Posted by c uppadhaya on Thursday, August 11, 2011 at 5:45am.
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)
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
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
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?
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 !
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
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.
Didn't see your previous post
Q.E.D.
I bow to you !!!
It was a freak coincidence that I saw it. I wouldn't have if you didn't point out my oversight. Thanks to you!!