 # Trigonometry Formulas For Class 11

Trigonometry is a branch of mathematics that studies the relationships between angles and lengths of triangles. It is a very important topic of mathematics just like statistics, linear algebra and calculus. In addition to mathematics, it also contributes majorly to engineering, physics, astronomy and architectural design. Trigonometry Formulas for class 11 play a crucial role in solving any problem related to this chapter. Also, check Trigonometry For Class 11 where students can learn notes, as per the CBSE syllabus and prepare for the exam.
Download the below PDF to get the formulas of class 11 trigonometry.

## List of Class 11 Trigonometry Formulas

Here is the list of formulas for Class 11 students as per the NCERT curriculum. All the formulas of trigonometry chapter are provided here for students to help them solve problems quickly.

 Trigonometry Formulas sin(−θ) = −sin θ cos(−θ) = cos θ tan(−θ) = −tan θ cosec(−θ) = −cosecθ sec(−θ) = sec θ cot(−θ) = −cot θ Product to Sum Formulas sin x sin y = 1/2 [cos(x–y) − cos(x+y)] cos x cos y = 1/2[cos(x–y) + cos(x+y)] sin x cos y = 1/2[sin(x+y) + sin(x−y)] cos x sin y = 1/2[sin(x+y) – sin(x−y)] Sum to Product Formulas sin x + sin y = 2 sin [(x+y)/2] cos [(x-y)/2] sin x – sin y = 2 cos [(x+y)/2] sin [(x-y)/2] cos x + cos y = 2 cos [(x+y)/2] cos [(x-y)/2] cos x – cos y = -2 sin [(x+y)/2] sin [(x-y)/2] Identities sin2 A + cos2 A = 1 1+tan2 A = sec2 A 1+cot2 A = cosec2 A

### Sign of Trigonometric Functions in Different Quadrants

 Quadrants→ I II III IV Sin A + + – – Cos A + – – + Tan A + – + – Cot A + – + – Sec A + – – + Cosec A + + – –

### Basic Trigonometric Formulas for Class 11

cos (A + B) = cos A cos B – sin A sin B

cos (A – B) = cos A cos B + sin A sin B

sin (A+B) = sin A cos B + cos A sin B

sin (A -B) = sin A cos B – cos A sin B

Based on the above addition formulas for sin and cos, we get the following below formulas:

• sin(π/2-A) = cos A
• cos(π/2-A) = sin A
• sin(π-A) = sin A
• cos(π-A) = -cos A
• sin(π+A)=-sin A
• cos(π+A)=-cos A
• sin(2π-A) = -sin A
• cos(2π-A) = cos A

If none of the angles A, B and (A ± B) is an odd multiple of π/2, then

• tan(A+B) = [(tan tan B)/(– tan tan B)]
• tan(A-B) = [(tan A – tan B)/(1 + tan tan B)]

If none of the angles A, B and (A ± B) is a multiple of π, then

• cot(A+B= [(coco− 1)/(cocoA)]
• cot(A-B= [(cocoB + 1)/(coB – coA)]

Some additional formulas for sum and product of angles:

• cos(A+B) cos(A–B)=cos2A–sin2B=cos2B–sin2A
• sin(A+B) sin(A–B) = sin2A–sin2B=cos2B–cos2A
• sinA+sinB = 2 sin (A+B)/2 cos (A-B)/2

Formulas for twice of the angles:

• sin2A = 2sinA cosA = [2tan A /(1+tan2A)]
• cos2A = cos2A–sin2A = 1–2sin2A = 2cos2A–1= [(1-tan2A)/(1+tan2A)]
• tan 2A = (2 tan A)/(1-tan2A)

Formulas for thrice of the angles:

• sin3A = 3sinA – 4sin3A
• cos3A = 4cos3A – 3cosA
• tan3A = [3tanA–tan3A]/[1−3tan2A]

### Video Lesson on Trigonometry Also check:

## Solved Examples

Example 1:

If sin 𝜃 = –4/5 and 𝜋 < 𝜃 < 3𝜋/2, find the value of all the other five trigonometric functions.

Solution:

Since, the value of theta ranges between 𝜋 < 𝜃 < 3𝜋/2, that means, 𝜃 lies in third quadrant.

