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Math Help - Anyone know how to solve this trig problem?

  1. #1
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    Anyone know how to solve this trig problem?

    tough question, thanks for your help

    tan[sin-1(4/5)+cos-1 (1)]

    can someone show me how to do it, I know how to using a calculator. thanks
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by mcdaking84 View Post
    tough question, thanks for your help

    tan[sin-1(4/5)+cos-1 (1)]

    can someone show me how to do it, I know how to using a calculator. thanks
    Here's one way to do it.

    See the diagram below.

    this problem is a lot simpler than it seems. at first it seems like you will need the double angle formula, but in fact you don't. remember that cos^-1(1) = 0

    so we want tan(sin^-1(4/5))
    let sin^-1(4/5) = x
    so we want tanx

    now if sin^-1(4/5) = x
    => sinx = 4/5 .............we can find tanx using several identities, or just draw the triangle, see below.

    we can find the that the missing side of the triangle is 3 using pythagoras' theorem, or we can simply recognize that this is the famous 3-4-5 triangle.

    so tanx is simply opposite/adjacent

    => tan[sin^-1(4/5)+cos^-1 (1)] = tan(sin^-1(4/5)) = tanx = 4/3
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  3. #3
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    Hello, mcdaking84!

    tan[arcsin(4/5) + arccos(1)]

    Let .A = arcsin(4/5) . . . then: .sin(A) = 4/5

    So A is an acute angle in a right triangle with opp = 4, hyp = 5.
    . . Using Pythagorus, we find that: adj = 3.
    Hence: .tan(A) = 4/3


    Let B = arccos(1) . . . then B = 0
    Hence: .tan(B) = 0

    . . . . . . . . . . . . . . . . . . . . . . . . . . .tan(A) + tan(B)
    We use the identity: .tan(A + B) .= .---------------------
    . . . . . . . . . . . . . . . . . . . . . . . . . .1 + tan(A)Ětan(B)

    . . . . . . . . . . . . . . . . . . . . . . . 4/3 + 0. . . . . 4
    Then we have: .tan(A + B) .= .-------------- .= .---
    . . . . . . . . . . . . . . . . . . . . . .1 + (4/3)(0) - - .3

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