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.i want the answer by quadratic method. can u help solving sum by that method
Oh well, if you do not like WLS' cool method then
cos T = sqrt(1 - sin^2 T)
6 sqrt(1 -sin^2 T) + 8 sin T = 9
let sin T = x
6 sqrt(1-x^2) + 8 x = 9
6 sqrt (1-x^2) = 9-8x
36 (1-x^2) = 81 - 144x + 64 x^2
36 -36 x^2 = 81 - 144x + 64 x^2
100 x^2 -144 x + 45 = 0
x = [ 144 +/- sqrt (20736 - 18000)]/200
x = [ 144 +/- 52.3 ]/200
x = .981 or .4585
but x is sin T so
T = 78.8 or T = 27.3
sum = 106.1
sin(106.1) = .961
Sure! To solve the equation 6cosθ + 8sinθ = 9 using the quadratic method, we need to rewrite it in the form of a quadratic equation.
Start by squaring both sides of the equation:
(6cosθ + 8sinθ)^2 = 9^2
Expanding the squared terms, we get:
36cos^2θ + 96cosθsinθ + 64sin^2θ = 81
Using the Pythagorean identity cos^2θ + sin^2θ = 1, we can make a substitution to simplify the equation:
36(1 - sin^2θ) + 96cosθsinθ + 64sin^2θ = 81
Rearranging the terms, we have:
100sin^2θ + 96cosθsinθ - 36 = 0
Now, we have a quadratic equation in terms of sinθ. In this case, let's use the substitution method to solve it.
Let's substitute sinθ with a variable, x:
100x^2 + 96cosθx - 36 = 0
Now, we can solve this quadratic equation for x:
Using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 100, b = 96cosθ, and c = -36.
Substituting these values into the quadratic formula, we get:
x = (-96cosθ ± √((96cosθ)^2 - 4(100)(-36))) / (2(100))
Simplifying further:
x = (-96cosθ ± √(9216cos^2θ + 14400)) / 200
Now, we have the values of x, which represent sinθ. To solve for the values of θ, we need to find the arcsine of x:
θ = arcsin(x), where x is the values we calculated.
Once you have the values of θ, you can substitute them in the expression sinα + sinβ to find the sum of their corresponding sine values.