Mathematics Questions and Answers – Trigonometric Functions of Sum and Difference of Two Angles-2

«
»

This set of Mathematics online test 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.
advertisement

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*\(\sqrt{3}\) = \(\sqrt{3}\).
advertisement
advertisement

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.
advertisement

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.
advertisement

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)
= \(\sqrt{3}\)/\(\sqrt{2}\).
advertisement

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.

Sanfoundry Global Education & Learning Series – Mathematics – Class 11.

To practice all areas of Mathematics for online tests, here is complete set of 1000+ Multiple Choice Questions and Answers.

Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

advertisement
advertisement
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter