Trigonometric Identities practice questions — AP Calculus BC

$\sin^2\dfrac{x}{2} = \dfrac{1 - 3/5}{2} = \dfrac{2/5}{2} = \dfrac{1}{5}$.

Since $x \in \left(0, \dfrac{\pi}{2}\right)$, we have $\dfrac{x}{2} \in \left(0, \dfrac{\pi}{4}\right)$, so $\sin\dfrac{x}{2} > 0$ and equals $\dfrac{1}{\sqrt{5}}$.

Since $x \in \!\left(\dfrac{\pi}{2}, \pi\right)$, $\cos x < 0$. From $\sin^2 x + \cos^2 x = 1$: $\cos^2 x = 1 - \dfrac{16}{25} = \dfrac{9}{25}$, so $\cos x = -\dfrac{3}{5}$.

Then $\sin 2x = 2 \cdot \dfrac{4}{5} \cdot \!\left(-\dfrac{3}{5}\right) = -\dfrac{24}{25}$.

By the identity $\cos(A-B) = \cos A\cos B + \sin A\sin B$, the expression equals $\cos(60^\circ) = \dfrac{1}{2}$.

$(\sin x + \cos x)^2 = \dfrac{1}{4}$. Expanding: $\sin^2 x + 2\sin x \cos x + \cos^2 x = \dfrac{1}{4}$, so $1 + 2\sin x \cos x = \dfrac{1}{4}$. Hence $\sin x \cos x = -\dfrac{3}{8}$.

$\sin 75^{\circ} + \sin 15^{\circ} = 2\sin\dfrac{90^{\circ}}{2}\cos\dfrac{60^{\circ}}{2} = 2 \sin 45^{\circ} \cos 30^{\circ} = 2 \cdot \dfrac{\sqrt 2}{2} \cdot \dfrac{\sqrt 3}{2} = \dfrac{\sqrt 6}{2}$.

$\cos 2x = 1 - 2\sin^2 x = 1 - 2 \cdot \dfrac{1}{9} = \dfrac{7}{9}$.

This is the standard sum identity: $\sin(x + y) = \sin x \cos y + \cos x \sin y$.

$\sin\dfrac{\pi}{3} = \dfrac{\sqrt{3}}{2}$ — a notable value from the unit circle.

$\sin\!\left(\pi + \dfrac{\pi}{6}\right) = -\sin\dfrac{\pi}{6} = -\dfrac{1}{2}$.

$2\sin x \cos x - \sqrt{3}\cos x = 0 \Rightarrow \cos x\,(2\sin x - \sqrt{3}) = 0$.

$\bullet$ $\cos x = 0 \Rightarrow x \in \!\left\{\dfrac{\pi}{2}, \dfrac{3\pi}{2}\right\}$

$\bullet$ $\sin x = \dfrac{\sqrt{3}}{2} \Rightarrow x \in \!\left\{\dfrac{\pi}{3}, \dfrac{2\pi}{3}\right\}$

Total: $4$ solutions on $[0, 2\pi)$.