Trigonometric Identities

Trigonometric identities are fundamental tools for simplifying and solving problems related to angles in a right triangle. These formulas simplify complex equations, making them easier to handle in algebra, calculus, as well as geometry. Whether you're confirming an identity, simplifying expressions, or solving trigonometric equations, these trigonometric identities’ formulas form a solid base. This is why, here, we'll look into types of identities and how to use them.

1.0Trigonometric Identities Definition

A trigonometric identity can be defined as an equation that involves various trigonometric functions that are valid for all values of the involved variables as long as they are within the functions' domains. Trigonometric identities are basic mathematical tools, especially when dealing with angles and their connections. All of these trigonometric identities’ formulas are typically based on the six trigonometric ratios – sine, cosine, tangent, cosecant, secant, and cotangent. 

2.0List of Trigonometric Identities

In trigonometry, there are a large number of trigonometric identities, each classified in different categories with a specific purpose. Every trigonometric identity is used not only for various problems but also for solving trigonometric identities themselves. Here is the list of some of the most important trigonometric identities: 

Reciprocal Identities

Reciprocal identities are simply the relation between the six basic trigonometric functions mentioned above and their reciprocals. These include: 

• $\sin \theta =\frac{1}{\mathrm{cosec}\theta }$
• $\cos \theta =\frac{1}{\sec \theta }$
• $\tan \theta =\frac{1}{\cot \theta }$
• $\cot \theta =\frac{1}{\tan \theta }$
• $\mathrm{cosec}\theta =\frac{1}{\sin \theta }$
• $\sec \theta =\frac{1}{\cos \theta }$

Quotient Identities

These identities express the tangent and cotangent functions in terms of sine and cosine, like this: 

• $\tan \theta =\frac{\sin \theta }{\cos \theta }$
• $\cot \theta =\frac{\cos \theta }{\sin \theta }$

Complementary and Supplementary Function Identities

As the name suggests, these identities relate the trigonometric functions to their complementary and supplementary angles. Note that two angles are complementary if their sum is 90, while two angles are supplementary if their sum equals 180. 

Complementary Function Identities 

$\begin{array}{rl}& \sin (90-\theta )=\cos \theta \\ & \cos (90-\theta )=\sin \theta \\ & \tan (90-\theta )=\cot \theta \\ & \mathrm{cosec}(90-\theta )=\sec \theta \\ & \sec (90-\theta )=\mathrm{cosec}\theta \\ & \cot (90-\theta )=\tan \theta \end{array}$

Supplementary Function Identities 

$\begin{array}{rl}& \sin (180-\theta )=\sin \theta \\ & \cos (180-\theta )=-\cos \theta \\ & \tan (180-\theta )=-\tan \theta \\ & \mathrm{cosec}(180-\theta )=\mathrm{cosec}\theta \\ & \sec (180-\theta )=-\sec \theta \\ & \cot (180-\theta )=-\cot \theta \end{array}$

Trigonometric Identities for Sum and Difference of Any Two Angles

The formulas for the sum and difference enable us to determine the sine, cosine, and tangent of the sum or difference of two angles, say A and B. These are:

$\begin{array}{rl}& \begin{array}{rl}& \sin (A+B)=\sin A\cos B+\cos A\sin B\\ & \sin (A-B)=\sin A\cos B-\cos A\sin B\\ & \cos (A+B)=\cos A\cos B-\sin A\sin B\\ & \cos (A-B)=\cos A\cos B+\sin A\sin B\\ & \tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}\end{array}\\ & \tan (A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}\end{array}$

Product–Sum Trigonometric Identities 

These identities are the compound of the product and the sum of trigonometric ratios with different angles, say A and B. 

