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Math Help - Solving of Trigonometry equation

  1. #1
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    Solving of Trigonometry equation

    Hi Guys !

    Am Currently stuck with this question :

    If sin(x+y) = 1 and if tanx = 9tany , find the value of tanx.

    Need help guys thanks !
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  2. #2
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    have you played with the fact that \sin (x+y) = \sin x \cos y + \sin y \cos x ?
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  3. #3
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    What if you write \tan{x}-9\tan{y}=0

    Does this help?

    Also, think about what the range of tan(x)?
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  4. #4
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    Hello, liukawa!

    \text{If }\sin(x+y) = 1\,\text{ and }\,\tan x \,=\, 9\tan y,\,\text{ find the value of }\tan x.

    I tried all kinds of gymnastics and got nowhere.
    Then my two brain cells got together . . .


    Since \sin(x+y) \,=\,1, then: . x + y \,=\,\frac{\pi}{2}
    . . That is: . y \:=\:\frac{\pi}{2} - x

    Then we have:
    . . \tan x \:=\:9\tan y \quad\Rightarrow\quad \tan x \:=\:9\tan(\frac{\pi}{2} - x) \quad\Rightarrow\quad \tan x \:=\:9\cot x
    . . . .
    I hope you followed that.


    Multiply both sides by \tan x\!:\;\;\tan^2\!x \:=\:9

    Therefore: . \tan x \:=\:\pm3

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  5. #5
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    Quote Originally Posted by liukawa View Post
    Hi Guys !

    Am Currently stuck with this question :

    If sin(x+y) = 1 and if tanx = 9tany , find the value of tanx.

    Need help guys thanks !
    Alternatively

    sin(x+y)=1\Rightarrow\ sin^2(x+y)+cos^2(x+y)=1\Rightarrow\ cos(x+y)=0

    \displaystyle\ tan(x+y)=\frac{sin(x+y)}{cos(x+y)}

    \displaystyle\ tan(x+y)=\frac{tanx+tany}{1-tanxtany}\Rightarrow\ tanxtany=1\Rightarrow\ tanx=\frac{1}{tany}

    \displaystyle\ 9tany=\frac{1}{tany}\Rightarrow\ 9tan^2y=1\Rightarrow\ tan^2y=\frac{1}{9}\Rightarrow\ tany=\pm\frac{1}{3}

    \Rightarrow\ tanx=9tany=\pm3
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