First slide

Extreme values of trigonometric functions in trigonometry

Question

If $u = \sqrt{a^{2} \cos^{2} θ + b^{2} \sin^{2} θ} + \sqrt{a^{2} \sin^{2} θ + b^{2} \cos^{2} θ}$ then the difference between the maximum and minimum
values of u² is given by

Moderate

A

$2 (a^{2} + b^{2})$
B

$2 \sqrt{a^{2} + b^{2}}$
C

$(a^{2} + b^{2})$
D

${(a - b)}^{2}$

Solution

$\begin{array}{l} u^{2} = a^{2} + b^{2} + 2 \sqrt{a^{2} \cos^{2} θ + b^{2} \sin^{2} θ} \times \sqrt{a^{2} \sin^{2} θ + b^{2} \cos^{2} θ} \\ = a^{2} + b^{2} + 2 \sqrt{\sin^{2} {θcos}^{2} θ (a^{4} + b^{4}) + a^{2} b^{2} (\sin^{4} θ + \cos^{4} θ)} \\ = a^{2} + b^{2} + 2 \sqrt{a^{2} b^{2} (1 - 2 \sin^{2} {θcos}^{2} θ) + (a^{4} + b^{4}) \sin^{2} {θcos}^{2} θ} \\ = (a^{2} + b^{2}) + 2 \sqrt{a^{2} b^{2} + {(a^{2} - b^{2})}^{2} \sin^{2} {θcos}^{2} θ} \\ = (a^{2} + b^{2}) + 2 \sqrt{a^{2} b^{2} + \frac{{(a^{2} - b^{2})}^{2}}{4} \sin^{2} 2 θ} \\ Max. u^{2} = (a^{2} + b^{2}) + 2 \sqrt{a^{2} b^{2} + \frac{{(a^{2} - b^{2})}^{2}}{4}} \end{array}$
$Min. u^{2} = (a^{2} + b^{2}) + 2 ab$
$\begin{matrix} \Rightarrow Difference = 2 \sqrt{a^{2} b^{2} + \frac{{(a^{2} - b^{2})}^{2}}{4}} - 2 ab \\ = \sqrt{4 a^{2} b^{2} + a^{4} + b^{4} - 2 a^{2} b^{2}} - 2 ab \\ = \sqrt{{(a^{2} + b^{2})}^{2}} - 2 ab \\ = a^{2} + b^{2} - 2 ab \\ = (a - b)^{2} \end{matrix}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App