Trigonometric identities

​​In a nutshell

Applying Pythagoras' theorem to the unit circle description of $$\sin$$​, $$\cos$$​ and $$\tan$$​ allows you to deduce an important identity relation the values of $$\sin(\theta)$$ and $$\cos(\theta)$$​ for any value of $$\theta$$​. This, combined with the identity $$\tan(\theta) \equiv \dfrac{\sin(\theta)}{\cos(\theta)}$$​ (whenever the fraction is well-defined) allows you to manipulate and simplify a wide range of expressions.

Pythagoras' theorem in the unit circle

​

Let $$P$$​ be a point on the unit circle centred at the origin such that the angle between the line $$OP$$​ and the positive $$x$$​-axis, measured counterclockwise from the positive $$x$$​-axis, is $$\theta$$​.

Recall that the coordinates of $$P$$​​ are given by $$(\cos(\theta), \sin(\theta))$$​, and the gradient of the line through $$O$$​ and $$P$$​ is $$\tan(\theta)$$.

Note: For all values of $$\theta$$ such that $$\cos(\theta) = 0$$, $$\tan(\theta)$$ is undefined, and it is well defined otherwise. In particular, $$\tan(\theta)$$ is undefined precisely whenever $$\theta$$ is an odd multiple of $$90^{\circ}$$.

Pythagoras' theorem tells you that the equation for the unit circle centred at the origin is $$x^2 + y^2 = 1$$. Since $$x=\cos(\theta)$$ and $$y=\sin(\theta)$$, this gives the identity:

$$\boxed{\sin^2(\theta)+\cos^2{\theta}\equiv 1}$$​​

Example 1

Given that $$\sin(\theta) = \dfrac{3}{5}$$, and that $$-90^{\circ} \leq \theta \leq 90^{\circ}$$, show that $$\cos(\theta) = \dfrac{4}{5}$$.

Substitute $$\sin(\theta) = \dfrac35$$​ into the equation $$\sin^2(\theta) + \cos^2(\theta) = 1$$, and rearrange to make $$\cos^2(\theta)$$​ the subject:

​$$\begin{aligned}\sin^2(\theta) + \cos^2(\theta) &= 1\\\left(\dfrac35\right)^2 + \cos^2(\theta) &= 1\\\cos^2(\theta) &= 1 - \dfrac{9}{25}\\\cos^2(\theta)&= \dfrac{16}{25}\end{aligned}$$​​

Now taking the square root tells you that $$\cos(\theta) = \pm\dfrac45$$. However, $$-90^{\circ} \leq \theta \leq 90^{\circ}$$ and using a $$\cos$$ graph, it can be seen that $$\cos(\theta) \geq 0$$.

Therefore, $$\underline{\cos(\theta) = \dfrac45.}$$​

An identity linking sin, cos and tan

When $$\cos(\theta) \neq 0$$​, the point $$(\cos(\theta), \sin(\theta))$$​ lies on the line $$y = \tan(\theta)x$$, so, substituting $$x= \cos(\theta)$$​ and $$y = \sin(\theta)$$​:

​$$\begin{aligned}y &= \tan(\theta)x\\\sin(\theta) &= \tan(\theta)\cos(\theta)\end{aligned}$$​​

This is more commonly written as:

​$$\boxed{\tan(\theta)\equiv \dfrac{\sin(\theta)}{\cos(\theta)}}$$​​

This holds for any value of $$\theta$$​ for which $$\tan(\theta)$$​ is well-defined​.

Example 2

Show that $$\tan^2(\theta) + 1 \equiv \dfrac{1}{\cos^2(\theta)}$$.

Apply the identities $$\tan(\theta) \equiv \dfrac{\sin(\theta)}{\cos(\theta)}$$ and $$1 \equiv \dfrac{\cos^2(\theta)}{\cos^2(\theta)}$$ so that $$\tan^2(\theta)$$ and $$1$$ have the same denominator:

$$\begin{aligned}\tan^2(\theta) +1 &= \left(\dfrac{\sin(\theta)}{\cos(\theta)}\right)^2 + \dfrac{\cos^2(\theta)}{\cos^2(\theta)}\\&= \dfrac{\sin^2(\theta)}{\cos^2(\theta)} + \dfrac{\cos^2(\theta)}{\cos^2(\theta)}\\&=\dfrac{\sin^2(\theta) + \cos^2(\theta)}{\cos^2(\theta)}\end{aligned}$$​​

Apply the identity $$\sin^2(\theta) + \cos^2(\theta) \equiv 1$$, and substitute this into the numerator:

$$\dfrac{\sin^2(\theta) + \cos^2(\theta)}{\cos^2(\theta)} = \dfrac{1}{\cos^2(\theta)}$$​​

The above holds whenever it is well-defined, therefore this is an identity.

Therefore, $$\underline{\tan^2(\theta) +1\equiv \dfrac{1}{\cos^2(\theta)}.}$$​