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Math Help - Are these correct? Part 2

  1. #1
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    Are these correct? Part 2

    16) if sin x= 1/5 then csc x =?
    my answer: 5

    17) sec x = 5/4 and value of tan x is positive what is sin x equal?
    my answer: 3/5

    18) cot x = 3/4 and cos x is negative. sin x =?
    my answer: -4/5

    19) cos x = -1/2 the value of of x can be in the form α*pi what is the value of α?
    my answer: 2/3

    20) the value of tan x is sqrt{3}. sin x is positive what is sec x?
    my answer: 2

    21) Evaluate: tan (sin^{-1}(1/sqrt{2}))
    my answer: 1

    22)  tan (cot^{-1}(1/3))=?
    My answer: 3

    23) sin (pi/4 + pi/6) can be expressed in the form [sqrt{2}(sqrt{3}+1)]/y. whats y?
    my answer: 4

    24) cos (pi/4 + pi/6) can be expressed in the form [sqrt{2}(sqrt{3}-1)]/z whats z?
    my answer: 4

    25) if sin x = sqrt{3}/2 and tan x= - sqrt{3} the vaule of x can be written in the form api. whats a?
    my answer: 2/3

    26) sin x = 1/4 and cos x > 0. cos x can be expressed in the form sqrt{a}/4 what is the value of a?
    my answer: 15

    27) sin x =1/4 and cos x>0. sin (2x) can be written in the form sqrt{15}/a what is the value of a?
    My answer: 8

    28) cos x= 1/3 and sin (x/2) > 0. The value of sin (x/2) can be written as 1/sqrt{a}. what is a?
    my answer: 3

    29) evaluate: 1/2 log_{1/4} 80 - 1/2 log_{1/4} 5
    my answer: -1
    Last edited by yoman360; June 17th 2009 at 08:52 PM.
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  2. #2
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    Lexington, MA (USA)
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    Hello, yoman360!

    You did even better this time!

    You did so well on every other problem,
    . . I assume you forgot to type your answer to #24.


    24)\;\;\cos \left(\tfrac{\pi}{4} + \tfrac{\pi}{6}\right) can be expressed in the form \frac{\sqrt{2}\left(\sqrt{3}-1\right)}{z} What is z?

    \text{We have: }\;\cos\left(\tfrac{\pi}{4}+\tfrac{\pi}{6}\right) \;=\;\underbrace{\cos\tfrac{\pi}{4}}_{\frac{\sqrt{  2}}{2}} \underbrace{\cos\tfrac{\pi}{6}}_{\frac{\sqrt{3}}{2  }} - \underbrace{\sin\tfrac{\pi}{4}}_{\frac{\sqrt{2}}{2  }}\underbrace{\sin\tfrac{\pi}{6}}_{\frac{1}{2}}

    . . . . . . . . = \;\frac{\sqrt{2}\sqrt{3} - \sqrt{2}}{4} \;=\;\frac{\sqrt{2}\left(\sqrt{3}-1\right)}{4}<br />

    Answer: 4

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  3. #3
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    Thank you very much!
    Last edited by yoman360; June 17th 2009 at 09:28 PM.
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