x}(\sin{x} - \cos{x})}$

$= \frac{\cos^2{x}}{\sin{x}(\cos{x} - \sin{x})} - \frac{\sin^2{x}}{\cos{x}(\cos{x} - \sin{x})}$

$= \frac{\cos^3{x}}{\sin{x}\cos{x}(\cos{x} - \sin{x})} - \frac{\sin^3{x}}{\sin{x}\cos{x}(\cos{x} - \sin{x})}$

$= \frac{\cos^3{x} - \sin^3{x}}{\sin{x}\cos{x}(\cos{x} - \sin{x})}$

$= \frac{\cos^2{x} + \sin{x}\cos{x} + \sin^2{x}}{\sin{x}\cos{x}}$

$= \frac{\cos^2{x}}{\sin{x}\cos{x}} + \frac{\sin{x}\cos{x}}{\sin{x}\cos{x}} + \frac{\sin^2{x}}{\sin{x}\cos{x}}$

$= \frac{\cos{x}}{\sin{x}} + 1 + \frac{\sin{x}}{\cos{x}}$

$= \cot{x} + \tan{x} + 1$ .

I DO get the same answer you provided.

adkinsjr · #3 $Old$ November 1st, 2009, 07:04 PM

Problem 2)

There are two solutions for $tan(x)=3$ . They're given by:

$Arctan(3)$ and $Arctan(3)+\pi$

Remember that $tan(x)=tan(x+\pi)$ since the function has a period $\pi$ .

My calculator gives $\approx 71.57$ in degrees for $Arctan(3)$ .

gexamb · #4 $Old$ November 1st, 2009, 08:28 PM

Quote:

Originally Posted by adkinsjr $View Post$

Problem 2)

There are two solutions for $tan(x)=3$ . They're given by:

$Arctan(3)$ and $Arctan(3)+\pi$

Remember that $tan(x)=tan(x+\pi)$ since the function has a period $\pi$ .

My calculator gives $\approx 71.57$ in degrees for $Arctan(3)$ .

but since the solution is limited to [0,2pi), tanx=3 is not in that period correct?

and we really have not gone over or studied arc functions, so i doubt that the professor is looking for an answer using that function. but is my logic correct, that 3 is not in the interval, therefore there is no solution?

Anyone want to take on Problem #3? Would really appreciate the help. Thank you to all that have helped me already, much obliged.