In Trigonometry , Let triangle $\Delta ABC$ be a right-angled triangle, right-angled at $\angle B$.
Let $\theta = \angle CAB$. Then $\theta$ is an acute angle. Then, we have
$\sin \theta =$ $ \frac{Opposite\;Side}{Hypotenuse} $ $=$ $ \frac{a}{b}$,
$\cos \theta =$ $ \frac{Adjacent\;Side}{Hypotenuse} $ $= $ $\frac{c}{b}$,
$\tan \theta =$ $ \frac{Opposite\;Side}{Adjacent\;Side}$ $ =$ $ \frac{a}{c}$,
$\cot \theta =$ $ \frac{Adjacent\;Side}{Opposite\;Side} $ $=$ $ \frac{c}{a}$,
$\csc \theta =$ $ \frac{Hypotenuse}{Opposite\;Side} $ $=$ $ \frac{b}{a}$, and
$\sec \theta =$ $ \frac{Hypotenuse}{Adjacent\;Side}$ $ =$ $ \frac{b}{c}$.
Reciprocal Relations between Certain Ratios
We have several relations which we use to solve trigonometry problems , let's look at them
$\sin \theta \times \csc \theta = $ $\frac{a}{b} \times \frac{b}{a} $ $ = 1$,
$\cos \theta \times \sec \theta = $ $\frac{c}{b} \times \frac{b}{c} $ $ = 1$, and
$\tan \theta \times \cot \theta = $ $\frac{a}{c} \times \frac{c}{a}$ $ = 1$.
The above results lead to the facts that
$\sin \theta$ and $\csc \theta$ are reciprocals,
$\cos \theta$ and $\sec \theta$ are reciprocals, and
$\tan \theta$ and $\cot \theta$ are reciprocals.
Therefore,
$\sin \theta \times \csc \theta = 1$ $\Longrightarrow \quad \sin \theta = $ $\frac{1}{\csc \theta}\quad$, and $\quad \csc \theta = $ $\frac{1}{\sin \theta}$. |
Similarly,
$\cos \theta \times \sec \theta = 1$ $\Longrightarrow \quad \cos \theta = $ $\frac{1}{\sec \theta} \quad$, and $\quad \sec \theta = $ $\frac{1}{\cos \theta}$ |
and
$\tan \theta \times \cot \theta = 1$ $\Longrightarrow \quad \tan \theta = $ $\frac{1}{\cot \theta}\quad $, and $\quad \cot \theta = $ $\frac{1}{\tan \theta}$. |
Expressing $; An Heta ;$ and $; Cot Heta ; $ in Terms of $; Sin Heta ;$ and $; Cos Heta$.
We have
$\frac{\sin \theta}{\cos \theta}$ $ = $ $\frac{\left(\frac{a}{b}\right)}{\left(\frac{c}{b}\right)}$ $=$ $\frac{a}{b}$$\times$$\frac{b}{c}$ $= $ $\frac{a}{c}$ $=\tan \theta$.
Similarly,
$\frac{\cos \theta}{\sin \theta}$ $=$ $\frac{\left(\frac{c}{b}\right)}{\left(\frac{a}{b}\right)}$ $=$ $\frac{c}{b}$ $\times$ $\frac{b}{a}$ $=$ $\frac{c}{a}$ $=\cot \theta$.
Therefore, we have
$\tan \theta$ $=$ $\frac{\sin \theta}{\cos \theta}$ $\Longrightarrow \quad \sin \theta = \tan \theta . \cos \theta \quad $ and $\quad \cos \theta = $ $\frac{\sin \theta}{\tan \theta}$. |
and
$\cot \theta$ $=$ $\frac{\cos \theta}{\sin \theta}$ $\Longrightarrow \quad \cos \theta = \cot \theta . \sin \theta \quad$ and $\quad \sin \theta = $ $\frac{\cos \theta}{\cot \theta}$. |
Euclid's Elements, Book 1, Proposition 47, Pythagorean Theorem
In right-angled triangles the square on the side subtending the right angle is equal to the squares on the sides containing the right angle.
