This note covers the symmetry of the sine, cosine and tangent graphs, and how it gives relationships such as \(\sin(\pi-\theta)=\sin\theta\) and \(\sin\left(\tfrac{\pi}{2}-\theta\right)=\cos\theta\). It matters because it lets you find every solution of a trigonometric equation and simplify expressions without a calculator.
Key ideas
Picture the unit circle. A point at angle \(\theta\) from the positive \(x\)-axis has coordinates \((\cos\theta,\sin\theta)\). The tangent is the ratio \(\sin\theta/\cos\theta\). Reflecting or rotating the point gives other angles whose coordinates differ only in sign or in which coordinate is which. This is the source of every relationship below.
Even and odd behaviour. The cosine graph is symmetric about the \(y\)-axis, so \(\cos(-\theta)=\cos\theta\). Sine and tangent are odd: their graphs look the same after a half-turn about the origin, so \(\sin(-\theta)=-\sin\theta\) and \(\tan(-\theta)=-\tan\theta\).
Reflection in the \(y\)-axis (supplementary angles). The angles \(\theta\) and \(\pi-\theta\) give points that are mirror images across the \(y\)-axis. Their heights are equal and their horizontal positions are opposites:
- \(\sin(\pi-\theta)=\sin\theta\)
- \(\cos(\pi-\theta)=-\cos\theta\)
- \(\tan(\pi-\theta)=-\tan\theta\)
Half-turn. Adding \(\pi\) moves the point to the opposite side of the circle, so both coordinates change sign:
- \(\sin(\pi+\theta)=-\sin\theta\)
- \(\cos(\pi+\theta)=-\cos\theta\)
- \(\tan(\pi+\theta)=\tan\theta\)
Periodicity. Sine and cosine repeat every \(2\pi\), so \(\sin(\theta+2\pi)=\sin\theta\). Tangent repeats every \(\pi\).
Complementary angles. The angles \(\theta\) and \(\tfrac{\pi}{2}-\theta\) give points that are mirror images in the line \(y=x\). The two coordinates swap, so \(\sin\left(\tfrac{\pi}{2}-\theta\right)=\cos\theta\) and \(\cos\left(\tfrac{\pi}{2}-\theta\right)=\sin\theta\). The name "cosine" comes from "complement's sine". The graph of cosine is the graph of sine shifted left by \(\tfrac{\pi}{2}\), so \(\cos\theta=\sin\left(\theta+\tfrac{\pi}{2}\right)\).
You can work with degrees in the same way: replace \(\pi\) by \(180^\circ\) and \(\tfrac{\pi}{2}\) by \(90^\circ\).
Using symmetry to solve equations. A calculator gives only one angle. To find the others in an interval, use the symmetry:
- If \(\sin x=k\), the solutions are \(x=\alpha\) and \(x=\pi-\alpha\), plus multiples of \(2\pi\).
- If \(\cos x=k\), the solutions are \(x=\alpha\) and \(x=-\alpha\), plus multiples of \(2\pi\). Equivalent: \(\alpha\) and \(2\pi-\alpha\).
- If \(\tan x=k\), the solutions are \(x=\alpha\) plus multiples of \(\pi\).
Here \(\alpha\) is the first value from your calculator. A quick sketch of the graph with a horizontal line at height \(k\) shows how many solutions to expect in the interval.
The special-angle values come from two triangles. The isosceles right triangle gives \(\sin45^\circ=\cos45^\circ=\tfrac{\sqrt2}{2}\) and \(\tan45^\circ=1\). Half an equilateral triangle gives \(\sin30^\circ=\tfrac12\), \(\cos30^\circ=\tfrac{\sqrt3}{2}\), \(\tan30^\circ=\tfrac{1}{\sqrt3}\), and the matching values at \(60^\circ\). Symmetry then extends them to all four quadrants, with the sign set by the quadrant: sine is positive in quadrants 1 and 2, cosine in quadrants 1 and 4, and tangent in quadrants 1 and 3.
Symmetry seen on the graphs. The sine graph on \(0\le x\le\pi\) is a single hump, and it is symmetric about the vertical line \(x=\tfrac{\pi}{2}\). That is exactly the statement \(\sin(\pi-\theta)=\sin\theta\). The same graph has half-turn symmetry about the point \((\pi,0)\), which is the statement \(\sin(2\pi-\theta)=-\sin\theta\). The cosine graph is symmetric about the \(y\)-axis and also about the line \(x=\pi\), which gives \(\cos(2\pi-\theta)=\cos\theta\). Looking for these features on a sketch is often quicker than remembering a list of formulae.
