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.

Trigonometry Formulas For Class 11 – PDF

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.

Sign of Trigonometric Functions in Different Quadrants

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

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

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

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:

• Trigonometry Formulas
• Trigonometry Formulas List
• Trigonometry Table

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( – 870^o)

Solution:

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

= cos ( 2 × 360^o + 150^o)

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

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

Example 3:

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

Solution:

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

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

= (1 + tan 11^o)/(1 – tan 11^o)

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

= (cos 11^o + sin 11^o)/(cos 11^o – sin 11^o) = 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.