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Question: If \(A = \cos^{2}\theta + \sin^{4}\theta,\) then for all values of θ...

If A=cos⁡2θ+sin⁡4θ,A = \cos^{2}\theta + \sin^{4}\theta, then for all values of θ

A

1≤A≤21 \leq A \leq 2

B

13/16≤A≤113/16 \leq A \leq 1

C

3/4≤A≤13/163/4 \leq A \leq 13/16

D

3/4≤A≤13/4 \leq A \leq 1

Answer

3/4≤A≤13/4 \leq A \leq 1

Explanation

Solution

A=cos⁡2θ+sin⁡4θA = \cos^{2}\theta + \sin^{4}\theta ⇒A=cos⁡2θ+sin⁡2θ.sin⁡2θA = \cos^{2}\theta + \sin^{2}\theta.\sin^{2}\theta

⇒ A≤cos⁡2θ+sin⁡2θA \leq \cos^{2}\theta + \sin^{2}\theta, [∵sin⁡2θ≤1]\lbrack\because\sin^{2}\theta \leq 1\rbrack

⇒ A≤1A \leq 1

Again A=cos⁡2θ+sin⁡4θ=(1−sin⁡2θ)+sin⁡4θA = \cos^{2}\theta + \sin^{4}\theta = (1 - \sin^{2}\theta) + \sin^{4}\theta

A=(sin⁡2θ−12)2+34≥34A = \left( \sin^{2}\theta - \frac{1}{2} \right)^{2} + \frac{3}{4} \geq \frac{3}{4}

Hence, 3/4≤A≤13/4 \leq A \leq 1.