trig

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tanx = 5/12 and sinx<0

Find sin2x and cos2x

How to get the answers of sin2x = 120/169 and cos 2x = 119/169?

  • trig -

    tanx = 5/12

    r^2 = x^2 + y^2
    = 12^2 + 5^2
    r = 13

    tanx >0 and sinx<0, so x must be in quad. III and
    and sinx = -5/13 , cosx = -12/13

    sin 2x = 2sinxcosx = 2(-5/13)(-12/13)
    = 120/169
    cos 2x = cos^2 x - sin^2 x
    = 144/169-25/169
    =119/169

    do the others the same way, or the way
    drwls showed you in the other post.

  • trig -

    tan x = 13 cot x

  • trig -

    tan x = 13 cot x
    tan x = 13/tan x
    tan^2 x = 13
    tan x = ±√13
    x = 74.5º,105.5º, 254.5º, 285.5º

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