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Math Help - can someone please chek my trig questions+ multiple chioce!

  1. #1
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    Thumbs up can someone please chek my trig questions+ multiple chioce!

    I attached them so yeah can you check them for me thanks!
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  2. #2
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    Hello, Justin!

    Sorry, only one of them is correct . . .


    1) What is the exact value of: .sin˛(π/6) - 2·sin(π/6)·cos(π/6) + cos˛(-π/6)
    . . . . . . . . _ . . . . - . - . - ._ . . . . . . . . . . . . _ . . . - . - . - . . . ._
    . . (A) 2 - √3 . . . (B) (2 - √3)/2 . . . (C) (2 + √3)/2 . . . (D) 2 + √3
    Since cos(-θ) = cos(θ), we have: .sin˛(π/6) + cos˛(π/6) .= .1

    Since 2·sinθ·cosθ .= .sin2θ, we have: .2·sin(π/6)·cos(π/6) .= .sin(π/3)
    . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . _ . . . . . . . . _
    The problem becomes: .1 - sin(π/3) .= .1 - √3/2 .= .(2 - √3)/2 . Answer (B)



    2. If cosθ = -sinθ, then sin˛θ is:
    . . . . . . . . . . . . - . . . - . . . . . _ . . . . . . . . . _
    . . (A) 1/4 . . . (B) 1/2 . . . (C) √2/2 . . . (D) -√2/2
    We have: .sinθ .= .-cosθ
    . . . . . . . . . . . . . . . . . . . . sinθ
    Divide both sides by cosθ: . ------ .= .-1 . . tanθ .= .-1 . . θ = π/4, 5π/4
    . . . . . . . . . . . . . . . . . . . . cosθ
    . . . - . . - . . - . . . . _
    Then: .sinθ .= .±1/√2
    . . . . . . . . . . . . . . . . . . . _
    Therefore: .sin˛θ .= .(±1/√2)˛ .= .1/2 . Answer (B)




    . . . . . . . . . . . . . . . . . . . .sin(A + B)·sin(A - B)
    3) What is another form of: ------------------------
    . . . . . . . . . . . . . . . . . . . . . .cos˛A·cos˛B

    (A) cos˛A - cos˛B . . (B) tan˛A + tan˛B . . (C) tan˛A - tan˛B . . (D) sin˛A - sin˛B
    . . . . . . . . (sinA·cosB + sinB·cosA)(sinA·cosB - sinB·cosA)
    We have: . -------------------------------------------------------
    . . . . . . . . . . . . . . . . . . . .cos˛A·cos˛B


    . . . . .sin˛A·cos˛B - sin˛B·cos˛A . . . .sin˛A·cos˛A . . sin˛B·cos˛A
    . . = . ------------------------------ . = . --------------- - ---------------
    . . . . . . . . . .cos˛A·cos˛B . . . . . . . . cos˛A·cos˛B . .cos˛A·cos˛B

    . - . - .sin˛A . .sin˛B
    . . = . ------- - ------- . = . tan˛A - tan˛B . . Answer (C) . . . Correct!
    . - . - .cos˛A . cos˛B



    . . . . . . - . . - . . - . . . . . . . . .__
    4. For the equation: .4·cosθ - √12 .= .0, where 0 < θ < 2π, determine the measure of θ.

    (A) π/3, 5π/3 . . (B) 2π/3, 4π/3 . . (C) 5π/6, 7π/6 . . (D) π/6, 11π/6
    . . . . . . . . . . . . . . . . __ . . . . . . . . . . . . _ . . . . . _
    We have: .4·cosθ .= .√12 . . cosθ .= .2√3/4 .= .√3/2

    Therefore: .θ .= .π/6, 11π/6 . Answer (D)

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