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March 30, 2017

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if alpha and beta are 2 different values of θ lying between 0 and 2π which satisfy the equation 6cosθ+8 sinθ=9 find the value of sin alpha + beta ...

  • Looking for a better way - math - ,

    8 sinØ = 9 - 6cosØ
    square both sides
    64 sin^2 Ø = 81 - 108 cosØ + 36 cos^2 Ø
    64(1 - cos^2 Ø) = 81 - 108 cosØ + 36 cos^2 ‚
    100 cos^2 Ø - 108cosØ + 17 = 0
    cosØ = .8887119 or .191288084 by the quad formula

    Ø = 27.288° or 332.712° or 78.972° or 281.023°
    since we squared the equation, all answers must be verified. I used my calculator and only
    27.288° and 78.972° satisfy the original equation.

    so let alpha be 27.288° and beta be 78.972°

    sin(alpha + beta) = sin(alpha)cos(beta) + cos(alpha)sin(beta)
    = sin(27.288+78.972) or sin(106.26°)
    = sin27.288cos78.972 + cos27.288sin78.972

    from above,
    if cos27.288° = .8887119 then sin27.288° = .45846606 by Pythagoras
    if cos 78.972° = .191288084 , then sin 78.972° = .9815339

    then sin27.288cos78.972 + cos27.288sin78.972
    = .96

    I get the same when I simply take
    sin(106.26° = .96

    (I retained all decimals that my calculator could hold, and used the calculators memory location to store all numbers,
    my last answer was .9599999999 and I will assume the 9 was repeating.
    It appears the answer might be exactly .96, which also suggests that there must be a better way than the above solution)

  • math - ,

    6cosθ+8 sinθ=9
    Divide by 10=√(6^2+8^2):
    (6/10)cosθ + (8/10)sinθ=9/10
    so if
    α=θ and
    β=asin(6/10), then
    sin(β)=6/10,
    cos(β)=8/10
    and
    sin(α+β)
    =(6/10)cosθ + (8/10)sinθ
    =0.9

  • However .... math - ,

    if sinβ = 6/10 then β = appr. 36.87° and if
    sin (α+β)) = .9
    α+β = 64.158° which makes α = 27.289° which was one of my angles.

    but both α and β were to be solutions in the original equation.
    27.289 works but 36.87 does not, the right side = 9.6

  • math - ,

    Yeah, you have a good point. I didn't show (nor check) that both α and β satisfy the original equation, and the answer is therefore... incorrect.

    All I had done was to show that φ=asin(6/10)=36.87° and
    sin(α+φ)=0.9,

    I should have continue this way:

    α+φ=64.158° or 115.842°.

    So α=asin(0.9)-asin(0.6)
    =64.158...-36.869...
    =27.288... as you had it.

    Taking the other value of α+φ and subtract φ
    β
    =115.842...-36.869...
    =78.972...
    So now sin(α+β)
    =sin(106.2602047...)
    =0.96 also as you had it.

    And I just noticed something amusing:
    sin(180-2φ)
    =sin(2φ)
    =0.96
    or
    sin(α+β)=sin(2φ)

    Is it just a coincidence or there's something behind it?

  • math - ,

    I just looked at it again, it is not a coincidence!

    Since sin(α+φ)=0.9
    and we got
    α=asin(0.9)-φ
    β=180-asin(0.9)-φ
    So by adding the two
    α+β = 180-2φ
    or
    sin(α+β)
    =sin(180-2φ)
    =sin(2φ)
    =sin(2 asin(0.6))
    =0.96 !

  • math - just got better! - ,

    The following shows that value of sin(α+β) in
    where α & β are solutions to
    6cosθ+8 sinθ=9 ...(1)
    does not depend on the right hand side, with the obvious stipulation that RHS ≤ √(6^2+8^2)

    Take the general case where
    Acosθ+Bsinθ = C
    where A^2+B^2 ≥ C^2

    We will divide both sides by √(A^2+B^2) to give
    sinφcosθ+cosφsinθ = sin(K), and where
    sinφ=A/√(A^2+B^2)
    cosφ=B/√(A^2+B^2)
    sin(K)=C/√(A^2+B^2) (K≤1)

    We therefore have
    sin(φ+θ)=sin(K)

    Substituting α & β for θ,
    sin(φ+α)=sin(K), and
    sin(φ+β)=sin(180-K)

    which means
    φ+α=K
    φ+β=180-K
    Adding the two equations
    2&phi+α+β=180
    α+β=180-2φ
    sin(α+β)
    =sin(180-2φ)
    =sin(2&phi)
    =sin(2*sin-1(A/√(A^2+B^2)))
    (independent of C!)
    =sin(2*sin-1(6/10)))
    =(6/10*8/10+8/10*6/10)
    =0.96 exactly

  • math - ,

    How did you ever notice that relationship ??

    The fact that the angles are certainly irrational, a coincidence is so highly unlikely.
    I cannot see any relation at this point.

    This looked like such a little innocent problem.
    I am not happy with my solution, since I relied on my calculator. The fact that the answer came out rather exact suggests that there has to be a more elegant solution.

  • math - ,

    Didn't see your previous post

    Q.E.D.

    I bow to you !!!

  • math - ,

    It was a freak coincidence that I saw it. I wouldn't have if you didn't point out my oversight. Thanks to you!!

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