Simplify f ( θ ) f(\theta) f ( θ )
Given
f ( θ ) = 3 ( sin 4 ( 3 π 2 − θ ) + sin 4 ( 3 π + θ ) ) − 2 ( 1 − sin 2 2 θ ) . f(\theta)=3\left(\sin^4\left(\frac{3\pi}{2}-\theta\right)+\sin^4(3\pi+\theta)\right)-2\left(1-\sin^2 2\theta\right). f ( θ ) = 3 ( sin 4 ( 2 3 π − θ ) + sin 4 ( 3 π + θ ) ) − 2 ( 1 − sin 2 2 θ ) . Use standard identities:
sin ( 3 π 2 − θ ) = − cos θ ⟹ sin 4 ( 3 π 2 − θ ) = cos 4 θ \sin\left(\frac{3\pi}{2}-\theta\right)=-\cos\theta \implies \sin^4\left(\frac{3\pi}{2}-\theta\right)=\cos^4\theta sin ( 2 3 π − θ ) = − cos θ ⟹ sin 4 ( 2 3 π − θ ) = cos 4 θ sin ( 3 π + θ ) = − sin θ ⟹ sin 4 ( 3 π + θ ) = sin 4 θ \sin(3\pi+\theta)=-\sin\theta \implies \sin^4(3\pi+\theta)=\sin^4\theta sin ( 3 π + θ ) = − sin θ ⟹ sin 4 ( 3 π + θ ) = sin 4 θ 1 − sin 2 2 θ = cos 2 2 θ 1-\sin^2 2\theta=\cos^2 2\theta 1 − sin 2 2 θ = cos 2 2 θ
So,
f ( θ ) = 3 ( cos 4 θ + sin 4 θ ) − 2 cos 2 2 θ . f(\theta)=3(\cos^4\theta+\sin^4\theta)-2\cos^2 2\theta. f ( θ ) = 3 ( cos 4 θ + sin 4 θ ) − 2 cos 2 2 θ . Now,
sin 4 θ + cos 4 θ = ( sin 2 θ + cos 2 θ ) 2 − 2 sin 2 θ cos 2 θ = 1 − 2 sin 2 θ cos 2 θ . \sin^4\theta+\cos^4\theta=(\sin^2\theta+\cos^2\theta)^2-2\sin^2\theta\cos^2\theta
=1-2\sin^2\theta\cos^2\theta. sin 4 θ + cos 4 θ = ( sin 2 θ + cos 2 θ ) 2 − 2 sin 2 θ cos 2 θ = 1 − 2 sin 2 θ cos 2 θ . Since
sin 2 2 θ = 4 sin 2 θ cos 2 θ , \sin^2 2\theta=4\sin^2\theta\cos^2\theta, sin 2 2 θ = 4 sin 2 θ cos 2 θ , we get
sin 4 θ + cos 4 θ = 1 − 1 2 sin 2 2 θ . \sin^4\theta+\cos^4\theta=1-\frac{1}{2}\sin^2 2\theta. sin 4 θ + cos 4 θ = 1 − 2 1 sin 2 2 θ . Also,
cos 2 2 θ = 1 − sin 2 2 θ . \cos^2 2\theta=1-\sin^2 2\theta. cos 2 2 θ = 1 − sin 2 2 θ . Hence
f ( θ ) = 3 ( 1 − 1 2 sin 2 2 θ ) − 2 ( 1 − sin 2 2 θ ) = 1 + 1 2 sin 2 2 θ . f(\theta)=3\left(1-\frac12\sin^2 2\theta\right)-2(1-\sin^2 2\theta)
=1+\frac12\sin^2 2\theta. f ( θ ) = 3 ( 1 − 2 1 sin 2 2 θ ) − 2 ( 1 − sin 2 2 θ ) = 1 + 2 1 sin 2 2 θ . So the simplified form is
f ( θ ) = 1 + 1 2 sin 2 2 θ . \boxed{f(\theta)=1+\frac12\sin^2 2\theta}. f ( θ ) = 1 + 2 1 sin 2 2 θ .
Differentiate
f ′ ( θ ) = 1 2 ⋅ d d θ ( sin 2 2 θ ) . f'(\theta)=\frac12\cdot \frac{d}{d\theta}(\sin^2 2\theta). f ′ ( θ ) = 2 1 ⋅ d θ d ( sin 2 2 θ ) . Using chain rule,
d d θ ( sin 2 2 θ ) = 2 sin 2 θ ⋅ ( 2 cos 2 θ ) = 4 sin 2 θ cos 2 θ = 2 sin 4 θ . \frac{d}{d\theta}(\sin^2 2\theta)=2\sin 2\theta\cdot (2\cos 2\theta)=4\sin 2\theta\cos 2\theta=2\sin 4\theta. d θ d ( sin 2 2 θ ) = 2 sin 2 θ ⋅ ( 2 cos 2 θ ) = 4 sin 2 θ cos 2 θ = 2 sin 4 θ . Therefore,
f ′ ( θ ) = 1 2 ( 2 sin 4 θ ) = sin 4 θ . f'(\theta)=\frac12(2\sin 4\theta)=\sin 4\theta. f ′ ( θ ) = 2 1 ( 2 sin 4 θ ) = sin 4 θ . We need
f ′ ( θ ) = − 3 2 ⟹ sin 4 θ = − 3 2 . f'(\theta)=-\frac{\sqrt3}{2} \implies \sin 4\theta=-\frac{\sqrt3}{2}. f ′ ( θ ) = − 2 3 ⟹ sin 4 θ = − 2 3 .
