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January 31, 2015

January 31, 2015

Posted by **c uppadhaya** on Thursday, August 11, 2011 at 5:45am.

- Looking for a better way - math -
**Reiny**, Thursday, August 11, 2011 at 8:33am8 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 -
**MathMate**, Thursday, August 11, 2011 at 9:43am6cosθ+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 -
**Reiny**, Thursday, August 11, 2011 at 4:13pmif 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 -
**MathMate**, Thursday, August 11, 2011 at 5:07pmYeah, 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 -
**MathMate**, Thursday, August 11, 2011 at 5:17pmI 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! -
**MathMate**, Thursday, August 11, 2011 at 5:39pmThe 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 -
**Reiny**, Thursday, August 11, 2011 at 5:46pmHow 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 -
**Reiny**, Thursday, August 11, 2011 at 5:48pmDidn't see your previous post

Q.E.D.

I bow to you !!!

- math -
**MathMate**, Thursday, August 11, 2011 at 6:13pmIt 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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