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March 25, 2017

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1)Find the exact value of cos 105 by using a half-angle formula.
A)sqrt 2 - sqrt 3 /2
B)-sqrt 2 - sqrt 3 /2
C)-sqrt 2 + sqrt 3 /2
D)sqrt 2 + sqrt 3 /2

cos 105
cos 105 = cos 210/2
sqrt 1 + 210/2
sqrt 1 + sqrt 3/2 /2
sqrt 2 + sqrt 3/2 which is D

2)Find the solution of sin2 theta = cos theta if 0 -< theta < 180
A)30 degrees and 90 degrees
B)30 degrees and 150 degrees
C)30 degrees, 90 degrees, 150 degrees
D)0 degrees, 90 degrees, and 150 degrees

sin2 theta = cos theta
sin2 theta - 1 = 0
(sin theta + 1)(sin theta - 1)= 0
sin theta = -1 or sin theta = 1 which is B

3)An insect population P in a certain area fluctuates with the seasons. It is estimated that P = 17,000 + 4500 sin pi(t)/52, where t is given in weeks. Determine the number of weeks it would take for the population to initially reach 20,000.
A)12 weeks
B)692 weeks
C)38 weeks
D)42 weeks

P = 17,000 + 4500 sin pi(t)/52
20,000 = 17,000 + 4500 sin pi(t)/52
3,000 = 4500 sin pi(t)/52
.666 = sin pi(t)/52
The closest thing is C

  • Trig(3 all with work) - ,

    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.

  • Trig(3 all with work) - ,

    #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

  • FOR REINY - ,

    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}

  • FOR REINY - ,

    Did you mean to direct this to me?
    I only answered question #3

  • FOR REINY - ,

    I apologize. I though you were talking about question 1.

    Sorry mate....

  • Trig(3 all with work) - ,

    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!

  • Trig(3 all with work) - ,

    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 sin2 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

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