Trigonometry & Inverse Trigonometry Question - None

Question

If $\cos x + \sin x = a$ where $-\frac{\pi}{2} < x < -\frac{\pi}{4}$, then $\cos 2x$ is equal to

Accepted Answer

Step 1: Given $-\frac{\pi}{2} < x < -\frac{\pi}{4}$, so $-\pi < 2x < -\frac{\pi}{2}$, meaning $2x$ is in the third quadrant.

Step 2: Square both sides of $\cos x + \sin x = a$:

$\cos^2 x + \sin^2 x + 2\cos x \sin x = a^2$

$1 + 2\sin x \cos x = a^2$

$1 + \sin 2x = a^2$

$\sin 2x = a^2 - 1$

Step 3: Use $\sin^2 2x + \cos^2 2x = 1$:

$\cos^2 2x = 1 - \sin^2 2x = 1 - (a^2 - 1)^2 = a^2(2 - a^2)$

Step 4: Since $2x$ is in the third quadrant, $\cos 2x < 0$:

$\cos 2x = -\sqrt{a^2(2 - a^2)} = -a\sqrt{2 - a^2}$

∴ Answer is (d) $a\sqrt{2 - a^2}$

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