Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.5

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Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.5

Question 1.
Find the value of cos 2A, A lies in the first quadrant, when

Solution:

Question 2.
If θ is an acute angle, then find

Solution:

Question 3.

Solution:

Question 4.
Prove that cos ⁵ θ = 16 cos⁵ θ – 20 cos³ θ + 5 cos θ.
Solution:
cos ⁵ θ = cos(2θ + 3θ) = cos 2θ cos 3θ – sin 2θ sin 3θ
= (2 cos² θ – 1) (4 cos³ θ – 3 cos θ) – 2 sin θ cos θ (3 sin θ – 4 sin³ θ)
= 8cos⁵ θ – 6 cos³ θ – 4 cos³ θ + 3 cos θ – 6 sin² θ cos θ + 8 cos θ sin⁴ θ
= 8 cos⁵ θ – 6 cos³ θ – 4 cos³ θ + 3 cos θ – 6(1 – cos² θ) cos θ + 8 cos θ (1 – cos² θ)²
= 8 cos⁵ θ – 6 cos³ θ – 4 cos³ θ + 3 cos θ – 6 cos θ + 6 cos³ θ + 8 cos 0(1+ cos⁴ θ – 2 cos² θ)
= 8 cos⁵ θ – 6 cos³ θ – 4 cos³ θ + 3 cos θ – 6 cos θ + 6 cos³ θ + 8 cos θ + 8 cos⁵ θ – 16 cos³ θ
= 16 cos⁵ θ – 20 cos³ θ + 5 cos θ = RHS

Question 5.

Solution:

Question 6.
If A + B = 45°, show that (1 + tanA) (1 + tanB) = 2.
Solution:

Now LHS = (1 + tan A) (1 + tan B)
= tan A + tan B + tan A tan B + 1
= (1 – tan A tan B) + (tan A tan B + 1) from (1)
= 2 = RHS

Question 7.
Prove that (1 + tan 1°)(1 + tan 2°)(1 + tan 3°)… (1 + tan 44°) is a multiple of 4.
Solution:
1 + tan 44° = 1 + tan (45° – 1°)

(1 + tan 1°)(1 + tan 44°) = 2
Similarly (1 + tan 2°) (1 + tan 43°) = 2
(1 + tan 3°) (1 + tan 42°) = 2
(1 + tan 22°) (1 + tan 23°) = 2
= (1 + tan 1°) (1 + tan 2°)… (1 + tan 44°) = 2 × 2 × … 22 times
It is a multiple of 4.

Question 8.

Solution:

Question 9.

Solution:

Question 10.
Prove that (1 + sec 2θ)(1 + sec 4θ)….. (1 + sec 2ⁿθ) = tan 2ⁿθ
Solution:

Question 11.

Solution:

Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.5 Additional Questions

Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.
Show that 4 sin A sin (60° + A). sin(60° – A) = sin 3A
Solution:
LHS = 4sinAsin(60° + A). sin(60° – A)
= 4 sin A{sin (60° + A). sin (60° – A)}
= 4 sin A {sin² 60° – sin² A)}