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Math Help - Trigonometric Relations/Identities help

  1. #1
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    Trigonometric Relations/Identities help

    Please help me solve these:

    1a) If tan(A) = 3/4, show that sin(A) = 3/4cos(A)

    1b) Therefore, evaluate:
    [4sin(A) - cos(A)] / [8sin(A) + 5cos(A)]

    2) Solve without a calculator, show working
    a) sin(60) x tan(45)
    b) 9tan^2(30)
    c) 6cos(60) x sin(30)

    3) If sin^2(x) = 7/8, find cos^2(x)

    4) If cos^2(x) = 0.36, fins sin^2(x)

    5) Find tan x if cos x = 8/17 and sin x = 15/17

    The last one is as an optional challenge, please help me if you can.

    6) ACD is a right angled triangle, where B is a point on AC, angle CAD = 30 deg, angle CBD = 60 deg and AB = 10.

    a) By considering angle BCD, prove CD = SQRT 3 BC
    b) By considering angle ACD, prove CD = [10 + BC] / [SQRT 3]
    c) Use these two results to find BC

    Thank you in advance!
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  2. #2
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    Hello,

    Quote Originally Posted by BG5965 View Post
    Please help me solve these:

    1a) If tan(A) = 3/4, show that sin(A) = 3/4cos(A)
    Use the definition of tan :

    \tan(A)=\frac{\sin(A)}{\cos(A)}

    1b) Therefore, evaluate:
    [4sin(A) - cos(A)] / [8sin(A) + 5cos(A)]
    Substitute \sin(A)=\frac 34 \cos(A)

    For example, 4 \sin(A)-\cos(A)=3 \cos(A)-\cos(A)=2 \cos(A)

    2) Solve without a calculator, show working
    a) sin(60) x tan(45)
    b) 9tan^2(30)
    c) 6cos(60) x sin(30)
    Use this unit circle : http://dcr.csusb.edu/LearningCenter/...UnitCircle.gif

    3) If sin^2(x) = 7/8, find cos^2(x)

    4) If cos^2(x) = 0.36, fins sin^2(x)
    Use identity \cos^2(x)+\sin^2(x)=1

    5) Find tan x if cos x = 8/17 and sin x = 15/17
    Use the definition of tan I gave you above

    The last one is as an optional challenge, please help me if you can.

    6) ACD is a right angled triangle, where B is a point on AC, angle CAD = 30 deg, angle CBD = 60 deg and AB = 10.

    a) By considering angle BCD, prove CD = SQRT 3 BC
    b) By considering angle ACD, prove CD = [10 + BC] / [SQRT 3]
    c) Use these two results to find BC
    Draw a sketch and use definitions of cosine, sine and tangent :

    cos=adjacent side/hypotenuse
    sin=opposit side/hypotenuse
    tan=opposit side/adjacent side

    If you don't know it, search on wikipedia
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  3. #3
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    Hi,

    2.)

    sin60*tan45

    =(sqrt3)/2 * 1

    =(sqrt3)/2

    b.)

    =9tan^2(30)

    =9(1/sqrt3)^2
    =9(1/3)
    =3

    c.)

    =6Cos60sin30
    =6(1/2)(1/2)
    =3/2

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