Let $\Delta ABC$ be a right-angled triangle, right-angled at $\angle B$.
Then Pythagorean Theorem says that the sum of the areas of squares on the sides $AB$ and $BC$ is equal to the area of the square on the side $AC$. That is,
$AB^2 + BC^2 = AC^2$
$\Longrightarrow \quad c^2 + a^2 = b^2$.
This theorem implies to the following trigonometric identities:
We have
$a^2 + c^2 = b^2$
Dividing on both sides by $b^2$, we have
$\Longrightarrow \quad $ $\frac{a^2}{b^2} $ $+$ $\frac{c^2}{b^2}$ $ = 1$
$\Longrightarrow \quad $ $\left(\frac{a}{b}\right)^2$ $+$ $\left(\frac{c}{b}\right)^2$ $ = 1$
$\Longrightarrow \quad \sin^2 \theta + \cos^2 \theta = 1.$
So, we have the following identities:
$\sin^2 \theta + \cos^2 \theta = 1$ This implies that $\sin^2 \theta = 1 - \cos^2 \theta \quad \Longrightarrow \quad \sin \theta = \sqrt{1 - \cos^2 \theta} \quad$, and $\cos^2 \theta = 1 - \sin^2 \theta \quad \Longrightarrow \quad \cos \theta = \sqrt{1 - \sin^2 \theta}$. |
Again, $a^2 + c^2 = b^2$ implies that $b^2 - a^2 = c^2$.
Dividing on both sides by $c^2$ we have
$\Longrightarrow \quad $ $\frac{b^2}{c^2} $ $-$ $\frac{a^2}{c^2}$ $ = 1$
$\Longrightarrow \quad $ $\left(\frac{b}{c}\right)^2$ $-$ $\left(\frac{a}{c}\right)^2$ $ = 1$
$\Longrightarrow \quad \sec^2 \theta - \tan^2 \theta = 1.$
So, we have the following identities:
$\sec^2 \theta - \tan^2 \theta = 1$ This implies that $\sec^2 \theta = 1 + \tan^2 \theta \quad \Longrightarrow \quad \sec \theta = \sqrt{1 + \tan^2 \theta} \quad$, and $\tan^2 \theta = \sec^2 \theta - 1 \quad \Longrightarrow \quad \tan \theta = \sqrt{\sec^2 \theta - 1}$. |
Again, $a^2 + c^2 = b^2$ implies that $b^2 - c^2 = a^2$.
Dividing on both sides by $a^2$ we have
$\Longrightarrow \quad $ $\frac{b^2}{a^2} $ $-$ $\frac{c^2}{a^2}$ $ = 1$
$\Longrightarrow \quad $ $\left(\frac{b}{a}\right)^2$ $-$ $\left(\frac{c}{a}\right)^2$ $ = 1$
$\Longrightarrow \quad \csc^2 \theta - \cot^2 \theta = 1.$
So, we have the following identities:
$\csc^2 \theta - \cot^2 \theta = 1$ This implies that $\csc^2 \theta = 1 + \cot^2 \theta \quad \Longrightarrow \quad \csc \theta = \sqrt{1 + \cot^2 \theta} \quad$, and $\cot^2 \theta = \csc^2 \theta - 1 \quad \Longrightarrow \quad \cot \theta = \sqrt{\csc^2 \theta - 1}$. |
We can express the trigonometric ratio of an angle in terms of any one of them.
To Express all the Trigonometrical Ratios in Terms of $; Sin Heta$.
We have
$\cos \theta = \sqrt{1 - \sin^2 \theta}$,
$\tan \theta = $ $\frac{\sin \theta}{\cos \theta} $ $=$ $\frac{\sin \theta}{\sqrt{1 - \sin^2 \theta}}$,
$\cot \theta = $ $ \frac{\cos \theta}{\sin \theta} $ $=$ $\frac{\sqrt{1 - \sin^2 \theta}}{\sin \theta}$,
$\sec \theta = $ $\frac{1}{\cos \theta}$ $=$ $\frac{1}{\sqrt{1- \sin^2 \theta}}$,
$\csc \theta = $ $\frac{1}{\sin \theta}$.