Solve for θ ∈ [ 0 , π ] \theta\in[0,\pi] θ ∈ [ 0 , π ]
Let
x = 4 θ . x=4\theta. x = 4 θ . θ ∈ [ 0 , π ] \theta\in[0,\pi] θ ∈ [ 0 , π ] x ∈ [ 0 , 4 π ] . x\in[0,4\pi]. x ∈ [ 0 , 4 π ] .
Now solve
sin x = − 3 2 . \sin x=-\frac{\sqrt3}{2}. sin x = − 2 3 . In [ 0 , 4 π ] [0,4\pi] [ 0 , 4 π ]
x = 4 π 3 , 5 π 3 , 10 π 3 , 11 π 3 . x=\frac{4\pi}{3},\; \frac{5\pi}{3},\; \frac{10\pi}{3},\; \frac{11\pi}{3}. x = 3 4 π , 3 5 π , 3 10 π , 3 11 π . Thus
θ = x 4 ⟹ θ ∈ { π 3 , 5 π 12 , 5 π 6 , 11 π 12 } . \theta=\frac{x}{4} \implies \theta\in \left\{\frac{\pi}{3},\frac{5\pi}{12},\frac{5\pi}{6},\frac{11\pi}{12}\right\}. θ = 4 x ⟹ θ ∈ { 3 π , 12 5 π , 6 5 π , 12 11 π } . So,
S = { π 3 , 5 π 12 , 5 π 6 , 11 π 12 } . S=\left\{\frac{\pi}{3},\frac{5\pi}{12},\frac{5\pi}{6},\frac{11\pi}{12}\right\}. S = { 3 π , 12 5 π , 6 5 π , 12 11 π } .
Find β \beta β
Given
4 β = ∑ θ ∈ S θ . 4\beta=\sum_{\theta\in S}\theta. 4 β = θ ∈ S ∑ θ . Now,
∑ θ ∈ S θ = π 3 + 5 π 12 + 5 π 6 + 11 π 12 . \sum_{\theta\in S}\theta=\frac{\pi}{3}+\frac{5\pi}{12}+\frac{5\pi}{6}+\frac{11\pi}{12}. θ ∈ S ∑ θ = 3 π + 12 5 π + 6 5 π + 12 11 π . Taking denominator 12 12 12
\frac{4\pi}{12}+\frac{5\pi}{12}+\frac{10\pi}{12}+\frac{11\pi}{12}= rac{30\pi}{12}=\frac{5\pi}{2}.
Hence
4 β = 5 π 2 ⟹ β = 5 π 8 . 4\beta=\frac{5\pi}{2}
\implies \beta=\frac{5\pi}{8}. 4 β = 2 5 π ⟹ β = 8 5 π .
Compute f ( β ) f(\beta) f ( β )
Using
f ( θ ) = 1 + 1 2 sin 2 2 θ , f(\theta)=1+\frac12\sin^2 2\theta, f ( θ ) = 1 + 2 1 sin 2 2 θ , we get
f ( β ) = 1 + 1 2 sin 2 ( 2 ⋅ 5 π 8 ) = 1 + 1 2 sin 2 ( 5 π 4 ) . f(\beta)=1+\frac12\sin^2\left(2\cdot \frac{5\pi}{8}\right)
=1+\frac12\sin^2\left(\frac{5\pi}{4}\right). f ( β ) = 1 + 2 1 sin 2 ( 2 ⋅ 8 5 π ) = 1 + 2 1 sin 2 ( 4 5 π ) . Now,
sin 5 π 4 = − 1 2 ⟹ sin 2 5 π 4 = 1 2 . \sin\frac{5\pi}{4}=-\frac{1}{\sqrt2}
\implies \sin^2\frac{5\pi}{4}=\frac12. sin 4 5 π = − 2 1 ⟹ sin 2 4 5 π = 2 1 . Therefore,
f ( β ) = 1 + 1 2 ⋅ 1 2 = 1 + 1 4 = 5 4 . f(\beta)=1+\frac12\cdot\frac12=1+\frac14=\frac54. f ( β ) = 1 + 2 1 ⋅ 2 1 = 1 + 4 1 = 4 5 . So the correct option is
C 5 4 . \boxed{\text{C }\frac54}. C 4 5 .
Comparison with stored answer
Stored correct answer: C
Our derived answer: C