Trigonometry! #27

Geometry Level 3

Given $A = \sin ^{2} \theta + \cos ^{4} \theta$ , then for all real values of $\theta$ , which of the following holds true?

This problem is part of the set Trigonometry .

\frac {3}{4} ≤ A ≤ 1

1≤A≤2

\frac {3}{4} ≤ A ≤ \frac {13}{16}

\frac {13}{16} ≤ A ≤ 1

1 solution

Sandeep Rathod
Feb 6, 2015

$A = 1 - \cos^2 A + \cos^4 A$

$A = 1 - \cos^2 A(1 - \cos^2 A)$

$A = 1 - \cos^2 A \sin^2 A$

$A = 1 - \dfrac{\sin^2 A}{4}$

$0 \leq \sin^2 A \leq 1$

$0 \leq \sin^2 A \leq \dfrac{1}{4}$

$-1 \leq - 1 + \sin^2 A \leq \dfrac{-3}{4}$

$\implies 1 \geq 1 - \dfrac{\sin^2 A}{4} \geq \dfrac{3}{4}$

$1 \geq A \geq \dfrac{3}{4}$

Perfecto! Btw, shouldn't the step before the first inequality be $A=1-\frac{\sin^{2}2\theta}{4}$ ?

And here's another way of doing the inequality.

For the minimum value of the expression, we need the maximum value of $sin^{2}2\theta$ , and for the maximum value of the expression we need the minimum value of $\sin^{2} 2\theta$ , since it is negative.

Hence for the minimum value, $\sin^{2}2\theta = 1 \Rightarrow A_{\text{min}}=\frac{3}{4}$ .

And for the maximum value, $\sin^{2}2\theta = 0\Rightarrow A_{\text{max}} = 1$

This is why we have the inequality $\boxed{\frac{3}{4}≤A≤1}$ .

Omkar Kulkarni - 6 years, 4 months ago

Trigonometry! #27

1 solution

0 pending reports