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Math Help - Conditional Trigonometry Proof?

  1. #1
    Super Member fardeen_gen's Avatar
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    Conditional Trigonometry Proof?

    If α + β + γ = π/2, show that,

    [(1 - tan α/2)(1 - tan β/2)(1 - tan γ/2)]/[(1 + tan α/2)(1 + tan α/2)(1 + tan α/2)]
    = (sin α + sin β + sin γ - 1)/(cos α + cos β + cos γ)

    I tried my level best for the past four hours. Couldnt do it. Can anybody help?
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by fardeen_gen View Post
    If α + β + γ = π/2, show that,

    [(1 - tan α/2)(1 - tan β/2)(1 - tan γ/2)]/[(1 + tan α/2)(1 + tan β/2)(1 + tan γ/2)]
    = (sin α + sin β + sin γ - 1)/(cos α + cos β + cos γ)

    I tried my level best for the past four hours. Couldnt do it. Can anybody help?
    Maybe this can help, but that's as far as I can go... :

    A=\frac{1-\tan \tfrac \alpha 2}{1+\tan \tfrac \alpha 2}

    Multiply by \frac{\cos \tfrac \alpha 2}{\cos \tfrac \alpha 2} :

    A=\frac{\cos \tfrac \alpha 2-\sin \tfrac \alpha 2}{\cos \tfrac \alpha 2+\sin \tfrac \alpha 2}

    Multiply by \frac{\cos \tfrac \alpha 2-\sin \tfrac \alpha 2}{\cos \tfrac \alpha 2-\sin \tfrac \alpha 2} :

    A=\frac{(\cos \tfrac \alpha 2-\sin \tfrac \alpha 2)^2}{(\cos \tfrac \alpha 2+\sin \tfrac \alpha 2)(\cos \tfrac \alpha 2-\sin \tfrac \alpha 2)}=\frac{\cos^2 \tfrac \alpha 2+\sin^2 \tfrac \alpha 2-2 \cos \tfrac \alpha 2\sin \tfrac \alpha 2}{\cos^2 \tfrac \alpha 2-\sin^2 \tfrac \alpha 2}

    ---------------------------
    Recall the following identities :
    \cos^2(x)+\sin^2(x)=1

    \sin(2x)=2\sin(x)\cos(x)

    \cos(2x)=\cos^2(x)-\sin^2(x)
    ----------------------------

    A=\frac{1-\sin \alpha}{\cos \alpha}

    This goes the same for B=\frac{1-\tan \tfrac \beta 2}{1+\tan \tfrac \beta 2}=\frac{1-\sin \beta}{\cos \beta} and C=\frac{1-\tan \tfrac \gamma 2}{1+\tan \tfrac \gamma 2}=\frac{1-\sin \gamma}{\cos \gamma}
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  3. #3
    Super Member fardeen_gen's Avatar
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    I think I managed to do it. Though it is big and clumsy. Here it is:

    We should start with the RHS as it is more complex

    Before we proceed let us use/deduce certain information as they shall be used. A B are used for generic ones and α β γ for these specified in problems)
    (actually they should be derived if required as required but I am deriving to keep the flow)
    α + β + γ = π/2
    so α + β = π/2 γ
    or α + β = π/2 γ 1

    cos a + cos b = 2 cos (a+b)/2 cos(a-b)/2 2

    taking sin of both sides of 1 we get
    sin (α + β) = sin (π/2 γ) = cos γ 3
    sin 2a = 2 sin a cos a . 4

    from 1
    (α + β)/2 = (π/2 γ)/2
    So sin (α + β)/2 = sin (π/2 γ)/2 = cos (π/2 + γ)/2 5

    Again as
    α + β + γ = π/2
    so α - β + γ = π/2 - 2 β
    so α - β + γ + π/2 = π - 2 β
    so (α - β + γ + π/2)/ 4 = π/4 - β/2 6