Why the shift works. Because \(\sin\theta\) and \(\cos\theta\) are the same curve shifted by a quarter of a period, any equation that mixes the two can often be rewritten using one function. For example, \(\cos\left(x-\tfrac{\pi}{2}\right)=\sin x\) because shifting the cosine curve right by \(\tfrac{\pi}{2}\) gives the sine curve. You can combine these with the double angle identities when an equation involves several related angles.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(\sin(-\theta)=-\sin\theta\), \(\tan(-\theta)=-\tan\theta\) | Sine and tangent are odd functions. |
| \(\cos(-\theta)=\cos\theta\) | Cosine is an even function. |
| \(\sin(\pi-\theta)=\sin\theta\), \(\cos(\pi-\theta)=-\cos\theta\) | Finds the second-quadrant partner of an angle. |
| \(\sin(\pi+\theta)=-\sin\theta\), \(\cos(\pi+\theta)=-\cos\theta\), \(\tan(\pi+\theta)=\tan\theta\) | Finds the third-quadrant partner; tangent period \(\pi\). |
| \(\sin\left(\tfrac{\pi}{2}-\theta\right)=\cos\theta\), \(\cos\left(\tfrac{\pi}{2}-\theta\right)=\sin\theta\) | Complementary angle relationships. |
| \(\tan\left(\tfrac{\pi}{2}-\theta\right)=\dfrac{1}{\tan\theta}\) | Tangent of the complement is the reciprocal. |
Worked examples
Example 1 Given that \(\cos32^\circ=0.848\) (3 s.f.), write down the values of (a) \(\cos148^\circ\), (b) \(\cos212^\circ\), (c) \(\sin58^\circ\), (d) \(\cos(-32^\circ)\).
(a) \(148^\circ=180^\circ-32^\circ\), so \(\cos148^\circ=-\cos32^\circ=-0.848\).
(b) \(212^\circ=180^\circ+32^\circ\), so \(\cos212^\circ=-\cos32^\circ=-0.848\).
(c) \(58^\circ=90^\circ-32^\circ\), so \(\sin58^\circ=\cos32^\circ=0.848\).
(d) Cosine is even, so \(\cos(-32^\circ)=\cos32^\circ=0.848\).
Answer: (a) \(-0.848\), (b) \(-0.848\), (c) \(0.848\), (d) \(0.848\).
Example 2 Solve \(\tan x=-1.5\) for \(0\le x\le2\pi\). Give answers in radians to 3 s.f.
Tangent is negative in quadrants 2 and 4. The reference angle is \(\arctan1.5=0.9828\).
Second quadrant: \(x=\pi-0.9828=2.1588\).
Fourth quadrant: \(x=2\pi-0.9828=5.3004\). (This equals \(2.1588+\pi\), because tangent has period \(\pi\).)
Both lie in the interval \([0,2\pi]\).
Answer: \(x=2.16\) or \(x=5.30\).
Example 3 Show that \(\sin(\pi+\theta)\cos\left(\tfrac{\pi}{2}-\theta\right)+\cos^2(\pi-\theta)=\cos2\theta\).
Replace each term: \(\sin(\pi+\theta)=-\sin\theta\), \(\cos\left(\tfrac{\pi}{2}-\theta\right)=\sin\theta\), and \(\cos(\pi-\theta)=-\cos\theta\).
The left side becomes \((-\sin\theta)(\sin\theta)+(-\cos\theta)^2=-\sin^2\theta+\cos^2\theta\).
This is the double angle form \(\cos2\theta=\cos^2\theta-\sin^2\theta\).
Answer: the left side equals \(\cos2\theta\), as required.
Common errors
- Using \(\sin(\pi-\theta)=-\sin\theta\). The sine stays the same; it is the cosine that changes sign.
- Forgetting that \((-\cos\theta)^2=\cos^2\theta\), and keeping a minus sign that should disappear.
- Giving only the calculator value for an equation, and missing the second solution (or the second and third for \(\tan\) over a long interval).
- Mixing degrees and radians, for example using \(\pi-\theta\) with \(\theta\) in degrees.
- Applying a period of \(2\pi\) to tangent. Its period is \(\pi\).
Exam tip: Sketch a quick graph or unit circle before solving. Mark the reference angle, then read off every solution inside the stated interval. This stops you missing solutions.
Check yourself
- Given \(\sin\tfrac{\pi}{5}=0.588\), write down \(\sin\tfrac{4\pi}{5}\).
Answer
\(\tfrac{4\pi}{5}=\pi-\tfrac{\pi}{5}\), so \(\sin\tfrac{4\pi}{5}=\sin\tfrac{\pi}{5}=0.588\). - Write \(\cos\left(\tfrac{\pi}{2}+\theta\right)\) in terms of \(\sin\theta\).
Answer
Use \(\cos\left(\tfrac{\pi}{2}+\theta\right)=\sin\left(\tfrac{\pi}{2}-\left(\tfrac{\pi}{2}+\theta\right)\right)=\sin(-\theta)=-\sin\theta\). - Find the exact value of \(\cos\tfrac{5\pi}{6}\).
Answer
\(\cos\tfrac{5\pi}{6}=\cos\left(\pi-\tfrac{\pi}{6}\right)=-\cos\tfrac{\pi}{6}=-\tfrac{\sqrt3}{2}\). - Solve \(\sin x=\tfrac{\sqrt3}{2}\) for \(0\le x\le2\pi\).
Answer
The reference angle is \(\tfrac{\pi}{3}\). Sine is positive in quadrants 1 and 2, so \(x=\tfrac{\pi}{3}\) or \(x=\pi-\tfrac{\pi}{3}=\tfrac{2\pi}{3}\).