Mathematics online test 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
Answer: a
Clarification: 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
Answer: c
Clarification: 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
Answer: b
Clarification: 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
Answer: a
Clarification: 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
Answer: d
Clarification: 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.
6. If sin x= 1/2, then sin 3x =_______________
a)(sqrt{3})/2
b) 1/2
c) 1/(sqrt{2})
d) 1
Answer: d
Clarification: 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
Answer: b
Clarification: 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})
Answer: b
Clarification: 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}})
Answer: a
Clarification: 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}})
Answer: d
Clarification: 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}})
Answer: a
Clarification: 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}})
Answer: d
Clarification: 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}})
Answer: c
Clarification: 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}})
Answer: b
Clarification: 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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