Last modified on January 29th, 2025

chapter outline

 

Equivalent Representations of Trigonometric Functions

Equivalent representations of trigonometric functions refer to different ways of expressing the functions. They are done using trigonometric identities and transformations (phase shifts).

For example, the sine function can be expressed as a phase-shifted cosine function:

sin x = cos(x – 90°)

This equivalence arises from the properties of the unit circle and periodicity of trigonometric functions. 

Such equivalences are used in simplifying equations, graphing functions, and analyzing their periodic behavior.  

For example,

sin x = 0.5 

⇒ x = sin-1 0.5 or x = π – sin-1 0.5

Using Trigonometric Identities

Trigonometric identities provide alternative forms of the same function that simplify expressions or calculations. 

Pythagorean Identities

  • sin2 θ + cos2 θ = 1
  • 1 + tan2 θ = sec2 θ
  • 1 + cot2 θ = cosec2 θ

By rewriting, sin2 θ can be expressed as sin2 θ = 1 – cos2 θ

Reciprocal Identities

  • ${\sin \theta =\dfrac{1}{\text{cosec}\, \theta }}$
  • ${\text{cosec}\, \theta =\dfrac{1}{\sin \theta }}$
  • ${\cos \theta =\dfrac{1}{\sec \theta }}$
  • ${\sec \theta =\dfrac{1}{\cos \theta }}$
  • ${\tan \theta =\dfrac{1}{\cot \theta }}$
  • ${\cot \theta =\dfrac{1}{\tan \theta }}$

Quotient Identities

  • tan θ = ${\dfrac{\sin \theta }{\cos \theta }}$
  • cot θ = ${\dfrac{\cos \theta }{\sin \theta }}$

Sum and Difference of Angles Identities

  • 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) = ${\dfrac{\tan A+\tan B}{1-\tan A\tan B}}$
  • tan (A – B) = ${\dfrac{\tan A-\tan B}{1+\tan A\tan B}}$

Double Angle Identities

  • sin 2θ = 2sin θ ⋅ cos θ = ${\dfrac{2\tan \theta }{1-\tan ^{2}\theta }}$
  • cos 2θ = cos2 θ – sin2 θ = ${\dfrac{1-\tan ^{2}\theta }{1+\tan ^{2}\theta }}$
  • cos 2θ = 2cos2 θ – 1 = 1 – 2 sin2 θ
  • tan 2θ = ${\dfrac{2\tan \theta }{1+\tan ^{2}\theta }}$
  • sec 2θ = ${\dfrac{\sec ^{2}\theta }{2-\sec ^{2}\theta }}$
  • cosec 2θ = ${\dfrac{\sec \theta \cdot \text{cosec}\, \theta }{2}}$
  • cot 2θ = ${\dfrac{\cot ^{2}\theta -1}{2\cot \theta }}$

Through Phase Shifts

All six trigonometric functions are related through phase shifts. Through a phase shift, one trigonometric function can be expressed in terms of another.

sin θ = cos(θ – 90°)

cos θ = sin(θ + 90°)

These relationships show that the sine and cosine functions are essentially the same curve, shifted by ${\dfrac{\pi }{2}}$ radians (90 degrees). 

Besides the functions being periodic, they repeat every 2π and thus are related as follows: 

sin θ = sin(2nπ + θ)

cos θ = cos (2nπ + θ)

tan θ = tan (nπ + θ)

cot θ = cot (nπ + θ)

sec θ = sec (2nπ + θ)

cosec θ = cosec (2nπ + θ) 

Solved Examples

What is the value of sin(3x + 45)° if sin x = 0.5 and cos x  = 0.866

Solution:

As we know, the sum of angles identities: sin (A + B) = sin A cos B + cos A sin B
Here, 
sin(3x + 45)° 
= sin 3x cos 45° + cos 3x sin 45° …..(i)
Since sin 45° = cos 45° = ${\dfrac{1}{\sqrt{2}}\approx 0.707}$
As we know, the triple angles identities: 
sin 3θ = 3 sin θ – 4 sin3 θ 
cos 3θ = 4 cos3 θ – 3 cos θ
Given sin x = 0.5 and cos x = 0.866
From (i), we get
sin(3x + 45)° 
= (3 sin x – 4 sin3 x)0.707 + (4 cos3 x – 3 cos x)0.707
= [(3)(0.5) – (4)(0.5)3]0.707 + [(4)(0.866)3 – (3)(0.866)]0.707
= [1.5 – 0.5]0.707 + [2.598 – 2.598]0.707
= 0.707 
Thus, sin(3x + 45)° = 0.707

Solve: cot x sin2 x = 3 cot x

Solution:

Given, cot x sin2 x = 3 cot x
⇒ cot x sin2 x – 3 cot x = 0
⇒ cot x(sin2 x – 3) = 0
Either cot x = 0 or sin2 x – 3 = 0
If cot x = 0, then
tan x = ∞
⇒ x = ${\dfrac{\pi }{2}+n\pi}$, n ∈ ℤ
If sin2 x – 3 = 0
⇒ sin2 x = 3, which is not possible since -1 ≤ sin x ≤ 1
Thus, x = ${\dfrac{\pi }{2}+n\pi}$, n ∈ ℤ

Evaluate ${\sin \left( \dfrac{\pi }{6}+\dfrac{\pi }{3}\right)}$ and express the answer in an equivalent algebraic form.

Solution:

As we know, the sum of angles identities: sin (A + B) = sin A cos B + cos A sin B
Here, 
${\sin \left( \dfrac{\pi }{6}+\dfrac{\pi }{3}\right)}$
= ${\sin \dfrac{\pi }{6}\cos \dfrac{\pi }{3}+\cos \dfrac{\pi }{6}\sin \dfrac{\pi }{3}}$ …..(i)
As we know, 
${\sin \dfrac{\pi }{6}}$ = ${\cos \dfrac{\pi }{3}}$ = ${\dfrac{1}{2}}$
${\cos \dfrac{\pi }{6}}$ = ${\sin \dfrac{\pi }{3}}$ = ${\dfrac{\sqrt{3}}{2}}$
Here, from (i), we get
= ${\dfrac{1}{2}\times \dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\times \dfrac{\sqrt{3}}{2}}$
= ${\dfrac{1}{4}+\dfrac{3}{4}}$
= ${1}$
Thus, an equivalent representation of ${\sin \left( \dfrac{\pi }{6}+\dfrac{\pi }{3}\right)}$ is 1

Last modified on January 29th, 2025