Trigonometric Identities practice questions — A-Level Maths

$\sin x+\cos x=\sqrt{2}\text{\hspace{0.17em}⁣}(\sin x\cdot {\displaystyle \frac{\sqrt{2}}{2}}+\cos x\cdot {\displaystyle \frac{\sqrt{2}}{2}})=\sqrt{2}\sin \text{\hspace{0.17em}⁣}(x+{\displaystyle \frac{\pi }{4}})$.

Since $\sin$ has maximum $1$, the maximum of $f$ is $\sqrt{2}$, attained at $x={\displaystyle \frac{\pi }{4}}$.

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

$\bullet$ $\cos x=0\Rightarrow x\in \text{\hspace{0.17em}⁣}\{{\displaystyle \frac{\pi }{2}},{\displaystyle \frac{3\pi }{2}}\}$

$\bullet$ $\sin x={\displaystyle \frac{\sqrt{3}}{2}}\Rightarrow x\in \text{\hspace{0.17em}⁣}\{{\displaystyle \frac{\pi }{3}},{\displaystyle \frac{2\pi }{3}}\}$

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

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

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

Since $x\in \text{\hspace{0.17em}⁣}({\displaystyle \frac{\pi }{2}},\pi )$, $\cos x<0$. From ${\sin }^{2}x+{\cos }^{2}x=1$: ${\cos }^{2}x=1-{\displaystyle \frac{16}{25}}={\displaystyle \frac{9}{25}}$, so $\cos x=-{\displaystyle \frac{3}{5}}$.

Then $\sin 2x=2\cdot {\displaystyle \frac{4}{5}}\cdot \text{\hspace{0.17em}⁣}(-{\displaystyle \frac{3}{5}})=-{\displaystyle \frac{24}{25}}$.

${\sin }^2{\displaystyle \frac{x}{2}}={\displaystyle \frac{1-3\mathrm{/}5}{2}}={\displaystyle \frac{2\mathrm{/}5}{2}}={\displaystyle \frac{1}{5}}$.

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

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

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

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

$\sin \text{\hspace{0.17em}⁣}(\pi +{\displaystyle \frac{\pi }{6}})=-\sin {\displaystyle \frac{\pi }{6}}=-{\displaystyle \frac{1}{2}}$.