Now, sin 𝜃 = –⅘ ⇒ cosec 𝜃 = 1/sin 𝜃 = – 5/4

∴ cot2 𝜃 = (cosec2 𝜃 – 1) = (25/16 – 1) = 9/16 ⇒ cot 𝜃 = ¾ (taking positive root as cot 𝜃 is positive in third quadrant)

tan 𝜃 = 1/cot 𝜃 = 4/3

Now, cos 𝜃 = cot 𝜃 sin 𝜃 = ¾ × (– ⅘ ) = – ⅗

∴ sec 𝜃 = 1/cos 𝜃 = – 5/3

Hence, all other trigonometric functions are cos 𝜃 = – ⅗, tan 𝜃 = 4/3, cot 𝜃 = ¾, sec 𝜃 = – 5/3 and cosec 𝜃 = – 5/4.

Example 2:

Evaluate: cos( – 870o)

Solution:

cos( – 870o) = cos(870o)   [as cos ( –𝜃) = cos 𝜃 ]

= cos ( 2 × 360o + 150o)

= cos 150o [as cos (2n𝜋 + 𝜃) = cos 𝜃 ]

= cos ( 180o – 30o) = – cos 30o = – √3/2

Example 3:

Prove that tan 56o = (cos 11o + sin 11o)/(cos 11o – sin 11o)

Solution:

We have, LHS tan 65o = tan (45o + 11o)

= (tan 45o + tan 11o)/(1 – tan 45o tan 11o) {since, 45o + 11o is not an odd multiple of 𝜋/2 }

= (1 + tan 11o)/(1 – tan 11o)

= {1 + (sin 11o/cos 11o)}/ {1 – (sin 11o/cos 11o)}

= (cos 11o + sin 11o)/(cos 11o – sin 11o) = RHS

Example 4:

Prove that sin x + sin 3x + sin 5x + sin 7x = 4sin 4x cox 2x cos x.

Solution:

Now, LHS = (sin 7x + sin x) + (sin 5x + sin 3x)

= 2 sin {(7x + x)/2} cos {(7x – x)/2} + 2 sin {(5x + 3x)/2} cos {(5x – 3x)/2}

= 2 sin 4x cos 3x + 2 sin 4x cos x

= 2 sin 4x (cos 3x + cos x)

= 2 sin 4x × 2 cos {(3x + x)/2} cos {(3x – x)/2}

= 2 sin 4x × 2 cos 2x cos x

= 4 sin 4x cos 2x cos x = RHS

## Practice Problems

1. Prove that (sin x – sin y)/(cos x + cos y) = tan {(x – y)/2}.
2. Prove that sin 𝜋/10 + sin 13𝜋/10 = – ½.
3. Prove that (1 + cos 𝜃)/(1 – cos 𝜃) = (cosec 𝜃 + cot 𝜃)2
4. If A + B + C = 𝜋, prove that sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C.

## Frequently Asked Questions on Trigonometry Formulas For Class 11

### What are the topics in trigonometry class 11?

The following are covered in CBSE Class 11 trigonometry:
Angles: Positive and Negative
System of Measuring angles: Sexagesimal (Degree Measure) and Circular System (Radian Measure); the relationship between both the systems
Trigonometric Functions (sine, cosine, tangent, co-tangent, secant, co-secant)
Trigonometric Identities
Sign of trigonometric functions in various quadrants
Values of some special angles of trigonometric functions
Trigonometric functions as sum and difference of angles
Trigonometric functions of multiples angles
Conditional identities
Trigonometric Equations
Sine formula, cosine formula, Napier’s Analogies

### What are the three basic identities of trigonometric functions?

The three basis trigonometric identities are
sin2 x + cos2 x = 1
1 + tan2 x = sec2 x
cosec2 x = 1 + cot2 x

### What are the signs of the trigonometric functions in various quadrants?

In the 1st quadrant, all trigonometric ratios are positive, in the 2nd quadrant, only sine and cosecant are positive, in the 3rd quadrant, tangent and co-tangent are positive, and in the 4th quadrant, only cosine and secant are positive.

### What is the value of tan 3x?

Tan 3x = [3 tan x – tan3 x]/[1 − 3 tan2 x]

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