Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that hold true for all values of the variables where the functions are defined.

• Describes the relationships among sine, cosine, tangent, and the other trigonometric functions.
• Fundamental tools for simplifying expressions, proving formulas, and solving trigonometric equations.
• Widely used in fields like geometry, engineering, and physics.

These are useful whenever trigonometric functions are involved in an expression or an equation. The six basic trigonometric ratios are sine, cosine, tangent, cosecant, secant, and cotangent. All these trigonometric ratios are defined using the sides of the right triangle, such as the adjacent side, the opposite side, and the hypotenuse.

List of Trigonometric Identities

There are a lot of identities in the study of Trigonometry, which involve all the trigonometric ratios. These identities are used to solve various problems throughout the academic landscape as well as the real life. Let us learn all the fundamental and advanced trigonometric identities.

Reciprocal Trigonometric Identities

In all trigonometric ratios, there is a reciprocal relation between a pair of ratios, which is given as follows:

• sin θ = 1/cosec θ
• cosec θ = 1/sin θ
   
• cos θ = 1/sec θ 
• sec θ = 1/cos θ
   
• tan θ = 1/cot θ
• cot θ = 1/tan θ

Pythagorean Trigonometric Identities

Pythagorean Trigonometric Identities are based on the right-triangle theorem or Pythagoras' theorem, and are as follows:

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

Trigonometric Ratio Identities

As tan and cot are defined as the ratio of sin and cos, which is given by the following identities:

• tan θ = sin θ/cos θ
• cot θ = cos θ/sin θ

Trigonometric Identities of Opposite Angles

In trigonometry, angles measured in the clockwise direction are measured in negative parity, and all trigonometric ratios defined for negative parity of angle are defined as follows:

• sin (-θ) = -sin θ
• cos (-θ) = cos θ
• tan (-θ) = -tan θ
• cot (-θ) = -cot θ
• sec (-θ) = sec θ
• cosec (-θ) = -cosec θ

Complementary Angles Identities

Complementary angles are a pair of angles whose measures add up to 90°. Now, the trigonometric identities for complementary angles are as follows:

• sin (90° – θ) = cos θ
• cos (90° – θ) = sin θ
• tan (90° – θ) = cot θ
• cot (90° – θ) = tan θ
• sec (90° – θ) = cosec θ
• cosec (90° – θ) = sec θ

Supplementary Angles Identities

Supplementary angles are a pair of angles whose measures add up to 180°. Now, the trigonometric identities for supplementary angles are:

• sin (180°- θ) = sinθ
• cos (180°- θ) = -cos θ
• cosec (180°- θ) = cosec θ
• sec (180°- θ)= -sec θ
• tan (180°- θ) = -tan θ
• cot (180°- θ) = -cot θ

Periodicity of Trigonometric Function

Trigonometric functions such as sin, cos, tan, cot, sec, and cosec are all periodic and have different periodicities. The following identities for the trigonometric ratios explain their periodicity.

• sin (n × 360° + θ) = sin θ
• sin (2nπ + θ) = sin θ
   
• cos (n × 360° + θ) = cos θ
• cos (2nπ + θ) = cos θ
   
• tan (n × 180° + θ) = tan θ
• tan (nπ + θ) = tan θ
   
• cosec (n × 360° + θ) = cosec θ
• cosec (2nπ + θ) = cosec θ
   
• sec (n × 360° + θ) = sec θ
• sec (2nπ + θ) = sec θ
   
• cot (n × 180° + θ) = cot θ
• cot (nπ + θ) = cot θ

Where, n ∈ Z, (Z = set of all integers)

Note: sin, cos, cosec, and sec have a period of 360° or 2π radians, and for tan and cot period is 180° or π radians.

Sum and Difference Identities

Trigonometric identities for the Sum and Difference of angles include formulas such as sin(A+B), cos(A-B), tan(A+B), etc.

• 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) = (tan A + tan B)/(1 - tan A tan B)
• tan (A-B) = (tan A - tan B)/(1 + tan A tan B) 

Note: Identities for sin (A+B), sin (A-B), cos (A+B), and cos (A-B) are called Ptolemy’s Identities.

