Question

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

Answer 1

NEED EASIER WAYS...
AS SHORT AS POSSIBLE...

my way:
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
since alpha and beta are roots of equation so these value of theta are alpha and beta or beta and alpha, whatever you say:

say,
cos (alpha) =0.88872 ; and
cos (beta) = 0.19128
;this implies
sin (alpha) =0.45845 ; and
sin (beta) =0.9815

sin(alpha + beta) = sin(alpha)cos(beta) + cos(alpha)sin(beta)

sin(alpha + beta) = [(0.45845)*(0.19128)]+[(0.88872)*(0.9815)]

sin(alpha + beta) = 0.959970996

difference in answers (accuracy)
0.96-0.959970996 = 2.9004 × 10^(-5)
ie. highly accurate answer .. would be more accurate if i used decimals up to greater places

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my teacher's way..

convert it to a quadratic equation of tanθ/2..
6 cosθ + 8 sinθ = 9
6 cosθ = 9-8 sinθ
9 tan^2 (θ/2) + 6 tan^2 (θ/2) -16 tan(θ/2) + 3 = 0
15 tan^2 (θ/2) -16 tan(θ/2) + 3 = 0
tan (α/2) & tan (β/2) are its roots
tan (α/2) + tan (β/2) = 16/15
tan (α/2) * tan (β/2) = 1/5
tan ((α+β)/2) = (16/15)/1- (1/5) = 4/3
sin (α +β) = [2 tan ((α+β)/2)] / [1+ tan^2 ((α+β)/2)]
=24/25 =0.96

DAMN!! THIS GUY NAILED IT!
(still need shorter way, if any)


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solutions on net:
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)


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another solution on net:

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

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0.6 cosθ + 0.8 sinθ = 0.9
Let sinx = 0.6; then
cosx = 0.8 and x = 36.87 degrees
sinx cosθ + cosx sinθ = 0.9
sin (x + θ) = 0.9
x + θ = 64.16 degrees or 115.84 degrees
alpha = θ1 = 27.29 degrees
beta = θ2 = 78.97

alpha + beta = 106.26 degrees

sin (alpha + beta) = 0.9600

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Found another way of doing this on internet:

0.6 cosθ + 0.8 sinθ = 0.9
Let sinx = 0.6; then
cosx = 0.8 and x = 36.87 degrees
sinx cosθ + cosx sinθ = 0.9
sin (x + θ) = 0.9
x + θ = 64.16 degrees or 115.84 degrees
alpha = θ1 = 27.29 degrees
beta = θ2 = 78.97

alpha + beta = 106.26 degrees

sin (alpha + beta) = 0.9600

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Found another way of doing this on internet:

0.6 cosθ + 0.8 sinθ = 0.9
Let sinx = 0.6; then
cosx = 0.8 and x = 36.87 degrees
sinx cosθ + cosx sinθ = 0.9
sin (x + θ) = 0.9
x + θ = 64.16 degrees or 115.84 degrees
alpha = θ1 = 27.29 degrees
beta = θ2 = 78.97
the use calc. to find
sin (alpha + beta)