Posted by Jon on Friday, February 15, 2008 at 3:17am.
3) Try again. t= 52/PI * arcsin .6666
2) the second line baffles me, how did you get sin^2T -1=0 from the line above?
1) Your use of the half angle formula seems to be missing. Your answer is wrong.
#1 of 1
The cosine of 105 has to be negative because it is in the second quadrant. The same applies to cos 210 in the third quadrant.
cos 210 = -(sqrt3)/2
|cos 105| = sqrt[(1/2)(1 - cos 210)]
= sqrt[(1/2)(1 - (sqrt3)/2]]
For cos 105, a minus sign must be put in front of that to make the answer negative. This agrees with none of your selections, but is the correct answer for cos 105, which is -0.25882
Reiny said:
"The same applies to cos 210 in the third quadrant."
When we deal with Half-Angle Formulas, the sign selected to be placed in front of the square root has to do with the ORIGINAL degree given.
He was given cos105 degrees (in quadrant 2) not 210 degrees (quadrant 3).
This is why cosine is NEGATIVE (in quadrant 2) and also why we use the formula cos(x/2) = -sqrt{(1 + cos(x))/2}
I will try question 1.
We need to use the correct Half-Angle Formula to find the exact value of cos105 degrees without using a calculator.
Because 105 degrees = cos(210)/2, we can use the Half-Angle Formula for
cos(x)/2 with x = 210 degrees. Also, because 105 degrees lies in quadrant 2, where cosine is negative, we choose the NEGATIVE SIGN when using the formula below.
The formula we want is this one:
cos(x/2) = - sqrt{(1 + cos(x)/2)}.
Do you see the negative IN FRONT of the square root?
Why? Like I said above, cosine is negative in quadrant 2.
cos(210)/2 = sqrt{(1 + cos210)/2)}
What is the value of cos210 degrees?
Cos210 degrees = - sqrt{3}/2.
We plug -sqrt{3}/2 in the radicand and simplify.
-sqrt{(1 - sqrt{3}/2)/2)}
After doing the algebra, we get
cos105 degrees = [sqrt{2) - sqrt {3}]/2...Choice A.
Done!
for #3 make sure that if you take arcsin your calculator is switched to radians.
I got an answer of 12 for bobpursleys expression
for #2 I too, like Bob, was puzzled
First of all is it
sin 2θ = cos θ or sin^{2} θ = cos θ ?
for the first one
2sinθcosθ - cosθ = 0
cosθ(1sinθ - 1) = 0
cosθ=0 or sinθ=1/2
so θ = 90º or θ = 30º or 150º for your domain, which is choice C
If you meant the second interpretation, change, change sin^2 θ to 1 - cos^2 θ and use the quadratic formula