Double Angle Identities

Using the trigonometric identities of the sum of angles, we can find a new identity, which is called the Double Angle Identities. To find these identities, we can put A = B in the sum of angle identities. For example,

a  we know, sin (A+B) = sin A cos B + cos A sin B

Substitute A = B = θ on both sides here, and we get:

sin (θ + θ) = sinθ cosθ + cosθ sinθ

• sin 2θ = 2 sinθ cosθ

Similarly,

• cos 2θ = cos²θ - sin ²θ = 2 cos ² θ - 1 = 1 - 2sin ² θ
• tan 2θ = (2tanθ)/(1 - tan²θ)

Half-Angle Formulas

Using double-angle formulas, half-angle formulas can be calculated. To calculate the angle formula, replace θ with θ/2, then,

• sin \frac{\theta}{2} = \pm\sqrt{\frac{1-cos \theta}{2}}

• \cos \frac{\theta}{2} = \pm \sqrt{\frac{1 + \cos \theta}{2}}

• \tan \frac{\theta}{2} = \pm \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}} = \frac{\sin \theta}{1 + \cos \theta} = \frac{1 - \cos \theta}{\sin \theta}

Other than the above-mentioned identities, there are some more half-angle identities, which are as follows:

• \sin \theta = \dfrac{2 \tan \tfrac{\theta}{2}}{1 + \tan^2 \tfrac{\theta}{2}}

• \cos \theta = \dfrac{1 - \tan^2 \tfrac{\theta}{2}}{1 + \tan^2 \tfrac{\theta}{2}}

• \tan \theta = \dfrac{2 \tan \tfrac{\theta}{2}}{1 - \tan^2 \tfrac{\theta}{2}}

Product-Sum Identities

The following identities state the relationship between the sum of two trigonometric ratios with the product of two trigonometric ratios.

• \sin A + \sin B = 2 \sin \frac{A + B}{2} \cos \frac{A - B}{2}

• \cos A + \cos B = 2 \cos \frac{A + B}{2} \cos \frac{A - B}{2}

• \sin A - \sin B = 2 \cos \frac{A + B}{2} \sin \frac{A - B}{2}

• \cos A - \cos B = -2 \sin \frac{A + B}{2} \sin \frac{A - B}{2}

Products Identities

Product Identities are formed when we add two of the sum and difference of angle identities, and are as follows:

• \sin A \cos B = \frac{\sin (A + B) + \sin (A - B)}{2}

• \cos A \cos B = \frac{\cos (A + B) + \cos (A - B)}{2}

• \sin A \sin B = \frac{\cos (A - B) - \cos (A + B)}{2}

Triple Angle Formulas

Other than double and half-angle formulas, there are identities for trigonometric ratios that are defined for triple angles. These triple-angle identities are as follows:

• \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}

Proof of the Trigonometric Identities

For any acute angle θ, prove that 

1. tanθ = sinθ/cosθ
2. cotθ = cosθ/sinθ
3. tanθ . cotθ = 1
4. sin²θ + cos²θ = 1
5. 1 + tan²θ = sec²θ
6. 1 + cot²θ = cosec²θ

Proof:

Consider a right-angled △ABC in which ∠B = 90°

Let AB = x units, BC = y units and AC = r units. 

Then, 

(1) tanθ = P/B = y/x = (y/r) / (x/r) 

∴ tanθ = sinθ/cosθ 

(2) cotθ = B/P = x/y = (x/r) / (y/r)

∴ cotθ = cosθ/sinθ 

(3) tanθ . cotθ  = (sinθ/cosθ) . (cosθ/sinθ) 

tanθ . cotθ = 1 

Then, by Pythagoras' theorem, we have x² + y² = r². 

Now, 

(4) sin²θ + cos²θ  = (y/r)² + (x/r)² = ( y²/r² + x²/r²)

                              = (x² + y²)/r² = r²/r² = 1 [x²+ y² = r²]

sin²θ + cos²θ = 1

(5) 1 + tan²θ = 1 + (y/x)² = 1 + y²/x² = (y² + x²)/x² = r²/x² [x² + y² = r²]

(r/x)² = sec²θ 

∴ 1 + tan²θ = sec²θ.

