Trigonometric Functions of Sum and Difference of Two Angles - Class 11 Maths MCQ

This set of Class 11 Maths Chapter 3 Multiple Choice Questions & Answers (MCQs) focuses on “Trigonometric Functions of Sum and Difference of Two Angles – 2”.

1. If sin x=1/2 and cos x=\(\sqrt{3}\)/2, then find sin 2x.
a) \(\sqrt{3}\)/2
b) 1/2
c) 1/\(\sqrt{2}\)
d) 1
View Answer

Answer: a
Explanation: We know, sin2x = 2 sin x cos x
sin 2x = 2*1/2*\(\sqrt{3}\)/2
sin 2x = \(\sqrt{3}\)/2.

2. If tan x=1/\(\sqrt{3}\), then sin 2x =___________________
a) 1/\(\sqrt{2}\)
b) 1/2
c) \(\sqrt{3}\)/2
d) 1
View Answer

Answer: c
Explanation: We know, sin 2x = (2 tan x) / (1 + tan²x)
sin 2x=(2*1/\(\sqrt{3}\)) / (1+1/3) = (2*3)/4*\(\sqrt{3}\) = \(\sqrt{3}\)/2.

3. If tan x = 1/\(\sqrt{3}\), then tan2x =_________________
a) 1
b) \(\sqrt{3}\)
c) 1/\(\sqrt{3}\)
d) 0
View Answer

Answer: b
Explanation: We know, sin 2x = (2 tan x) / (1 – tan²x)
tan 2x = (2*1/\(\sqrt{3}\)) / (1-1/3) = (2*3)/2*\(\sqrt{3}\) = \(\sqrt{3}\).

4. Which is correct?
a) sin 3x = 3sinx – 4sin³x
b) sin 3x = 4sinx – 3sin³x
c) sin 3x = 3sin³x – 4sinx
d) sin 3x = 4sin³x – 3sinx
View Answer

Answer: a
Explanation: sin 3x = sin (2x + x)
= sin 2x cos x + cos 2x sin x
= (2 sin x cos x) cos x + (1 – 2sin² x) sin x
= 2 sin x (1 – sin² x) + sin x – 2 sin³ x
= 2 sin x – 2 sin³ x + sin x – 2 sin³ x
= 3 sin x – 4 sin³ x.

5. Which is correct?
a) cos 3x = 3cosx – 4cos³x
b) cos 3x = 4cosx – 3cos³x
c) cos 3x = 3cos³x – 4cosx
d) cos 3x = 4cos³x – 3cosx
View Answer

Answer: d
Explanation: cos 3x = cos (2x +x)
= cos 2x cos x – sin 2x sin x
= (2cos² x – 1) cos x – 2sin x cos x sin x
= (2cos² x – 1) cos x – 2cos x (1 – cos² x)
= 2cos³ x – cos x – 2cos x + 2 cos³ x
= 4cos³ x – 3cos x.

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6. If sin x= 1/2, then sin 3x =_______________
a)\(\sqrt{3}\)/2
b) 1/2
c) 1/\(\sqrt{2}\)
d) 1
View Answer

Answer: d
Explanation: We know, sin 3x = 3sin x – 4sin³x
= 3(1/2) – 4(1/2)³
= 3/2 – 4/8 = 3/2 – 1/2 = 1.

7. If cos x=1/2, then cos 3x =_______________
a) 0
b) -1
c) 1/\(\sqrt{2}\)
d) 1
View Answer

Answer: b
Explanation: We know, cos 3x=4cos³x – 3cos x
= 4(1/2)³–3(1/2)
= 4/8 – 3/2 = 1/2 – 3/2 = -1.

8. Find tan 3x if tan x= 1.
a) 1
b) -1
c) 1/\(\sqrt{3}\)
d) \(\sqrt{3}\)
View Answer

Answer: b
Explanation: We know, tan 3x = (3tanx – tan³x)/(1-3tan²x)
= (3*1 – 1)/(1-3)
= (2)/(-2) = -1.

