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Math Help - Some sort of trig question ???

  1. #1
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    Some sort of trig question ???

    Hi chaps, I got a Question here that has me confused , any thoughts or help is appreciated (how do we even use a calculator to find the answer?)
    Thanks

    Given that cos
    θ = -4/5 and π/2 ≤ θ ≤ π

    find, in surd form (do not use calculators)

    A) tan θ

    B) Sin θ

    C) sin (θ + 2π/3)

    D) cos 2 θ
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Burnt Flower View Post
    Hi chaps, I got a Question here that has me confused , any thoughts or help is appreciated (how do we even use a calculator to find the answer?)
    Thanks

    Given that cos
    θ = -4/5 and π/2 ≤ θ ≤ π

    find, in surd form (do not use calculators)
    \frac {\pi}2 \le \theta \le \pi means we are in the second quadrant. so we should know that only sine (and cosecant) is positive.

    recall your trig ratios. \mbox{cosine} = \frac {\mbox {Adjacent}}{\mbox {Hypotenuse}} etc.

    knowing the cosine ratio, we can draw the right-triangle as you see below, with one acute angle \theta, with the adjacent side 4 and the hypotenuse 5. we can find the missing side by Pythagoras' theorem. now we can use the trig ratios to find the value of the other trig functions.

    A) tan θ
    \tan \theta = \frac {\mbox {Opposite}}{\mbox {Adjacent}}

    remember, the answer should be negative


    B) Sin θ
    \sin \theta = \frac {\mbox {Opposite}}{\mbox {Hypotenuse}}

    remember, the answer should be positive
    C) sin (θ + 2π/3)
    use the addition formula for sine: \sin (A + B) = \sin A \cos B + \sin B \cos A

    D) cos 2 θ
    use the double angle formula for cosine: \cos 2 \theta = \cos^2 \theta - \sin^2 \theta



    EDIT: or you could use formulas. for instance, recall that: \sin^2 \theta + \cos^2 \theta = 1. you can use that to solve for sine if given cosine.
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    Last edited by Jhevon; September 29th 2007 at 09:44 PM.
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