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Math Help - expressing sin(x) in terms of tan(x)

  1. #1
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    expressing sin(x) in terms of tan(x)

    I saw this in a list of previous examination questions today and became intrigued. It asks:

    "The acute angle x radians is such that tan(x) = k. Express, in terms of k, sin(x)."

    The obvious answer is k cos(x). But as a 2 mark question I get the feeling that something else is required (and I'm not sure if cos(x) counts as 'in terms of x').

    I have feeling there's something deceptively obvious here than I'm missing. Can anyone give me hints, or am I overthinking this?
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  2. #2
    Member Traveller's Avatar
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    Hints :

    sin^2(x) + cos^2(x) = 1

    tan^2(x) + 1 = sec^2(x)
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  3. #3
    MHF Contributor chisigma's Avatar
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    Are well known the identities...

    \displaystyle \sin x = \frac{2\ \tan \frac{x}{2}} {1+\tan^{2} \frac{x}{2}} (1)

    \displaystyle \cos x = \frac{1- \tan^{2} \frac{x}{2}} {1+\tan^{2} \frac{x}{2}} (2)

    If You can use \cos x and \sin x as function of \tan \frac{x}{2} instead of \tan x very well!... otherwise You can devide (1) and (2) and obtain...

    \displaystyle \tan x = \frac {2 \tan \frac{x}{2}}{1- \tan^{2} \frac{x}{2}} (3)

    ... the use (3) to find \tan \frac{x}{2} as function of \tan x and then substitute in (1) and (2)... a very tedius job!...

    Kind regards

    \chi \sigma
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  4. #4
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    Hello, EvilKitty!

    \text{The acute angle }x\text{ radians is such that: }\:\tan x  \,=\, k

    \text{Express }\sin x\text{ in terms of }k,

    Your answer, k\cos x, is in terms of \,k and \,x.


    We are given: . \tan x \:=\:\dfrac{k}{1} \:=\:\dfrac{opp}{adj}


    Angle \,x is in a right triangle with: . opp = k,\; adj = 1

    . . Pythagorus says: . hyp = \sqrt{k^2+1}


    Therefore: . \sin x \;=\;\dfrac{opp}{hyp} \;=\;\dfrac{k}{\sqrt{k^2+1}}
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  5. #5
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    Thank you very much, all!

    Although I managed to work it out with identities, Soroban's answer may have been the 'intended' one, as neither double angle formulas or sec(x), cosec(x) etc. are part of my syllabus.
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