9. cos 75° + cos 15° =___________________
a) \(\frac{\sqrt{3}}{\sqrt{2}}\)
b) \(\frac{\sqrt{2}}{\sqrt{3}}\)
c) \(\frac{\sqrt{3}}{2}\)
d) \(\frac{1}{\sqrt{2}}\)
View Answer

Answer: a
Explanation: We know, cos x + cos y = 2 cos (x+y)/2 cos (x-y)/2
cos 75° + cos 15° = 2 cos (75°+15°)/2 cos(75°-15°)/2
= 2 * cos 45° * cos30°
= 2*(1/\(\sqrt{2}\))*(\(\sqrt{3}\)/2)
= \(\sqrt{3}\)/\(\sqrt{2}\).

10. cos 75° – cos 15° =___________________
a) \(\frac{\sqrt{3}}{\sqrt{2}}\)
b) \(\frac{\sqrt{2}}{\sqrt{3}}\)
c) \(\frac{1}{\sqrt{2}}\)
d) –\(\frac{1}{\sqrt{2}}\)
View Answer

Answer: d
Explanation: cos x – cos y = – 2 sin (x+y)/2 sin (x-y)/2
cos 75° – cos 15° = – 2 sin (75°+15°)/2 sin(75°-15°)/2
= – 2 sin 45° sin 30°
= – 2*1/\(\sqrt{2}\)*1/2= – 1/\(\sqrt{2}\).

11. sin 75° + sin 15° = _________________
a) \(\frac{\sqrt{3}}{\sqrt{2}}\)
b) \(\frac{\sqrt{2}}{\sqrt{3}}\)
c) \(\frac{\sqrt{3}}{2}\)
d) \(\frac{1}{\sqrt{2}}\)
View Answer

Answer: a
Explanation: We know, sin x + sin y = 2 sin (x+y)/2 cos (x-y)/2
sin 75° + sin 15° = 2 sin (75°+15°)/2 cos(75°-15°)/2
= 2 sin 45° cos 30°
= 2*1/\(\sqrt{2}\)*\(\sqrt{3}\)/2 = \(\sqrt{3}\)/\(\sqrt{2}\).

12. sin 75° – sin 15° =_________________
a) \(\frac{\sqrt{3}}{\sqrt{2}}\)
b) \(\frac{\sqrt{2}}{\sqrt{3}}\)
c) \(\frac{\sqrt{3}}{2}\)
d) \(\frac{1}{\sqrt{2}}\)
View Answer

Answer: d
Explanation: We know, sin x – sin y = 2 cos (x+y)/2 sin (x-y)/2
sin 75° – sin 15° = 2 cos (75°+15°)/2 sin(75°-15°)/2
= 2 cos 45° sin 30°
= 2*1/\(\sqrt{2}\)*1/2 = 1/\(\sqrt{2}\).

13. 2 sin 75° sin15° =_____________________________
a) \(\frac{\sqrt{3}}{2}\)
b) \(\frac{\sqrt{2}}{\sqrt{3}}\)
c) \(\frac{1}{2}\)
d) \(\frac{1}{\sqrt{2}}\)
View Answer

Answer: c
Explanation: We know, 2 sin x sin y = cos (x – y) – cos (x + y)
2 sin 75° sin15° = cos (75°-15°) – cos (75°+15°)
= cos 60° – cos 90° = 1/2 – 0 = 1/2.

14. 2 cos 75° cos 15° =____________________
a) \(\frac{\sqrt{3}}{\sqrt{2}}\)
b) \(\frac{1}{2}\)
c) \(\frac{\sqrt{3}}{2}\)
d) \(\frac{1}{\sqrt{2}}\)
View Answer

Answer: b
Explanation: We know, 2 cos x cos y = cos (x + y) + cos (x – y)
2 cos 75° cos 15° = cos (75°+15°) + cos (75°-15°)
= cos 90° + cos 60°
= 0 + 1/2 = 1/2.

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