(6) 1 + cot²θ = 1 + (x/y)² = 1 + x²/y² = (x² + y²)/y² = r²/y² [x² + y² = r²]

(r²/y²) = cosec²θ

∴ 1 + cot²θ = cosec²θ

Also Check

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• Trigonometry Practice Questions Medium
  
• Trigonometry Practice Questions Hard

Summarizing Trigonometric Identities

Suggested Quiz

10 Questions

If sec⁡ A + tan ⁡A = x, then the value of sec⁡ A − tan ⁡A is

• A

  1/x

• B

  x

• C

  1/x²

• D

  x²

Explanation:

As we know, sec²A − tan²A = 1

Also, a² - b² = (a + b)(a - b)

Combining both, we get

(sec A + tan A)(sec A − tan A) = 1

⇒ x · (sec A − tan A) = 1

⇒ sec A − tan A = 1/x

If sin⁡ θ + cos ⁡θ = √2​, then sin⁡³θ + cos⁡³θ is equal to:

• A

  3√2/2

• B

  √2

• C

  √2/2

• D

  2

Explanation:

Given sin⁡ θ + cos ⁡θ = √2

⁡Recall that sin²θ + cos⁡²θ = 1.

Now, [Tex]\sin^3 \theta + \cos^3 \theta = (\sin \theta + \cos \theta)(\sin^2 \theta - \sin \theta \cos \theta + \cos^2 \theta) = (\sqrt{2})(1 - \sin \theta \cos \theta)[/Tex]

Now, sin⁡ θ + cos ⁡θ = √2

Square both sides;

sin^⁡2⁡θ + cos²θ + 2 sin⁡θ cos⁡θ = 2

⇒ 1 + 2sin⁡θcos⁡θ = 2

⇒ sin⁡ θ cos ⁡θ = 1/2

Thus, [Tex]\sin^3 \theta + \cos^3 \theta = \sqrt{2}(1 - \frac{1}{2}) = \sqrt{2} \cdot \frac{1}{2} = \frac{\sqrt{2}}{2}[/Tex]

Find the value of sin⁶ 𝛉 + cos⁶ 𝛉 in terms of cos⁡2θ.

• A

  [Tex]\frac{1}{4}(1 - 3\sin^2 2\theta)[/Tex]

• B

  [Tex]\frac{1}{4}(1 - 3\cos^2 2\theta)[/Tex]

• C

  [Tex]1 + \frac{3}{4} \cos^2 2\theta[/Tex]

• D

  [Tex]1 - \frac{3}{8} \cos^2 2\theta[/Tex]

Explanation:

Use the identity: [Tex]\sin^6 \theta + \cos^6 \theta = (\sin^2 \theta)^3 + (\cos^2 \theta)^3 = (\sin^2 \theta + \cos^2 \theta)^3 - 3\sin^2 \theta \cos^2 \theta (\sin^2 \theta + \cos^2 \theta)[/Tex]

Since [Tex]\sin^2 \theta + \cos^2 \theta = 1[/Tex],

Thus, [Tex]\sin^6 \theta + \cos^6 \theta = 1 - 3\sin^2 \theta \cos^2 \theta[/Tex]

Now, [Tex]\sin^2 \theta \cos^2 \theta = \frac{1}{4} \sin^2 2\theta = \frac{1}{4} \cdot (1 - \cos^2 2\theta) [/Tex]

But to express in terms of cos⁡ 2θ,

[Tex]\sin^2 2\theta = 1 - \cos^2 2\theta[/Tex]

Thus, [Tex]\sin^6 \theta + \cos^6 \theta = 1 - 3 \cdot \frac{1}{4} \cdot (1 - \cos^2 2\theta) = 1 - \frac{3}{4} + \frac{3}{4}\cos^2 2\theta = \frac{1}{4} + \frac{3}{4}\cos^2 2\theta[/Tex]

[Tex]\Rightarrow \sin^6 \theta + \cos^6 \theta = \frac{1}{4}(1 + 3\cos^2 2\theta)[/Tex]

Value of tan 75°

• A

  2 + √3​

• B

  1 + √3​

• C

  2 - √3​

• D

  1 - √3​

Explanation:

We know, [Tex]\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}[/Tex]

​Let A = 45∘ and B = 30∘:

tan⁡75° = tan⁡(45° + 30°)= [Tex]\frac{\tan 45^\circ + \tan 30^\circ}{1 - \tan 45^\circ \tan 30^\circ}[/Tex]

We know, tan⁡45° = 1 and tan⁡30° = 1/√3

Now, [Tex]\tan 75^\circ = \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1 \cdot \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}[/Tex]

[Tex]\Rightarrow \tan 75^\circ = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \cdot \frac{\sqrt{3} + 1}{\sqrt{3} + 1}[/Tex]