    Again as
    α + β + γ = π/2
    so α - β - γ = π/2 - 2 β 2 γ
    so α - β - γ - π/2 = - 2 β 2 γ
    so (α - β - γ - π/2)/ 4 = - (β + γ)/2
    so cos (α - β - γ - π/2)/ 4 = cos (β + γ)/2 as cos - A= cos A
    = cos (π/2 α)/2
    = cos (π/4 α/2) .. 7
    And sin (α - β - γ - π/2)/ 4 = - sin (β + γ)/2 = - sin (π/4 α/2)
    Or sin (-α + β +γ + π/2)/ 4 = sin (π/4 α/2) 8


    Now let us find he denominator
    cos α + cos β + cos γ
    = 2 cos (α + β)/2 cos(α - β )/2 + cos γ (using 2)
    = 2 cos (α + β)/2 cos(α - β )/2 + sin (α + β) (using 3)
    = 2 cos (α + β)/2 cos(α - β )/2 + 2 cos (α + β)/2 sin (α + β)/2 ( using 4)
    = 2 cos (α + β)/2 (cos(α - β )/2 + sin (α + β)/2)
    = 2 cos (π/2 - γ)/2 (cos(α - β )/2 + cos (π/2 + γ)/2)) (using 1 and 5)
    = 2 cos (π/4 γ/2)*2 cos (α - β + π/2 + γ)/2 cos (α - β - π/2 - γ)/2 (using 2)
    = 4 cos (π/4 γ/2 ) cos (π/4 - β/2) cos (α - β - π/2 - γ)/2 (using 6)
    = 4 cos (π/4 γ/2 ) cos (π/4 - β/2) cos (π/4 α/2) using 7

    So cos α + cos β + cos γ = 4 cos (π/4 γ/2 ) cos (π/4 - β/2) cos (π/4 α/2) (A)
    Now numerator

    For the additional identities

    Sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2 9
    sin (α + β + γ) = sin π/2 = 1 . 10

    sin A sin B = 2 sin (A-B)/2 cos (A+B)/2 11


    cos A cos B =2 sin (A+B)/2 sin (B-A)/2 12

    tan (π/4- A)= (1- tan π/4 tan A) / (tan π/4 + tan A) = ( 1- Tan A)/(1+tan A) 13

    based on above numerator

    sin α + sin β + sin γ 1
    = 2 sin (α + β)/2 cos(α - β )/2 + sin γ sin (α + β + γ) (using 9 and 10)
    = 2 sin (π/4 γ/2) cos(α - β )/2 (sin (α + β + γ) - sin γ) (using 5)
    = 2 sin (π/4 γ/2) cos(α - β )/2 - 2 sin (α + β)/2 cos (α + β + γ) + γ)/2 (using 11)
    = 2 sin (π/4 γ/2) cos(α - β )/2 - 2 sin (π/4 γ/2) cos (π/2+ γ)/2 (using 5)
    = 2 sin (π/4 γ/2) (cos(α - β )/2 - cos (π/2+ γ)/2)
    = 2 sin (π/4 γ/2) (2 sin (α - β + π/2+ γ)/4 sin (-α + β +π/2-+γ)/4 (using 12)
    = 4 sin (π/4 γ/2) sin (π/4 β/2) sin (π/4- α/2) (using 6 and 8)

    sin α + sin β + sin γ 1 = 4 sin (π/4 γ/2) sin (π/4 β/2) sin (π/4- α/2) (B)

    from A and B we get
    RHS =
    (sin α + sin β + sin γ 1)/ (cos α + cos β + cos γ)
    = (4 sin (π/4 γ/2) sin (π/4 β/2) sin (π/4- α/2))/ (4 cos (π/4 γ/2 ) cos (π/4 - β/2) cos (π/4 α/2))
    = tan (π/4 γ/2) tan (π/4 β/2) tan (π/2- α/2)
    = tan (π/4- α/2) tan (π/4 β/2) tan (π/4 γ/2)
    = [(1 - tan α/2)(1 - tan β/2)(1 - tan γ/2)]/[(1 + tan α/2)(1 + tan β/2)(1 + tan γ/2)](using 13 and rearranging the terms)

    = LHS
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