Relationships Between Trigonometric Functions
Symmetries of Trigonometric Functions
Sine Function Symmetry
The sine function has symmetry:
$$\sin(\pi - \theta) = \sin(\theta)$$
This relationship is derived from the fact that the sine function is even about the line $x = \frac \pi2 $ graphically.
This means that
$$\sin(\frac{\pi}{2}-u)=\sin(\frac{\pi}{2}+u)$$
For all $u \in \mathbb{R}$
Using the substitution $ = x - \frac{\pi}{2}$:
$$\sin(\frac{\pi}{2} - (x - \frac{\pi}{2}))=\sin(\frac{\pi}{2}+ x - \frac{\pi}{2})$$
$$\sin(\pi - x)=\sin(x)$$
For instance, $\sin(60°) = \sin(120°)$ because $120° = 180° - 60° = \pi - 60°$.
This symmetry property of sine is related to its odd function nature, where $\sin(-\theta) = -\sin(\theta)$.
Cosine Function Symmetry
The cosine function has a similar symmetry property:
$$\cos(\pi - \theta) = -\cos(\theta)$$
This relationship is derived from the fact that the cosine function is odd about the line $x = \frac{\pi}{\2}$ in its graph.
Therefore, we can say that for all $u\in \mathbb{R}$:
$$\cos(\frac{\pi}{2}-u)=-\cos(\frac{\pi}{2}+u)$$
Therefore, using the substitution $u = x-\frac{\pi}{2}$
$$\cos(\frac{\pi}{2}-(x-\frac{\pi}{2}))=-\cos(\frac{\pi}{2}+(x-\frac{\pi}{2}))$$
$$\cos(\pi - x)=-\cos(x)$$
hence the identity for symmetry is proven
For example, $\cos(60°) = -\cos(120°)$, as $120° = 180° - 60° = \pi - 60°$.
The cosine function is an even function, meaning $\cos(-\theta) = \cos(\theta)$, which is related to but distinct from this symmetry property.
Tangent Function Symmetry
The tangent function exhibits the following symmetry:
$$\tan(\pi - \theta) = -\tan(\theta)$$
This property indicates that the tangent function is odd about the line $x=\frac{\pi}{2}$ in its graph.
We can use the previous two identities to prove this identity
$$\tan(\pi - \theta)= \frac{\sin \pi - \theta}{\cos \pi - \theta}$$
$$\tan(\pi - \theta)= \frac{\sin \theta}{-\cos \theta}$$
$$\tan(\pi - \theta)= -\tan \theta$$
Hence proving the established identity.
As an illustration, $\tan(60°) = -\tan(120°)$, since $120° = 180° - 60° = \pi - 60°$.
The tangent function is an odd function, where $\tan(-\theta) = -\tan(\theta)$, which is consistent with this symmetry property.
Connection to the Unit Circle
These symmetry properties are connected to the geometry of the unit circle. When we consider a point $(\cos\theta, \sin\theta)$ on the unit circle, the point corresponding to $\pi - \theta$ is $(-\cos\theta, \sin\theta)$.
Visualizing these relationships on the unit circle can greatly aid in remembering and applying the symmetry properties.
Problem: Simplify $\sin(150°) + \cos(30°)$
Solution:
- Recognize that $150° = 180° - 30° = \pi - 30°$
- Apply the symmetry property: $\sin(150°) = \sin(30°)$
- We know $\cos(30°) = \frac{\sqrt{3}}{2}$ and $\sin(30°) = \frac{1}{2}$
- Therefore, $\sin(150°) + \cos(30°) = \frac{1}{2} + \frac{\sqrt{3}}{2}$
Students often forget to consider these symmetry properties when solving trigonometric equations, leading to unnecessary complications or missed solutions.
Calculation trick
If you're in a rush and you don't remember these identities, I am going to provide a simple trick for simplifying trigonometric functions.
For for all sine and cosine fucntions, you can add or subtract $\pi$ from the input and then change the sign of the function. For example:
$$\cos \theta = - \cos(\theta + \pi)$$
$$\sin \theta = - \sin(\theta - \pi)$$
For tangent functions, you can add or subtract $\pi$ from the input without changing its value. This comes from the fact that the tangent function is periodic every $\pi$ units.
$$\tan \theta = \tan (\theta + \pi) $$
$$\cot \theta = \cot (\theta - \pi) $$