[Tex]\Rightarrow \tan 75^\circ = \frac{(\sqrt{3} + 1)^2}{(\sqrt{3})^2 - (1)^2}[/Tex]

[Tex]\Rightarrow \tan 75^\circ = \frac{3 + 2\sqrt{3} + 1}{3 - 1} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}[/Tex]

[Tex]\frac{\sin ^2 A-\sin ^2 B}{\sin A \cos A-\sin B \cos B}=[/Tex]

• A

  tan (A - B)

• B

  tan (A + B)

• C

  cot (A - B)

• D

  cot (A + B)

Explanation:

Formula Used:

• [Tex]\sin ^2 A-\sin ^2 B = \sin (A+B) \sin (A-B)[/Tex],
• [Tex]2 \sin X \cos X = \sin 2X[/Tex]
• [Tex]\sin X -\sin Y = 2 \cos \left(\frac{(X+Y)}{2}\right) \sin \left(\frac{(X-Y)}{2}\right)[/Tex]

Now, [Tex]\frac{\sin ^2 A-\sin ^2 B}{\sin A \cos A-\sin B \cos B}=\frac{2 \sin (A+B) \sin (A-B)}{\sin 2 A-\sin 2 B} =\frac{2 \sin (A+B) \sin (A-B)}{2 \cos (A+B) \sin (A-B)}=\frac{\sin (A+B)}{\cos (A+B)}=\tan (A+B)[/Tex]

The sum of the solutions x∈R of the equation [Tex]\frac{3\cos 2x + \cos^3 2x} {\cos 6x −\sin 6x} = x^3 − x^2 + 6[/Tex] is: [JEE Main 2024 29 January Evening Shift]

• A

  3

• B

  1

• C

  0

• D

  -1

Explanation:

[Tex] \frac{3\cos⁡ 2x + \cos^3 ⁡2x}{\cos^6⁡x −\sin^6⁡x} = x^3 - x^2 + 6[/Tex]

[Tex]\Rightarrow  \frac{\cos 2x(3 + \cos^2 ⁡2x}{(\cos^2⁡x)^3 −(\sin^2⁡x)^3} = x^3 - x^2 + 6[/Tex]

[Tex]\Rightarrow \text{LHS} =  \frac{\cos 2x(3 + \cos^2 ⁡2x}{(\cos^2⁡x − \sin^2⁡x)[(\cos^2 2x)^2 + (\sin^2 2x)^2 +\sin^2 2x \cdot \cos^2 2x]}[/Tex]

[Tex]\Rightarrow \text{LHS} =  \frac{\cos 2x(3 + \cos^2 ⁡2x)}{ (\cos^2⁡x − \sin^2⁡x)[(\cos^2 x)^2 + (\sin^2 x)^2 +2\sin^2 x \cdot \cos^2 x - \sin^2 x \cdot \cos^2 x]}[/Tex]

[Tex]\Rightarrow \text{LHS} =  \frac{\cos 2x(3 + \cos^2 ⁡2x)}{ \cos 2⁡x[(\cos^2 x + \sin^2 x)^2 - \sin^2 x \cdot \cos^2 x]}[/Tex]

If [Tex]\sin A+\sin B=C, \cos A+\cos B=D, then \ the \ value \ of \sin (A+B)=[/Tex]

• A

  CD

• B

  [Tex]\frac{C D}{C^2+D^2}[/Tex]

• C

  [Tex]\frac{C^2+D^2}{2 C D}[/Tex]

• D

  [Tex]\frac{2 C D}{C^2+D^2}[/Tex]

Explanation:

As given [Tex]\frac{\sin A+\sin B}{\cos A+\cos B}=\frac{C}{D}[/Tex]

[Tex]\Rightarrow \frac{2 \sin \frac{A+B}{2} \cdot \cos \frac{A-B}{2}}{2 \cos \frac{A+B}{2} \cdot \cos \frac{A-B}{2}}=\frac{C}{D}[/Tex]

[Tex]\Rightarrow \tan \frac{A+B}{2}=\frac{C}{D}[/Tex],

As we know, [Tex]\sin 2\theta = \frac{2\tan \theta}{1 + \tan^2\theta}[/Tex]

