Answer :
Answer:
Step-by-step explanation:
A. :
Recall that sin 2Θ = 2sinΘcosΘ
So,
2sinΘcosΘ = sinΘ
cosΘ = [tex]\frac{sin(O)}{2sin(O)}[/tex] = 1/2
Recalling special right triangles or simply memorizing trigonometric values, cos (π/3) = 1/2 or Θ = π/3 or 5π/3
B. :
Recall that cos2Θ = 1 - 2sin^2(Θ) = 2cos^2(Θ) - 1
For our cases let's rewrite cos2Θ in the same terms as sinΘ.
1 - 2sin^2(Θ) + sinΘ = 0
-1 + 2sin^2(Θ) - sinΘ = 0
(multiplied both sides by -1 to make sin^2 positive)
2sin^2(Θ) - sin(Θ) - 1 = 0 (rearranged)
Hey, this looks like a quadratic equation!
We can make the factoring easier by assigning sinΘ = x,
2x^2 - x - 1 = 0 (now that's better!)
(2x+1)(x-1) = 0
x = -1/2 or 1
sinΘ = -1/2 or sinΘ = 1
So, Θ = 11π/6 or π/2
C. :
Plugging the formula of sin2Θ into the equation given, we get,
2sinΘcosΘ = √3cosΘ
2sinΘ = [tex]\frac{\sqrt{3}cosO)}{cos(O)}[/tex]
2sinΘ = √3
sinΘ = (√3)/2
So, Θ = π/3 or 2π/3
D. :
Again, plugging the formula of sin2Θ into the equation given, we get,
cosΘ = sinΘ(2sinΘcosΘ)
cosΘ = 2sin^2(Θ)cosΘ (sinΘ*sinΘ is like x*x = x^2)
cosΘ/cosΘ = 2sin^2(Θ) (divide both sides by cosΘ)
1 = 2sin^2(Θ)
1/2 = sin^2(Θ) (divide both sides by 2)
√(1/2) = sinΘ (square root both sides to get rid of the ^2 on the right side)
√2/2 = sinΘ
So, Θ = π/4 or 3π/4
E. :
Plugging the formula of cos2Θ into the given equation, we get,
2cos^2(Θ) - 1 = cosΘ
2cos^2(Θ) - cosΘ - 1 = 0 (subtract cosΘ both sides)
cosΘ = x
2x^2 - x - 1 = 0 (same quadratic as B)
cosΘ = -1/2 or cosΘ = 1
So, Θ = 2π/3 or Θ = 0/2π
(we're dealing with the cosine function so, the values Θ will be different from B's)
Let me know if you have any questions!
