Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.6

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Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.6

Question 1.
Express each of the following as a sum or difference
(i) sin 35° cos 28°
(ii) sin 4x cos 2x
(iii) 2 sin 10θ cos 2θ
(iv) cos 5θ cos 2θ
(v) sin 5θ sin 4θ.
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.6 1
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Question 2.
Express each of the following as a product
(i) sin 75° – sin 35°
(ii) cos 65° + cos 15°
(iii) sin 50° + sin 40°
(iv) cos 35° – cos 75°.
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.6 4

Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.6

Question 3.
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Solution:
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Question 4.
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Solution:
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Question 5.
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Solution:
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Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.6

Question 6.
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Solution:
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Question 7.
Prove that sin x + sin 2x + sin 3x = sin 2x (1 + 2 cos x).
Solution:
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Question 8.
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Solution:
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Question 9.
Prove that 1 + cos 2x + cos 4x + cos 6x = 4 cos x cos 2x cos 3x.
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.6 20
LHS = 1 + cos 2x + cos 4x + cos 6x
= (1 + cos 6x) + (cos 2x + cos 4x)
= 2cos2 3x + 2cos 3x cos x
= 2 cos 3x (cos 3x + cos x)
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Question 10.
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Solution:
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Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.6

Question 11.
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Solution:
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Question 12.
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Solution:
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Question 13.
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Solution:
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Question 14.
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Solution:
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Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.6 Additional Questions

Question 1.
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Solution:
Let cos 20° cos 40° cos 60° cos 80° = x
Multiply by 2 sin 20° on both sides.
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Question 2.
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Solution:
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Question 3.
Prove that sin 50° – sin 70° + cos 80° = 0.
Solution:
LHS = sin 50° – sin 70° + cos 80°
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Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.6

Question 4.
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Solution:
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Question 5.
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Solution:
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Question 6.
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Solution:
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Question 7.
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Solution:
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Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.6

Question 8.
Prove that tan 70° – tan 20° – 2 tan 40° = 4 tan 10°
Solution:
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Question 9.
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Solution:
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Question 10.
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Solution:
Multiplying numerator and denominator by 2, we have
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