3.8 Unit circle and trigonometric identitiesIB Maths: Applications and Interpretation HL: Revision notes

The unit circle has centre $O(0,0)$ and radius 1. Measure an angle $\theta$ anticlockwise from the positive $x$-axis. The point $P$ where the angle's arm meets the circle has coordinates $P=(\cos\theta,\ \sin\theta).$ So $\cos\theta$ is the $x$-coordinate and $\sin\theta$ is the $y$-coordinate. This definition works for every angle, not just acute ones, and it explains the signs:

• $0^\circ<\theta<90^\circ$: $\cos>0$, $\sin>0$
• $90^\circ<\theta<180^\circ$: $\cos<0$, $\sin>0$
• $180^\circ<\theta<270^\circ$: $\cos<0$, $\sin<0$
• $270^\circ<\theta<360^\circ$: $\cos>0$, $\sin<0$

Example: $\theta=210^\circ$ is in the third quadrant, so $P=(\cos210^\circ,\sin210^\circ)=(-0.866,-0.5)$. Because the radius is 1, both coordinates always lie between $-1$ and $1$.

The point $(\cos\theta,\sin\theta)$ is on a circle of radius 1, so $x^2+y^2=1$, which gives the Pythagorean identity $\cos^2\theta+\sin^2\theta=1.$ The tangent is defined by $\tan\theta=\frac{\sin\theta}{\cos\theta},\qquad\cos\theta\ne0.$ $\tan\theta$ is undefined at $90^\circ$ and $270^\circ$, where $\cos\theta=0$.

Example: $\cos\theta=-0.28$ and $\theta$ is in the third quadrant. Then $\sin^2\theta=1-0.0784=0.9216$, so $\sin\theta=-0.96$ (negative in the third quadrant) and $\tan\theta=\frac{-0.96}{-0.28}=3.43$.

The identity turns one known ratio into the other two; the quadrant tells you which sign to choose.

Imagine $P$ moving anticlockwise round the unit circle. Plot its $y$-coordinate against the angle $\theta$ and you trace the graph of $y=\sin\theta$. Plot its $x$-coordinate against $\theta$ and you trace $y=\cos\theta$.

Reading the circle gives the key features:

• Both graphs repeat every $360^\circ$ (the period) and lie between $-1$ and $1$ (amplitude 1).
• $\sin\theta$ is 0 at $0^\circ,180^\circ,360^\circ$, has a maximum of 1 at $90^\circ$ and a minimum of $-1$ at $270^\circ$.
• $\cos\theta$ is 1 at $0^\circ$, 0 at $90^\circ$ and $270^\circ$, and $-1$ at $180^\circ$.
• The cosine graph is the sine graph shifted $90^\circ$ to the left.

Exact values such as $\sin30^\circ=\frac12$, $\cos60^\circ=\frac12$ and $\sin60^\circ=\frac{\sqrt3}{2}$ are not examined, but they help you see the shapes. In the exam, use your GDC.

The sine rule $\frac{a}{\sin A}=\frac{b}{\sin B}$ finds an angle from $\sin B$. But $\sin B=\sin(180^\circ-B)$, so two angles in $0^\circ<B<180^\circ$ share the same sine. This gives the ambiguous case: when you are given two sides and a non-included angle (SSA), there may be two different triangles.

Example: $a=7$, $b=10$, $A=30^\circ$. $\sin B=\frac{10\sin30^\circ}{7}=0.714\ldots\Rightarrow B=45.6^\circ\ \text{or}\ B=180^\circ-45.6^\circ=134.4^\circ.$ Both are valid, as $30^\circ+134.4^\circ<180^\circ$. Then $C=104.4^\circ$ or $15.6^\circ$, and each gives a different third side $c$.

Two triangles occur when $A$ is acute and $b\sin A<a<b$. If $a\ge b$ the obtuse angle does not fit and there is one triangle. If $a<b\sin A$ then $\sin B>1$ and there is none.

To solve a trigonometric equation in a finite interval, such as $0^\circ\le x\le360^\circ$, use your GDC: graph each side as a function and find the intersections within the interval.

Example: solve $3\sin x=2$ for $0^\circ\le x\le360^\circ$. Plot $y=3\sin x$ and $y=2$: the intersections are at $x=41.8^\circ$ and $x=138^\circ$.

You can check by symmetry: $\sin x=\frac23$ gives $x=41.8^\circ$ and $180^\circ-41.8^\circ=138.2^\circ$. A larger interval, such as $0^\circ\le x\le720^\circ$, gives more solutions: add $360^\circ$ to each.

Always set the GDC to degree mode, set the window to match the interval, and give answers to 3 significant figures.

Real periodic situations (tides, daylight hours, a Ferris wheel) are modelled by $y=a\sin(bx)+d$ or $y=a\cos(bx)+d$, where

• $|a|$ is the amplitude,
• the period is $\frac{360}{b}$ (in degrees),
• $d$ is the midline, so the maximum is $d+|a|$ and the minimum is $d-|a|$.

Example: a Ferris wheel passenger's height is $h=10+8\sin(20t^\circ)$ metres. The period is $\frac{360}{20}=18$ minutes. To find when $h=14$: $\sin(20t^\circ)=0.5$, so $20t=30$ or $150$, giving $t=1.5$ and $t=7.5$ minutes. You can find these on the GDC as intersections of $y=h$ and $y=14$.

Interpret answers in context: say what the time or height means, with units.