$\begin{array}{rl}& \sin A+\sin B=2\sin \left(\frac{A+B}{2}\right)\cdot \cos \left(\frac{A-B}{2}\right)\\ & \cos A+\cos B=2\cos \left(\frac{A+B}{2}\right)\cdot \cos \left(\frac{A-B}{2}\right)\\ & \sin A-\sin B=2\cos \left(\frac{A+B}{2}\right)\cdot \sin \left(\frac{A-B}{2}\right)\\ & \cos A-\cos B=-2\sin \left(\frac{A+B}{2}\right)\cdot \sin \left(\frac{A-B}{2}\right)\end{array}$

Trigonometric Identities of Products

$\begin{array}{rl}& 2\sin A\sin B=\cos (A-B)-\cos (A+B)\\ & 2\sin A\cos B=\sin (A-B)-\sin (A+B)\\ & 2\cos A\cos B=\cos (A+B)+\cos (A+B)\end{array}$

Pythagorean Identities 

These are the most commonly used trigonometric identities derived from the Pythagorean theorem. These include: 

$\begin{array}{rl}& {\sin }^2\theta +{\cos }^2\theta =1\\ & 1+{\tan }^2\theta ={\sec }^2\theta \\ & 1+{\cot }^2\theta ={\mathrm{cosec}}^2\theta \end{array}$

Double Angle Trigonometric Identities 

These are the identities used when the angle of trigonometric ratios is doubled, these identities are: 

$\begin{array}{rl}& \sin 2\theta =2\sin \theta \cos \theta =\frac{2\tan \theta }{1+{\tan }^2\theta }\\ & \cos 2\theta ={\cos }^2\theta -{\sin }^2\theta =2{\cos }^2\theta -1=1-{\sin }^2\theta =\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\\ & \tan 2\theta =\frac{2\tan \theta }{1-\tan \theta }\end{array}$

Half Angle Identities

These are the identities used when the angle of trigonometric ratios is halved; these identities are:

$\begin{array}{rl}& \sin \left(\frac{\theta }{2}\right)=\pm \sqrt{\frac{1-\cos \theta }{2}}\\ & \cos \left(\frac{\theta }{2}\right)=\pm \sqrt{\frac{1+\cos \theta }{2}}\\ & \tan \left(\frac{\theta }{2}\right)=\pm \sqrt{\frac{1-\cos \theta }{1+\cos \theta }}\end{array}$

Triple Angle Identities 

These are the identities used when the angle of trigonometric ratios are tripled, these identities are:

$\begin{array}{rl}\sin 3\theta & =3\sin \theta -4{\sin }^3\theta \\ \cos 3\theta & =4{\cos }^3\theta -3\cos \theta \\ \tan 3\theta & =\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }\end{array}$

3.0Proving Trigonometric Identities

In this section, we will focus mainly on proving the trigonometric identities of Pythagorean identity functions. These are the most important and frequently used identities, appearing in almost every complex problem of trigonometry.  

To derive these identities, first, we will take a right-angled triangle ABC, as shown in the figure here: 

Apply Pythagoras' theorem in triangle ABC, which states: 

$A{C}^2=A{B}^2+B{C}^2\dots ..(\mathrm{a})$

For Pythagorean Identity 1: 

Divide equation (a) by AC²; we will have: 

$\begin{array}{rl}& \frac{A{C}^2}{A{C}^2}=\frac{A{B}^2}{A{C}^2}+\frac{B{C}^2}{A{C}^2}\\ & \frac{A{B}^2}{A{C}^2}+\frac{B{C}^2}{A{C}^2}=1\\ & {\left(\frac{AB}{AC}\right)}^2+{\left(\frac{BC}{AC}\right)}^2=1\end{array}$

Since, 

$\begin{array}{rl}& \sin \theta =\frac{\text{ Perpendicular }}{\text{ Hypotenuse }}=\frac{BC}{AC}\\ & \cos \theta =\frac{\text{ Base }}{\text{ Hypotenuse }}=\frac{AB}{AC}\end{array}$

Hence, equation 1 will become: 

$\begin{array}{rl}& (\cos \theta {)}^2+(\sin \theta {)}^2=1\\ & \text{ Or, }{\sin }^2\theta +{\cos }^2\theta =1\\ & (\cos \theta {)}^2+(\sin \theta {)}^2=1\end{array}$

For Pythagorean Identity 2: 

