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Math Help - Some help with trig

  1. #1
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    Some help with trig

    I am terrible at proofs and trig-so this is like a double whammy, and i have tried these ones for hours to no avail
    1. Prove that sin 1/6pi=1/2 , cos 1/6pi=1/2 sq.rt.(3)
    ...it says hint: use previous example, which asks to prove sin3x=3sinx-4sin^3 x and cos3x=cosx -4sin^2 x cosx....prove also cos3x=4cos^3 x -3cosx

    (im not too worried about the hint part---just posted it incase someone found it helpful while prooving)
    2. prove that sin 1/3pi=1/2sq.rt. 3, cos 1/3pi=1/2

    3. prove that sine 1/4pi=cos 1/4pi=1/2 sq.rt. 2

    any explanations and help is greatly apprecaited....thanks in advance
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  2. #2
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    Quote Originally Posted by twostep08 View Post
    1. Prove that sin 1/6pi=1/2 , cos 1/6pi=1/2 sq.rt.(3)
    Can't you just use an equilateral triangle with length of 2?

    Quote Originally Posted by twostep08 View Post

    ...it says hint: use previous example, which asks to prove sin3x=3sinx-4sin^3 x and cos3x=cosx -4sin^2 x cosx....prove also cos3x=4cos^3 x -3cosx
    Did you prove these?
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  3. #3
    Senior Member pacman's Avatar
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    1. Prove that sin 1/6pi=1/2.

    Costruct a tringle, with hypotenuse equal to 2, opposite side 1 and adjacent side = sqrt 3.

    a) sin pi/6 = sin 30 = opposite side/hypotenuse = 1/2

    b) cos pi/6 = cos 30 = adjacent side/hypotenuse = (sqrt 3)/2

    use this identity:

    i) sin (A + B) = sin A cos B + cos A sin B

    ii) cos (A + B) = cos A cos B - sin A sin B

    iii) sin^2 x + cos^2 x = 1; sin^2 x = (1-cos^2 x); cos^2 x = (1-sin^2 x)

    iV) cos 2x = cos^2 x - sin^2 x = 1 - 2sin^2 x = 2cos^2 x - 1

    ----------------------------------------------------------------------

    c) Then, we havesin 3x = sin (2x + x) = sin 2x cos x + cos 2x sin x

    sin 3x = [(sin x cos x + cos x sin x) cos x] + [(cos x cos x - sin x sin x) sin x]

    sin 3x = 2(sin x cos x)cos x + (cos^2 x - sin^2 x) sin x

    sin 3x = (2sin x cos^2 x + cos^2 x sin x) - sin^3 x

    sin 3x = 3 sin x cos^2 x - sin^3 x

    sin 3x = 3 sin x (1 - sin^2 x) - sin^3 x

    sin 3x = 3 sin x - 3sin^3 x - sin^3 x

    sin 3x = 3 sin x - (3sin^3 x + sin^3 x)

    sin 3x = 3 sin x - 4sin^3 x

    d) Prove: cos 3x = cos 3x = 4cos^3 x - 3 cosx

    cos 3x = cos (2x + x) = cos 2x cos x - sin 2x sin x

    cos 3x = (2cos^2 x - 1)(cos x) - (2sin x cos x)(sin x)

    cos 3x = 2cos^3 x - cos x - 2sin^2 x cos x

    cos 3x = 2 cos^3 x - cos x - 2(1 - cos^2 x)(cos x)

    cos 3x = 2 cos^3 x - cos x - 2cos x + 2 cos^3 x

    cos 3x = 4 cos^3 x - 3 cos x

    ----------------------------------------------------------------

    2) You try the rest . . . .


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