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Math Help - Addition Formulae (iii) trigonometry identity

  1. #1
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    Addition Formulae (iii) trigonometry identity

    OK, after help from here, I tried 4 more questions and thought i got the hang of this. Now i have another dilemma. :S

    By expressing: tan(alpha + beta) = sin(alpha + beta) / cos(alpha + beta)

    Prove that: tan(alpha + beta) = tan alpha + tan beta / 1 - tan alpha tan beta


    Here are my first steps:
    = sin(alpha + beta) / cos(alpha + beta)
    = sin alpha cos beta + cos alpha sin beta / cos alpha cos beta - sin alpha sin beta

    now im stuck,
    should i cancel out the cos and sin from top and bottom?
    i tried that, but that didn't work. after i thought, since i was lookin for tan in my answer, i would sin / cos = tan
    that also did not work out.... (or maybe i did something wrong)
    please guide me.
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  2. #2
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    You have,
    \tan (x+y)=\frac{\sin(x+y)}{\cos(x+y)}
    \frac{\sin x\cos y+\cos x\sin y}{\cos x\cos y-\sin x\sin y}
    Divide through when non-zero by \cos x\cos y the numerator and denominator,
    \frac{\tan x+\tan y}{1-\tan x\tan y}
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  3. #3
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    Divide through when non-zero by \cos x\cos y the numerator and denominator,

    ^

    youv lost me there, can u explain clearer,
    what do you mean by "Divide through when non-zero"
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  4. #4
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    Quote Originally Posted by r_maths View Post
    youv lost me there, can u explain clearer,
    \frac{\sin x\cos y+\cos x\sin y}{\cos x\cos y-\sin x\sin y}
    Divide through,
    \frac{\frac{\sin x \cos y}{\cos x\cos y}+\frac{\cos x\sin y}{\cos x\cos y}}{\frac{\cos x\cos y}{\cos x\cos y}-\frac{\sin x\sin y}{\cos x\cos y}}
    Thus, you are going to get some canceling.
    And you also need to use the fact that,
    \frac{\sin x}{\cos x}=\tan x
    what do you mean by "Divide through when non-zero"
    I am not saying if the cosines are not zero you can divide through, thus the formula for tangent addition will hold true.
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  5. #5
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    thanks,
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