Trigonometry
Defining Sine, Cosine, and Tangent.
Table of contents
1. Sine & Cosecant
Consider a right triangle such that the side $AB$ stands perpendicular on side $BC$ while the hypotenuse $AC$ makes an angle $\theta$ with the base.
By definition, the sine of $\theta$ in the given right triangle is the ratio of the side opposite to the angle ($AB$ in this case) to the hypotenuse of the triangle ($AC$ in this case): $$\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}} = \frac{AB}{AC}$$
The reciprocal of sine is called cosecant, denoted $\csc\theta$: $$\begin{align} \csc \theta &= \frac{1}{\sin\theta} \\ &= \frac{\text{hypotenuse}}{\text{opposite}}\\ &= \frac{AC}{AB} \end{align}$$ $$\boxed{\therefore \csc \theta = \frac{AC}{AB}}$$
2. Cosine & Secant
By definition, the cosine of $\theta$ in the above right triangle is the ratio of the side adjacent to the angle ($BC$ in this case) to the hypotenuse of the triangle ($AC$ in this case): $$\Rightarrow \cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}} = \frac{BC}{AC}$$
The Secant of $\theta$ is the reciprocal of cosine: $$\begin{align} \sec \theta &= \frac{1}{\cos \theta}\\ &= \frac{\text{hypotenuse}}{\text{adjacent}}\\ &= \frac{AC}{BC} \end{align}$$ $$\boxed{\therefore \sec \theta = \frac{AC}{BC}}$$
3. Tangent & Cotangent
By definition, the tangent of $\theta$ in the above right triangle is the ration of the side opposite to the angle to the side adjacent to the angle: $$\Rightarrow \tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{\text{height}}{\text{base}}$$ $$\Rightarrow \tan \theta = \frac{AB}{BC}$$ The above fraction can be rewritten as: $$\begin{align} \tan \theta &= \frac{AB}{BC}\\ &= \frac{AB/AC}{BC/AC}\\ &= \frac{\sin \theta}{\cos \theta} \end{align}$$ $$\boxed{\therefore \tan \theta = \frac{\sin \theta}{\cos \theta}}$$
The Cotangent of $\theta$ is simply the reciprocal of tangent: $$\Rightarrow \cot \theta = \frac{1}{\tan \theta}$$ Therefore, in terms of the sides of the triangle: $$\cot \theta = \frac{BC}{AB}$$ And, in terms of sine and cosine: $$\cot \theta = \frac{\cos \theta}{\sin \theta}$$
4. Trigonometric identities
$\Rightarrow \sin^2 \theta + \cos^2 \theta = 1:$
Using the Pythagorean Theorem on the right triangle in Fig 1, we have: $$AB^2 + BC^2 = AC^2$$ Let us divide both sides of the above equation by $AC^2$. This gives: $$\frac{AB^2}{AC^2} + \frac{BC^2}{AC^2} = \frac{AC^2}{AC^2}$$ $$\therefore \left (\frac{AB}{AC}\right )^2 + \left (\frac{BC}{AC}\right )^2 = 1$$ From the definitions of sine and cosine, we know that $\frac{AB}{AC} = \sin \theta$ and $\frac{BC}{AC} = \cos \theta$. Substituting these into the above equation, we get: $$\boxed{\sin^2 \theta + \cos^2 \theta = 1}$$ The above identity can be rearranged to get the following forms: $$\boxed{\cos^2 \theta = 1 - \sin^2 \theta}$$ $$\boxed{\sin^2\theta = 1 - \cos^2\theta}$$
$\Rightarrow \tan^2 \theta + 1 = \sec^2\theta:$
Starting from the Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$, we can derive the identity involving tangent and secant by diving each side by $\cos^2 \theta$ as shown below: $$\begin{align} \sin^2 \theta + \cos^2 \theta &= 1\\ \therefore \frac{\sin^2 \theta + \cos^2 \theta}{\cos^2 \theta} &= \frac{1}{\cos^2\theta}\\ \therefore \frac{\sin^2 \theta}{\cos^2 \theta} + \frac{\cos^2 \theta}{\cos^2 \theta} &= \frac{1}{\cos^2 \theta} \end{align}$$ $$\therefore \boxed{\tan^2 \theta + 1 = \sec^2 \theta}$$
$\Rightarrow 1 + \cot^2\theta = \csc^2\theta:$
In the previous derivation we divided the Pythagorean identity by $\cos^2 \theta$. Let us now divide the same identity by $\sin^2 \theta$. This gives: $$\frac{\sin^2 \theta}{\sin^2 \theta} + \frac{\cos^2 \theta}{\sin^2 \theta} = \frac{1}{\sin^2 \theta}$$ $$\therefore \boxed{1 + \cot^2 \theta = \csc^2 \theta}$$