To Express all the Trigonometrical Ratios in Terms of $;cos Heta$.
We have
$\sin \theta = \sqrt{1 - \cos^2 \theta}$,
$\tan \theta = $ $ \frac{\sin \theta}{\cos \theta} $ $=$ $\frac{\sqrt{1 - \cos^2 \theta}}{\cos \theta}$,
$\cot \theta = $ $\frac{\cos \theta}{\sin \theta} $ $=$ $\frac{\cos \theta}{\sqrt{1 - \cos^2 \theta}}$,
$\sec \theta = $ $\frac{1}{\cos \theta}$,
$\csc \theta = $ $\frac{1}{\sin \theta}$ $=$ $\frac{1}{\sqrt{1- \cos^2 \theta}}$.
To Express all the Trigonometrical Ratios in Terms of $; An Heta$.
We have
$\sec \theta = \sqrt{1 + \tan^2 \theta}$,
$\cos \theta = $ $\frac{1}{\sec \theta}$ $=$ $\frac{1}{\sqrt{1+ \tan^2 \theta}}$,
$\sin \theta = \tan \theta . \cos \theta = $ $\frac{\tan \theta}{\sqrt{1 + \tan^2 \theta}}$,
$\cot \theta = $ $\frac{1}{\tan \theta}$,
$\csc \theta = $ $\frac{1}{\sin \theta}$ $=$ $\frac{\sqrt{1 + \tan^2 \theta}}{\tan \theta}$.
To Express all the Trigonometrical Ratios in Terms of $; Cot Heta$.
We have
$\csc \theta = \sqrt{1 + \cot^2 \theta}$,
$\sin \theta = $ $\frac{1}{\csc \theta}$ $=$ $\frac{1}{\sqrt{1+ \cot^2 \theta}}$,
$\cos \theta = \cot \theta . \sin \theta = $ $\frac{\cot \theta}{\sqrt{1 + \cot^2 \theta}}$,
$\tan \theta = $ $\frac{1}{\cot \theta}$,
$\sec \theta = $ $\frac{1}{\cos \theta}$ $=$ $\frac{\sqrt{1 + \cot^2 \theta}}{\cot \theta}$.
To Express all the Trigonometrical Ratios in Terms of $; Sec Heta$.
We have
$\tan \theta = \sqrt{\sec^2 \theta - 1}$,
$\cot \theta = $ $\frac{1}{\tan \theta}$ $=$ $\frac{1}{\sqrt{\sec^2 \theta - 1}}$,
$\cos \theta = $ $\frac{1}{\sec \theta}$,
$\sin \theta = \tan \theta . \cos \theta = $ $\frac{\sqrt{\sec^2 \theta -1}}{\sec \theta}$,
$\csc \theta = $ $\frac{1}{\sin \theta}$ $=$ $\frac{\sec \theta}{\sqrt{\sec^2 \theta - 1}}$.
To Express all the Trigonometrical Ratios in Terms of $; Csc Heta$.
We have
$\cot \theta = \sqrt\csc^2 \theta - 1$,
$\tan \theta = $ $\frac1\cot \theta$ $=$ $\frac1\sqrt\csc^2 \theta - 1$,
$\sin \theta = $ $\frac1\csc \theta$,
$\cos \theta = \cot \theta . \sin \theta = $ $\frac\sqrt\csc^2 \theta -1\csc \theta$ ,
$\sec \theta = $ $\frac1\cos \theta$ $=$ $\frac\csc \theta\sqrt\csc^2 \theta - 1$ .
The above illustrations are included in the following table :
Related Tags
Explain Basic Trigonometric Identities , What are the Basic Trigonometric Identities , Introduction to Basic Trigonometric Identities