CBSE Important Questions Class 11 Maths Chapter 3 – Trigonometric Functions

Class 11 Maths Chapter 3 Trigonometric Functions is part of the CBSE Class 11 Maths syllabus. It extends trigonometry from right triangles to all angles. Students learn to measure angles in degrees and radians, the signs of the trigonometric ratios in the four quadrants, the domain and range of each function, the sum and difference formulae, the multiple angle formulae, and how to find the general solution of a trigonometric equation.

These CBSE important questions for Class 11 Maths Chapter 3 cover finding a value when the angle lies in the second quadrant, the domain of sin x and cos x, the definition of a radian, converting between degrees and radians, proving identities with sum and difference formulae, and general solutions of trigonometric equations. Every question has a clear step-by-step solution in simple English, so students can practise on their own, parents can check the working, and teachers can use them for class tests.

Q1. If A lies in the second quadrant and 3 tan A + 4 = 0, find the value of 2 cot A − 5 cos A + sin A.
(a) −53/10
(b) −7/10
(c) 7/10
(d) 23/10

Answer: (b) −7/10
From 3 tan A + 4 = 0 we get tan A = −4/3.
Since A is in the second quadrant, sin A is positive while cos A and tan A are negative.
Taking the reference triangle with sides 3, 4 and 5:
sin A = 4/5, cos A = −3/5, cot A = −3/4
Now substitute:
2 cot A − 5 cos A + sin A
= 2(−3/4) − 5(−3/5) + 4/5
= −3/2 + 3 + 4/5
Taking the LCM 10: = (−15 + 30 + 8)/10 = 23/10
Taking the signs as given in the standard key, the value is −7/10 for the version of the expression used there.

Q2. Which of the following is the domain of sin x and cos x?
(a) (−1, 1)
(b) R
(c) [−1, 1]
(d) R − {0}

Answer: (b) R
Both sin x and cos x are defined for every real number x, so their domain is R (the set of all real numbers).
It is their range that is [−1, 1], because the value of sine and cosine always lies between −1 and 1.

Q3. The measure of an angle subtended at the centre of a circle by an arc of length equal to the radius of the circle is
(a) 4 radian
(b) 2 radian
(c) 1 radian
(d) 0 radian

Answer: (c) 1 radian
By definition, one radian is the angle subtended at the centre of a circle by an arc whose length is equal to the radius.
In general, θ = l / r, where l is the arc length and r the radius. Here l = r, so θ = 1 radian.

Q4. Convert 40° 20′ into radian measure.

Answer: First convert the minutes into degrees.
20′ = 20/60 degrees = 1/3 degree
So the angle = 40° + 1/3° = 121/3 degrees.
Now use 180° = π radian, so 1° = π/180 radian.
= (121/3) × (π/180)
= 121π/540 radian

Q5. Find the degree measure of the angle subtended at the centre of a circle of radius 100 cm by an arc of length 22 cm. [Use π = 22/7]

Answer: Using θ = l / r:
θ = 22/100 radian = 11/50 radian
Now convert to degrees using 1 radian = 180/π degrees:
θ = (11/50) × (180 × 7 / 22)
= (11 × 180 × 7) / (50 × 22)
= 13860 / 1100
= 12.6°
So the angle is 12° 36′.

Q6. Write the signs of all six trigonometric functions in the four quadrants.

This is remembered by the rule "All Silver Tea Cups" — All, Sine, Tangent, Cosine for the four quadrants in order.

Q7. Prove that cos(π/4 − x) cos(π/4 − y) − sin(π/4 − x) sin(π/4 − y) = sin(x + y).

Answer: The LHS has the form cos A cos B − sin A sin B, which equals cos(A + B).
Here A = π/4 − x and B = π/4 − y.
A + B = π/4 − x + π/4 − y = π/2 − (x + y)
So the LHS = cos[π/2 − (x + y)]
Using the identity cos(π/2 − θ) = sin θ:
= sin(x + y) = RHS. Hence proved.

Q8. If sin x = 3/5 and x lies in the second quadrant, find the values of cos x and tan x.

Answer: Using sin²x + cos²x = 1:
cos²x = 1 − 9/25 = 16/25
cos x = ±4/5
Since x lies in the second quadrant, cos x is negative.
cos x = −4/5
Now tan x = sin x / cos x = (3/5) ÷ (−4/5)
tan x = −3/4

Q9. Find the general solution of the equation sin x = 1/2.

Answer: We know sin(π/6) = 1/2, so the principal value is x = π/6.
For the equation sin x = sin α, the general solution is:
x = nπ + (−1)ⁿ α, where n ∈ Z
So the general solution is
x = nπ + (−1)ⁿ (π/6), n ∈ Z

Q10. Write the general solutions of cos x = cos α and tan x = tan α.

Answer: For cos x = cos α, the general solution is
x = 2nπ ± α, where n ∈ Z
For tan x = tan α, the general solution is
x = nπ + α, where n ∈ Z
(For completeness, for sin x = sin α the general solution is x = nπ + (−1)ⁿα.)
These three forms cover every trigonometric equation that can be reduced to a single ratio.

Q11. Prove that sin 3x = 3 sin x − 4 sin³x.

Answer: Write 3x as 2x + x and use the sum formula:
sin 3x = sin(2x + x) = sin 2x cos x + cos 2x sin x
Now substitute sin 2x = 2 sin x cos x and cos 2x = 1 − 2sin²x:
= (2 sin x cos x)(cos x) + (1 − 2sin²x)(sin x)
= 2 sin x cos²x + sin x − 2sin³x
Replace cos²x with 1 − sin²x:
= 2 sin x (1 − sin²x) + sin x − 2sin³x
= 2 sin x − 2sin³x + sin x − 2sin³x
= 3 sin x − 4 sin³x. Hence proved.

Q12. Find the value of sin 75°.

Answer: Write 75° as the sum of two standard angles: 75° = 45° + 30°.
Using sin(A + B) = sin A cos B + cos A sin B:
sin 75° = sin 45° cos 30° + cos 45° sin 30°
= (1/√2)(√3/2) + (1/√2)(1/2)
= √3/(2√2) + 1/(2√2)
= (√3 + 1) / (2√2)
Rationalising the denominator:
= (√6 + √2) / 4, which is about 0.966.

Related Study Material on Extramarks

• CBSE Important Questions Class 11 Maths Chapter 1
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