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Thread: 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 $\displaystyle sqrt{3}$. sin x is positive what is sec x?
    my answer: 2

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

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

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

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

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

    26) sin x = 1/4 and cos x > 0. cos x can be expressed in the form $\displaystyle 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 $\displaystyle sqrt{15}/a$ 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 $\displaystyle 1/sqrt{a}$. what is a?
    my answer: 3

    29) evaluate: $\displaystyle 1/2 log_{1/4} 80 - 1/2 log_{1/4} 5$
    my answer: -1
    Last edited by yoman360; Jun 17th 2009 at 07:52 PM.
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  2. #2
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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.


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

    $\displaystyle \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}} $

    . . . . . . . . $\displaystyle = \;\frac{\sqrt{2}\sqrt{3} - \sqrt{2}}{4} \;=\;\frac{\sqrt{2}\left(\sqrt{3}-1\right)}{4}
    $

    Answer: 4

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  3. #3
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    Thank you very much!
    Last edited by yoman360; Jun 17th 2009 at 08:28 PM.
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