[mysto]

I'm working on TRIG right now and up until now it has been, well I wouldn't say easy but not that difficult... However, I have slammed head first into proving angles and expressions, I get confused. The last 2 problems on here I have absolutely no clue what I am doing with these... If anyone can help me I would greatly appreciate it. :confused: It makes my head hurt...


1) If cos alpha=2/5 and sin Beta=1, where alpha is in quadrant IV and beta is on the positive Y axis, then csc(alpha + beta) is equal to...

I came up with... 2/5 other options are -2/5, 5/2, -5/2 [cos=2/5=.999 sin=1=.01745]

2) If angle = 150 deg, then sin 2angle is equal to...

I came up with... - sqrt 3/2 other options are: sqrt3/4, -1/2, sqrt3/2 [ I came up with 2*150= 300 sin 300= -0.866025]

3) What is the value of tan angle/2, if angle= 5*pi/3...

here are the options: 1/3, -sqrt3, sqrt3/3, -sqrt3/3 [ I came up with 5*pi/3= 5.236/2= 2.618 tan2.618=.0457]

4) What is the value of sin angle/2, when cot angle = 5/12, csc angle is negative and 0 deg < angle < 360 deg?

here are the options: 9/13, -2*sqrt13/13, -3*sqrt13/13, 3*sqrt13/13. [No Idea! ]

Thank You

[SirOracle]

Hey, I know how messy trig can be, so you just have to remember to keep working at it, because a lot of practice really is the only way to improve.

1) using a and b for alpha and beta, respectively
csc(a + b) = 1/(sin(a + b)) = 1/(sin(a)cos(b) + sin(b)cos(a)) using the trig identities
since sin(b)=1 we know that b = 90 degrees or pi/2 radians and thus cos(b)=0
this makes the first term of the denominator 0 and so the whole expression is now
1/(sin(b)cos(a))
sin(b)=1 and cos(a)=2/5, so
1/(sin(b)cos(a)) = 1/((1)(2/5)) = 5/2

2) I also got -sqrt(3)/2

3) using c as the angle, so c=5*pi/3
tan(c/2) = tan(5*pi/6) = sin(5*pi/6)/cos(5*pi/6)
which from the unit circle is
sin(5*pi/6)/cos(5*pi/6) = (1/2)/(-sqrt(3)/2) = 1/-sqrt(3) = -sqrt(3)/3

4) using d as the angle
since we are given that cot(d) = 5/12, we need to draw out a right triangle and pick one of the non-right angles to be d
since cotangent is adjacent/opposite, we know the two respective legs of the right triangle are 5 and 12, and Pythagorean Theorem shows us that the hypotenuse is 13... set this aside, we will need it later

since the cotangent is positive and the cosecant is negative, we know we are dealing with the third quadrant, in which both sine and cosine are negative

sin(d/2) = (+/-)sqrt((1-cos(d))/2) based on the half-angle trig identity (keep track of the parentheses, I know it is kind of hard to tell what the equations are)
from the triangle we created we can see that cos(d)=-5/13 (negative in the third quadrant)
(+/-)sqrt((1-cos(d))/2) = (+/-)sqrt((1-(-5/13))/2) = (+/-)sqrt(9/13) = (+/-)3*sqrt(13)/13
half of an angle in the third qudrant will fall somewhere between 90 and 135, so sin(d/2) should be positive, and the final answer is
3*sqrt(13)/13

I hope this was understandable and best of luck in the rest of trig!

[mysto]

I understand how csc is 1/sin. However I don't see how: 1/(sin(a + b)) = 1/(sin(a)cos(b) + sin(b)cos(a))? What happens to the cos a and sin b?
Anyway I appreciate your help.

[SirOracle]

well, it is a trig identity that sin(a + b) = sin(a)cos(b) + sin(b)cos(a)... I do not know the proof off the top of my head

in that particular example, cos(b) = 0, so the sin(a)cos(b) term disappears
and I just plugged in for the cos(a) and sin(b)