Trigonometric graphs and exact valuesAQA A-Level Maths: Revision notes

The graphs of $y=\sin x$ and $y=\cos x$ are smooth waves that repeat every $2\pi$ radians ($360^\circ$): $\sin(x+2\pi)=\sin x$ and $\cos(x+2\pi)=\cos x$. This repeat distance is the period. Both functions take values only between $-1$ and $1$ (the range).

• $y=\sin x$ starts at the origin, reaches $1$ at $x=\frac{\pi}{2}$, crosses the axis at $\pi$, reaches $-1$ at $\frac{3\pi}{2}$ and returns to $0$ at $2\pi$. Zeros at $x=n\pi$.
• $y=\cos x$ starts at $(0,1)$, crosses the axis at $\frac{\pi}{2}$, reaches $-1$ at $\pi$, crosses again at $\frac{3\pi}{2}$ and returns to $1$ at $2\pi$. Zeros at $x=\frac{\pi}{2}+n\pi$. The cosine curve is the sine curve translated $\frac{\pi}{2}$ to the left: $\cos x=\sin\left(x+\frac{\pi}{2}\right)$.

$\tan x=\frac{\sin x}{\cos x}$ is undefined wherever $\cos x=0$, so the graph has vertical asymptotes at $x=\frac{\pi}{2}+n\pi$ ($x=90^\circ+180^\circ n$). Between asymptotes it increases through the axis, with zeros at $x=n\pi$. Its range is all real numbers, and its period is $\pi$ ($180^\circ$): $\tan(x+\pi)=\tan x$. Near an asymptote the graph climbs without limit, so it has no maximum or minimum value.

The graphs have symmetries you can use to relate values at different angles (shown in radians; replace $\pi$ with $180^\circ$ for degrees):

• Odd and even: $\sin(-x)=-\sin x$ and $\tan(-x)=-\tan x$ (rotational symmetry about the origin), while $\cos(-x)=\cos x$ (reflection in the $y$-axis).
• Reflection about $x=\frac{\pi}{2}$: $\sin(\pi-x)=\sin x$ and $\cos(\pi-x)=-\cos x$.
• Shift by $\pi$: $\sin(x+\pi)=-\sin x$, $\cos(x+\pi)=-\cos x$ and $\tan(x+\pi)=\tan x$. These give the sign of each function in each quadrant: all positive in the first, only sine in the second, only tangent in the third, only cosine in the fourth.

You must know these exact values for the angles $0,\frac{\pi}{6},\frac{\pi}{4},\frac{\pi}{3},\frac{\pi}{2}$ (that is $0^\circ,30^\circ,45^\circ,60^\circ,90^\circ$):

• $\sin$: $0,\ \frac12,\ \frac{\sqrt2}{2},\ \frac{\sqrt3}{2},\ 1$
• $\cos$: $1,\ \frac{\sqrt3}{2},\ \frac{\sqrt2}{2},\ \frac12,\ 0$
• $\tan$: $0,\ \frac{1}{\sqrt3},\ 1,\ \sqrt3$, and undefined at $\frac{\pi}{2}$. To rebuild them, use a $45^\circ$ right-angled triangle (sides $1,1,\sqrt2$) and half an equilateral triangle (sides $1,2,\sqrt3$, with angles $30^\circ$ and $60^\circ$).

For an angle outside the first quadrant, find its reference angle (the acute angle to the $x$-axis), take the exact value for that angle, then fix the sign from the quadrant. Example: $\sin\frac{5\pi}{6}$. The reference angle is $\pi-\frac{5\pi}{6}=\frac{\pi}{6}$ and sine is positive in the second quadrant, so $\sin\frac{5\pi}{6}=\frac12$. Example: $\tan\frac{5\pi}{4}$. The reference angle is $\frac{5\pi}{4}-\pi=\frac{\pi}{4}$ and tangent is positive in the third quadrant, so $\tan\frac{5\pi}{4}=1$. Example: $\cos\frac{2\pi}{3}=-\cos\frac{\pi}{3}=-\frac12$.

Because each curve repeats, a solution is repeated every period (every $2\pi$ for sine and cosine, every $\pi$ for tangent). Within one period, symmetry gives the second solution: if $\sin x=k$ has solution $\alpha$, the other in $[0,2\pi]$ is $\pi-\alpha$. Worked example: the curves $y=\sin x$ and $y=\cos x$ meet at $x=\frac{\pi}{4}$, where both equal $\frac{\sqrt2}{2}$. Because $\sin(x+\pi)=-\sin x$ and $\cos(x+\pi)=-\cos x$, they also meet at $x=\frac{5\pi}{4}$, and so at $x=\frac{\pi}{4}+n\pi$, giving two intersections in every interval of $2\pi$.