Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.7

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Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.7

Question 1.
If A + B + C = 180°, prove that
(i) sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C

(iii) sin² A + sin² B + sin² C = 2 + 2 cos A cos B cos C
(iv) sin² A + sin² B – sin² C = 2 sin A sin B cos C

(vi) sin A + sin B + sin C = 4 cos \(\frac{\mathbf{A}}{2}\) cos \(\frac{\mathbf{B}}{2}\) cos \(\frac{\mathbf{c}}{2}\)
(vii) sin(B + C – A) + sin(C + A – B) + sin(A + B – C) = 4 sin A sin B sin C.
Solution:
(i) LHS = (sin 2A + sin 2B) + sin 2C
= 2 sin (A + B) cos (A – B) + 2 sin C cos C
[sin (A + B) = sin (180° – C) = sin C]
= 2 sin C cos (A – B) + 2 sin C cos C
= 2 sin C [ cos (A – B) + cos C]
{cos C = cos [180° – (A + B)] = – cos (A + B)}
= 2 sin C [cos (A – B) – cos (A + B)]

(ii)

(iii)

[cos (180° –  C) – cos C – cos C]
= 2 + cos C [cos (A – B) + cos (A + B)]
= 2+ cos C[2 cos A cos B]
= 2 + 2 cos A cos B cos C = RHS

(iv)

(v)

(vi)

(vii) Now A + B + C = 180°
So B + C = 180° – A
sin (B + C – A) = sin (180° – A – A)
= sin(180° – 2A) = sin 2A
Now LHS = sin 2A + sin 2B + sin 2C
= 4 sin A sin B sin C (from (i) ans) = RHS

Question 2.
If A + B + C = 2s, then prove that sin(s – A) sin(s – B) + sin s sin(s – C) = sin A sin B.
Solution:

Question 3.

Solution:

⇒ A+B+C = 180°
⇒ A + B = 180° – C
multiply 2 on both sides
⇒ 2A + 2B = 360° – 2C
⇒ 2(A + B) = 360° – 2C
⇒ tan(2A + 2B) = tan(360° – 2C) = -tan 2C

⇒ tan 2A + tan 2B = -tan2C[1 – tan 2A tan 2B]
⇒ tan 2A + tan 2B = -tan 2C + tan 2A tan 2B tan 2C
⇒ tan 2A + tan 2B + tan 2C = tan 2A tan 2B tan 2C

Question 4.
If A + B + C = \(\frac{\pi}{2}\), prove the following
(i) sin 2A + sin 2B + sin 2C = 4 cos A cos B cos C
(ii) cos 2A + cos 2B + cos 2C = 1 + 4 sin A sin B sin C.
Solution:
(i) LHS = (sin 2A + sin 2B) + sin 2C
= 2 sin (A + B) cos (A – B) + 2 sin C cos C = 2 sin (90° – C) cos (A – B) + 2 sin C cos C
= 2 cos C [cos (A – B) + sin C] + cos (A + B) ( ∴ A + B = π/2 – C)
= 2 cos C [cos (A – B) + cos (A + B)]
= 2 cos C [2 cos A cos B]
= 4 cos A cos B cos C = RHS

(ii) LHS = (cos 2A + cos 2B) + cos 2C
= 2 cos (A + B) cos (A – B) + 1 – 2 sin² C
= 1 + 2 sin C (cos (A – B) – 2 sin² C)
{∴ cos (A + B) = cos (90° – C) = sin C}
= 1 + 2 sin C [cos (A- B) – sin C]
= 1 + 2 sin C [cos (A – B) – cos (A + B)]
= 1 + 2 sin C [2 sin A sin B]
= 1 + 4 sin A sin B sin C
= RHS

Question 5.
If ∆ABC is a right triangle and if ∠A = \(\frac{\pi}{2}\), then prove that
(i) cos² B + cos² C = 1
(ii) sin² B + sin² C = 1

Solution:

Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.7 Additional Questions

Question 1.
A + B + C = π, prove that sin 2A – sin 2B + sin 2C = 4 cos A sin B cos C
Solution:
LHS = sin 2A – sin 2B + sin 2C
= sin 2A + sin 2C – sin 2B

= 2 sin (A + C) cos (A – C) – 2 sin B cos B
= 2 sin (180° – B) cos (A-C) – 2 sin B cos B
= 2 sin B cos (A – C) – 2 sin B cos B
= 2 sin B [cos (A – C) – cos B]
= 2 sin B [cos (A – C) – cos (180° – (A + C))]
= 2 sin B [cos (A – C) + cos (A + C)]
= 2 sin B [2 cos A cos C]
= 4 cos A sin B cos C
= RHS

Question 2.

Solution:


substitute in (1) we get,