solve the equation 5sin(θ -pi/6)=8cosθ for values of 0<=θ <=2pi
5sin(θ-π/6) = 8cosθ
5(sinθcosπ/6-cosθsinπ/6) = 8cosθ
5√3/2 sinθ - 5/2 cosθ = 8cosθ
5√3 sinθ = 21 cosθ
75 sin^2 θ = 441 cos^2θ
516 sin^2 θ = 441
sinθ = ±21/√516 = ±0.924
θ = 1.18 or 1.96 or 4.32 or 5.10
that is,
θ = .37π or .62π or 1.37π or 1.62π
so,
θ-π/6 = .21π or .46π or 1.21π or 1.46π
Now, we squared things to solve for θ, so we need to check for extraneous solutions. Note that the 2nd and 4th solutions are in QII and QIV, where sin and cos have opposite signs. So, only
θ = 1.18 or 4.32
To verify, see
http://www.wolframalpha.com/input/?i=solve+5sin%28%CE%B8-%CF%80%2F6%29+%3D+8cos%CE%B8+for+%CE%B8+%3D+0..2pi
or
from Steve's
5√3 sinθ = 21 cosθ
sinØ/cosØ = 21/5√3
tanØ = 21/5√3 , and the tangent is + in I and III
Ø = 1.18 in I
or
Ø = π + 1.18 = 4.32 in III
To solve the equation 5sin(θ - π/6) = 8cosθ for values of 0 ≤ θ ≤ 2π, we will use trigonometric identities and properties to simplify and solve the equation.
Step 1: Rewrite the equation using trigonometric identities.
5sin(θ - π/6) = 8cosθ
Step 2: Expand the equation using the difference of angles formula.
5[sinθcos(π/6) - cosθsin(π/6)] = 8cosθ
Step 3: Simplify the equation.
5[(sinθ * √3/2) - (cosθ * 1/2)] = 8cosθ
Step 4: Distribute and rearrange the terms.
5(sinθ * √3/2) - 5(cosθ * 1/2) = 8cosθ
Step 5: Simplify the equation further.
(5sinθ * √3/2) - (5cosθ * 1/2) = 8cosθ
Step 6: Multiply through by 2 to get rid of the denominators.
(10sinθ * √3) - (10cosθ) = 16cosθ
Step 7: Move all the terms with cosθ to one side.
10sinθ√3 - 16cosθ + 10cosθ = 0
Step 8: Combine like terms.
10sinθ√3 - 6cosθ = 0
Step 9: Divide through by 2 to simplify.
5sinθ√3 - 3cosθ = 0
We have simplified the equation to 5sinθ√3 - 3cosθ = 0.
To continue solving the equation, we can now divide both sides of the equation by sqrt( 3 ) to get rid of the coefficient in front of sinθ.
(5sinθ√3 - 3cosθ) / sqrt( 3 ) = 0 / sqrt( 3 )
5sinθ - (3cosθ / sqrt( 3 ) ) = 0
Now, we can rewrite - (3cosθ / sqrt( 3 ) ) as - ( cosθ * sqrt( 3 ) / sqrt ( 3 ) ) to rationalize the denominator.
5sinθ - ( cosθ * sqrt( 3 ) / sqrt ( 3 ) ) = 0
5sinθ - cosθ * sqrt( 3 ) / sqrt( 3 ) = 0
Next, let's combine the terms with sinθ and cosθ.
sinθ - (sqrt( 3 ) / 5) cosθ = 0
Now we can use the identity sin(A - B) = sinA * cosB - cosA * sinB, to rewrite the equation.
sinθ - (sqrt( 3 ) / 5 ) cosθ = 0
sinθ - cos(π/3) cosθ = 0
sinθ - ( 1/2 ) cosθ = 0
Lastly, we can use the identity sin^2θ + cos^2θ = 1, to solve for values of θ.
sinθ - ( 1/2 ) cosθ = 0
sinθ = ( 1/2 ) cosθ
Now, we know that sinθ = ( 1/2 ) and cosθ = ( 1/2 ). This can only occur in two quadrants: the first and second quadrants.
In the first quadrant, sinθ > 0 and cosθ > 0.
Therefore, we have sinθ = ( 1/2 ) and cosθ = ( 1/2 ).
In the second quadrant, sinθ > 0 and cosθ < 0.
Therefore, we have sinθ = ( 1/2 ) and cosθ = -( 1/2 ).
Using these values of sinθ and cosθ, we can find the values of θ.
In the first quadrant:
θ = π/6
and
θ = π/2 - π/6 = π/3
In the second quadrant:
θ = π - π/6 = 5π/6
and
θ = π - π/3 = 2π/3
Therefore, the values of θ that satisfy the equation 5sin(θ - π/6) = 8cosθ for the given range 0 ≤ θ ≤ 2π are:
θ = π/6, π/3, 5π/6, and 2π/3.