First slide

Trigonometric Identities

Question

$\begin{array}{l} I f \sin θ (1 + \sin θ) + \cos θ (1 + \cos θ) = x \\ a n d \sin θ (1 - \sin θ) + \cos θ (1 - \cos θ) = y t h e n \end{array}$

Easy

A

$x^{2} - 2 x = \sin 2 θ$
B

$y^{2} + 2 y = \sin 2 θ$
C

$x y = \sin 2 θ$
D

$x - y = 2$

Solution

Given
$\begin{array}{l} \sin θ (1 + \sin θ) + \cos θ (1 + \cos θ) = x \Rightarrow \sin θ + \sin^{2} θ + + \cos θ + \cos^{2} θ = x \Rightarrow \sin θ + \cos θ = x - 1 \\ a n d \sin θ (1 - \sin θ) + \cos θ (1 - \cos θ) = y \\ \Rightarrow \sin θ + \cos θ = x - 1 = y + 1 \\ ∴ x - y = 2 \end{array}$
$x y = {(\sin θ + \cos θ)}^{2} - 1 = \sin (2 θ)$

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