Divide the equation (a) by AB² 

$\begin{array}{rl}& \begin{array}{rl}& \frac{A{C}^2}{A{B}^2}=\frac{A{B}^2}{A{B}^2}+\frac{B{C}^2}{A{B}^2}\\ & {\left(\frac{AC}{AB}\right)}^2=1+{\left(\frac{BC}{AC}\right)}^2\dots \end{array}\\ & \text{ Since, }\\ & \begin{array}{rl}& \tan \theta =\frac{\text{ Perpendicular }}{\text{ Base }}=\frac{BC}{AB}\\ & \sec \theta =\frac{\text{ Hypotenuse }}{\text{ Base }}=\frac{AC}{AB}\end{array}\end{array}$

For Pythagorean Identity 3: 

Divide equation (a) by BC²

$\begin{array}{rl}& \begin{array}{rl}& \frac{A{C}^2}{B{C}^2}=\frac{A{B}^2}{B{C}^2}+\frac{B{C}^2}{B{C}^2}\\ & {\left(\frac{AC}{BC}\right)}^2=1+{\left(\frac{AB}{BC}\right)}^2\dots \end{array}\\ & \text{ Since, }\\ & \begin{array}{rl}& \mathrm{cosec}\theta =\frac{\text{ Hypotenuse }}{\text{ Perpendicular }}=\frac{AC}{BC}\\ & \cot \theta =\frac{\text{ Base }}{\text{ Perpendicular }}=\frac{AB}{BC}\end{array}\end{array}$

Hence, equation 3 will become, 

${\mathrm{cosec}}^2\theta =1+{\cot }^2\theta$

4.0Trigonometric Identities Derivatives

The derivatives of trigonometric identities are another set of important formulas in trigonometry as well as calculus. See the below-mentioned trigonometric identities table for these derivatives to understand them better: 

5.0Solved Examples of Trigonometric Identities 

Problem 1: If $\sec \theta +\tan \theta =p$, then find the value of $\sec \theta -\tan \theta$. 

Solution: From Identity 2, we have: 

$\begin{array}{rl}& {\sec }^2\theta =1+{\tan }^2\theta \\ & {\sec }^2\theta -{\tan }^2\theta =1\\ & (\sec \theta +\tan \theta )(\sec \theta -\tan \theta )=1\\ & p(\sec \theta -\tan \theta )=1\\ & \text{ Hence, }\sec \theta -\tan \theta =\frac{1}{p}\end{array}$

Problem 2: Prove that: $\frac{1-{\cos }^2\theta }{\sin \theta }=\sin \theta$

Solution: From identity 1, we have: 

$\begin{array}{rl}& {\sin }^2\theta +{\cos }^2\theta =1\\ & {\sin }^2\theta =1-{\cos }^2\theta \end{array}$

Taking the LHS of the equation

$\frac{1-{\cos }^2\theta }{\sin \theta }$

From equation 1: 

$\frac{{\sin }^2\theta }{\sin \theta }=\sin \theta$

LHS = RHS, hence proved. 

Problem 3: Simplify the Expression $\frac{1+{\tan }^2\theta }{1+{\cot }^2\theta }$.

Solution: From identities 2 and 3: 

$\begin{array}{rl}& 1+{\tan }^2\theta ={\sec }^2\theta \\ & 1+{\cot }^2\theta ={\mathrm{cosec}}^2\theta \end{array}$

Now, taking the comparison of the equation with these identities, we get: 

$\begin{array}{rl}& \frac{{\sec }^2\theta }{{\mathrm{cosec}}^2\theta }=\frac{\frac{1}{{\cos }^2\theta }}{\frac{1}{{\sin }^2\theta }}=\frac{{\sin }^2\theta }{{\cos }^2\theta }\\ & ={\left(\frac{\sin \theta }{\cos \theta }\right)}^2\dots \dots (1)\end{array}$

From quotient identities: 

$\tan \theta =\frac{\sin \theta }{\cos \theta }\dots (2)$

From equations 1 and 2: 

$\frac{1+{\tan }^2\theta }{1+{\cot }^2\theta }={\tan }^2\theta$