AHL 3.11—Relationships between trig functions

In this subtopic we explore how the sine, cosine, and tangent functions behave when their arguments are transformed in the form $\pi -\theta$. These symmetry relationships arise naturally from the geometry of the unit circle and the graphical properties of each function. Mastering them allows you to simplify expressions, solve equations more efficiently, and spot patterns in trigonometric problems.

The three key identities are:

$\sin (\pi -\theta )=\sin (\theta )$
$\cos (\pi -\theta )=-\cos (\theta )$
$\tan (\pi -\theta )=-\tan (\theta )$

Each of these can be derived in two ways: using compound angle identities (AHL 3.10) or using geometric symmetry arguments from the unit circle. Understanding both derivation routes , not just memorising the results , is what IB HL questions will test.

This subtopic also covers the related co-function identities ($\frac{\pi }{2}\pm \theta$) and the extension of these symmetries to reciprocal trigonometric functions.

Deriving $\sin (\pi -\theta )=\sin (\theta )$:

Method 1 , Compound angle identity (primary method):

Using the compound angle formula $\sin (A-B)=\sin A\cos B-\cos A\sin B$ with $A=\pi$ and $B=\theta$:

$\sin (\pi -\theta )=\sin \pi \cos \theta -\cos \pi \sin \theta =(0)\cos \theta -(-1)\sin \theta =\sin \theta ✓$

This is the method IB mark schemes most commonly credit.

Method 2 , Graph symmetry argument:

The graph of $y=\sin x$ is symmetric about the vertical line $x=\frac{\pi }{2}$. Formally, the function satisfies:

$\sin \left(\frac{\pi }{2}-u\right)=\sin \left(\frac{\pi }{2}+u\right)\text{for all }u\in \mathbb{R}$

This holds because $\sin \left(\frac{\pi }{2}-u\right)=\cos u$ and $\sin \left(\frac{\pi }{2}+u\right)=\cos u$ (both follow from the compound angle formula). Substituting $u=x-\frac{\pi }{2}$:

$\sin (\pi -x)=\sin (x)✓$

We say sine is symmetric about $x=\frac{\pi }{2}$.