Question
Mathematics Question on Trigonometric Functions
Let f(?)=(1+sin2?)(2−sin2?). Then for all values of ?
A
f(?)>49
B
f(?)<2
C
f(?)>411
D
2≤f(?)≤49
Answer
2≤f(?)≤49
Explanation
Solution
Given f(θ)=(1+sin2θ)(2−sin2θ)
=2+2sin2θ−sin2θ−sin4θ
=−sin4θ+sin2θ+2
=−(sin4θ−sin2θ−2)
=-\left\\{\sin ^{4} \theta-\sin ^{2} \theta+\frac{1}{4}-\frac{9}{4}\right\\}
=+49−(sin2θ−21)2 ....(i)
∵−1≤sinθ≤1
⇒0≤sin2θ≤1
⇒−21≤sin2θ−21≤21
⇒0≤(sin2θ−21)2≤41
⇒0≥−(sin2θ−21)2≥−41
⇒49≥49−(sin2θ−21)2≥49−41
⇒2≤f(θ)≤49 [from E(i)]