First slide

Trigonometric Identities

Question

$x = \sqrt{a^{2} \cos^{2} α + b^{2} \sin^{2} α} + \sqrt{a^{2} \sin^{2} α + b^{2} \cos^{2} α}$ then $x^{2} = a^{2} + b^{2} + 2 \sqrt{p (a^{2} + b^{2}) - p^{2}}$ , where p is equal to

Moderate

A

$a^{2} \cos^{2} α + b^{2} \sin^{2} α$
B

$a^{2} \sin^{2} α + b^{2} \cos^{2} α$
C

$\frac{1}{2} [a^{2} + b^{2} + (a^{2} - b^{2}) \cos 2 α]$
D

$\frac{1}{2} [a^{2} + b^{2} - (a^{2} - b^{2}) \cos 2 α]$

Solution

$x = \sqrt{a^{2} \cos^{2} α + b^{2} \sin^{2} α} + \sqrt{a^{2} \sin^{2} α + b^{2} \cos^{2} α}$
$\Rightarrow x^{2} = a^{2} + b^{2} + 2 \sqrt{(a^{2} \cos^{2} α + b^{2} \sin^{2} α) (a^{2} \sin^{2} α + b^{2} \cos^{2} α)}$
= a²+b²+2k,
where k= $\sqrt{[(a^{2} + b^{2}) - (a^{2} \sin^{2} α + b^{2} \cos^{2} α)] \times (a^{2} \sin^{2} α + b^{2} \cos^{2} α)}$
$∴ x = a^{2} + b^{2} + 2 \sqrt{(a^{2} + b^{2}) p - p^{2}}$
where $p = a^{2} \sin^{2} α + b^{2} \cos^{2} α$
$\begin{array}{l} = \frac{a^{2}}{2} (1 - \cos 2 α) + \frac{b^{2}}{2} (1 + \cos 2 α) \\ = \frac{1}{2} [a^{2} + b^{2} - (a^{2} - b^{2}) \cos 2 α] \end{array}$
Also, $x^{2} = a^{2} + b^{2} + 2 \sqrt{(a^{2} \cos^{2} α + b^{2} \sin^{2} α) (a^{2} \sin^{2} α + b^{2} \cos^{2} α)}$
$= a^{2} + b^{2} + 2 \sqrt{(a^{2} + b^{2}) p - p^{2}}$
where, $p = a^{2} \cos^{2} α + b^{2} \sin^{2} α$
or $p = \frac{1}{2} [a^{2} + b^{2} + (a^{2} - b^{2}) \cos 2 α]$

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