Thus, [Tex]\sin (A+B)=\frac{2 \tan \frac{A+B}{2}}{1+\tan ^2 \frac{A+B}{2}} = \frac{2\frac{C}{D}}{1 -\frac{C^2}{D^2}}=\frac{\frac{2C}{D}}{\frac{1+ C^2}{D^2}} = \frac{2CD}{C^2 + D^2}[/Tex]

[Tex]=\frac{2 \frac{C}{D}}{1+\frac{C^2}{D^2}}=\frac{2 C D}{\left(C^2+D^2\right)}[/Tex]

If x + 1/x = 2 cos θ, then x³ + 1/x³ =

• A

  cos 3θ

• B

  2 cos 3θ

• C

  (1/2) cos 3θ

• D

  (1/3) cos 3θ

Explanation:

We have x + 1/x = 2 cos θ,

Now [Tex]x^3+\frac{1}{x^3} = \left(x+\frac{1}{x}\right)^3 - 3x \frac{1}{x}\left(x+\frac{1}{x}\right)\left|=(2 \cos \theta)^3-3(2 \cos \theta)=8 \cos ^3 \theta-6 \cos \theta\right|= 2\left(4 \cos ^3 \theta-3 \cos \theta\right)=2 \cos 3 \theta[/Tex]

Trick: Put x = 1 [Tex]\Rightarrow \theta=0^{\circ}[/Tex] then [Tex]x^3+\frac{1}{x^3}=2=2 \cos 3 \theta[/Tex],

If y=(1 + tan A)(1 - tan B) where A - B = π/4, then (y+1)^y+1 is equal to

• A

  9

• B

  4

• C

  27

• D

  81

Explanation:

A - B = π/4

[Tex]\Rightarrow \tan (A-B)=\tan \frac{\pi}{4}[/Tex]

[Tex]\Rightarrow \frac{\tan A-\tan B}{1+\tan A \tan B} = 1[/Tex]

[Tex]\Rightarrow \tan A-\tan B-\tan A \tan B = 1,[/Tex]

[Tex]\Rightarrow \tan A-\tan B-\tan A \tan B + 1 = 2[/Tex]

[Tex]\Rightarrow(1+\tan A)(1-\tan B)=2[/Tex]

As y = (1 + tan A)(1 - tan B)

Thus, y = 2

Hence, (y+1)^y+1 = (2+1)²⁺¹ = 3³ = 27

If equal to a sin² x + b cos² x = c, b sin² y + a cos² y = d, and a tan x = b tan y, then what is a²/b² ?

• A

  [Tex]\frac{(b-c)(d-b)}{(a-d)(c-a)}[/Tex]

• B

  [Tex]\frac{(a-d)(c-a)}{(b-c)(d-b)}[/Tex]

• C

  [Tex]\frac{(d-a)(c-a)}{(b-c)(d-b)}[/Tex]

• D

  [Tex]\frac{(b-c)(b-d)}{(a-c)(a-d)}[/Tex]

Explanation:

[Tex]a \sin ^2 x+b \cos ^2 x=c \\ \Rightarrow(b-a) \cos ^2 x=c-a\\ \Rightarrow(b-a) = \frac{c - a}{\cos ^2 x} = (c - a)\sec^2 x\\ \Rightarrow(b-a)=(c-a)\left(1+\tan ^2 x\right)[/Tex]

Thus, [Tex]\tan ^2 x=\frac{b-c}{c-a}[/Tex]

and

[Tex]b \sin ^2 y+a \cos ^2 y=d\\\Rightarrow(a-b) \cos ^2 y=d-b\\\Rightarrow(a-b) =\frac{d-b}{\cos ^2 y} = (d - b)sec^2 y\\\Rightarrow(a-b)=(d-b)\left(1+\tan ^2 y\right)[/Tex]

Thus, [Tex]\tan ^2 y=\frac{a-d}{d-b}[/Tex]

[Tex]\therefore \frac{\tan ^2 x}{\tan ^2 y}=\frac{(b-c)(d-b)}{(c-a)(a-d)}[/Tex] . . .(i)

Given: a tan x=b tan y,

[Tex]\Rightarrow \frac{\tan x}{\tan y}=\frac{b}{a} . . . (ii)[/Tex]

Using equation (i) and (ii), we get

[Tex]\frac{b^2}{a^2}=\frac{(b-c)(d-b)}{(c-a)(a-d)}[/Tex]

[Tex]\Rightarrow \frac{a^2}{b^2}=\frac{(c-a)(a-d)}{(b-c)(d-b)}[/Tex]

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