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    Hi, i ma having trouble with the questions below, if anyone could help that would be really useful. Is there an easy way to do them.....i find them really tricky! (i especially struggle on the proofs!):confused:

    1. Prove that tanθ +cotθ = 2 cosec 2θ

    2. Solve by getting the exact answers in terms of π, 2(1- cos 2θ) = tan θ

    3. Express sin^2θ in the form a + b cos 2θ, where a and b are constants.

    4. Solve the equation 3 – 3 cos 2θ = 2 cosec θ, giving all solutions in radians in the interval 0 < θ < 2π

    Thanks
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    1. think about sin^2+cos^2 = 1 and sin2x=2sinxcosx... that made it possible for me to solve it...
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    TSR Group Staff
    Q1:

    \displaystyle\huge \text{RHS} \equiv 2\csc 2\theta
    \displaystyle\huge \equiv \frac{2}{\sin 2\theta}
    \displaystyle\huge \equiv \frac{2}{2\sin \theta \cos \theta}
    \displaystyle\huge \equiv \frac{1}{\sin \theta \cos \theta}
    \displaystyle\huge \equiv \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta}
    \displaystyle\huge \equiv \frac{\sin^2 \theta}{\sin \theta \cos \theta} + \frac{\cos^2 \theta}{\sin \theta \cos \theta}
    \displaystyle\huge \equiv \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta}
    \displaystyle\huge \equiv \tan \theta + \cot \theta \equiv \text{LHS}

    TSR's LaTeX system is crap... Those silly floating twos are supposed to be exponents, but for some reason decided they'd be happier riding on top of their trig functions instead of squaring them. :rolleyes:
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    (Original post by Dez)
    Q1:TSR's LaTeX system is crap... Those silly floating twos are supposed to be exponents, but for some reason decided they'd be happier riding on top of their trig functions instead of squaring them. :rolleyes:
    Thanks a lot
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    (Original post by nota bene)
    1. think about sin^2+cos^2 = 1 and sin2x=2sinxcosx... that made it possible for me to solve it...
    thanks for the help
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    3. Express sin^2θ in the form a + b cos 2θ, where a and b are constants.
    cos 2θ = 1 - 2sin²θ

    Re-arranging:

    2sin²θ = 1 - cos2θ

    sin²θ = 1/2 - (cos2θ)/2

    So a = 1/2, b = -1/2.

    Hope this helps,

    ~~Simba
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    (Original post by Simba)
    cos 2θ = 1 - 2sin²θ

    Re-arranging:

    2sin²θ = 1 - cos2θ

    sin²θ = 1/2 - (cos2θ)/2

    So a = 1/2, b = -1/2.

    Hope this helps,

    ~~Simba
    thanks for that - just looking at you signature - that is amazing getting 100 in all modules
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    4. Solve the equation 3 – 3 cos 2θ = 2 cosec θ, giving all solutions in radians in the interval 0 < θ < 2π

    \tex \large 3-3cos 2\theta = 2 cosec \theta

    \tex \large 3 - 3(1-2sin^2\theta) = \frac{2}{sin\theta}

    \tex \large 6sin^2\theta = \frac{2}{sin\theta}

    \tex \large sin^3\theta = \frac{1}{3}

    \tex \large sin \theta = \frac{1}{^3\sqrt{3}}

    \tex \large \theta = 0.77, 2.38, 3.91, 5.32

    The root 3 is the cubic root of 3 not 3root3
 
 
 
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