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Math Help - trionometrey

  1. #1
    bay
    bay is offline
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    Red face trionometrey

    Can someone please check my triq questions thanks!
    Thanks!
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  2. #2
    Super Member

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    Hello, bay!

    1) The exact value of: 6Ěcos(-61π/6) - 2Ěsin(13π/4) is:
    . . . . . . _ . . ._ . . . . . . . . ._ . . . . . . . . _ . . . . . . _ . . _
    . . A. 3√3 + √2 . . B. -3 + √2 . . C. -3 - √2 . . D. 3√3 - √2
    Since -61π/6 = -π/6 .and .13π/4 = 5π/4

    we have: .6Ěcos(-π/6) - 2Ěsin(5π/4)
    . . . . . . . . . _ . - . . - . _ . - . . - . . ._ . . ._
    . . . . = . 6(√3/2) - 2(-√2/2) . = . 3√3 + √2 . answer A


    I agree with your answers to #2, 3, and 4.



    5. Another form of .cos(30░ - A) + sin(60░ + A) is:
    . . . . . . . . . . . . . _ . . . . . . . . . . . . . . . _ . . . . . . . . . . . . . . . . . _
    . . A. sin A . . B. √3Ěcos A + sin A . . C. (√3/2)Ěsin A + cos A . . D. √2Ěsin A + cos A
    . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ._
    cos(30░ - A) .= .cos 30░Ěcos A + sin 30░Ěsin A .= .(√3/2)Ěcos A + (1/2)Ěsin A
    . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ._
    sin(60░ + A) .= .sin 60░Ěcos A + cos 60░Ěsin A .= .(√3/2)Ěcos A + (1/2)Ěsin A

    . . . . . . . . . . . . . . . . . . . . . . . . . . . ._
    Add: .cos(30░ - A) + sin(60░ + A) .= .√3ĚcosA + sin A . answer B

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