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Trigonometry

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Find the exact value of tan(a-b)

sin a = 4/5, -3pi/2<a<-pi;
tan b = -sqrt2, pi/2<b<pi

identity used is:

tan(a-b)=(tan a-tan b)/1+tan a tan b

simplify answer using radicals.

(a is alpha, b is beta)

  • Trigonometry - ,

    Both the sine and cosine curves are the same for
    -3pi/2<a<-pi as they are for π/2<a<π (2nd quad)
    so if sina = 4/5, then cosa = -3/5
    and tana = -4/3

    then tan(a-b)
    = (tana - tanb)/(1 + tanatanb)
    = ((-4/3) - (-√2))/( 1 + (-4/3)(-√2))
    = (√2 - 4/3)/(1 + 4√2/3)
    = (3√2 - 4)/(3 + 4√2)

    I don't know if you have to rationalize that, if you do carefully multiply top and bottom by (3 - 4√2)

  • Trigonometry - ,

    I put that in but it said the answer was
    36-25ã2/23
    any idea how that works?

  • Trigonometry - ,

    That is exactly what my answer works out to if you rationalize it.
    I had suggested to do that.

  • Trigonometry - ,

    8 cot(A) − 8/
    1 + tan(−A)

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