Example 1 :
Find the value of cos 2A, A lies in the first quadrant, when
(i) cos A = 15/17
Solution :
We have three formulas for cos 2A,
cos 2A = cos2A - sin2A
cos 2A = 1 - 2 sin2A
cos 2A = 2cos2A - 1
In the above formula, let us choose the 3rd formula
cos 2A = 2cos2A - 1
= 2(15/17)2 - 1
= 2(225/289) - 1
= (450 - 289)/289
cos 2A = 161/289
(ii) sin A = 4/5
Solution :
cos 2A = 1 - 2sin2A
= 1 - 2(4/5)2
= 1 - 2(16/25)
= 1 - (32/25)
= (25 - 32)/25
cos 2A = -7/25
(iii) tan A = 16/63
Solution :
cos 2A = (1 - tan2A) / (1 + tan2A)
= (1 - (16/63)2) / (1 + (16/63)2)
= (1 - (256/3969)) / (1 + (256/3969))
= (3713/3969) / (4225/3969)
= 3713/4225
Example 2 :
If θ is an acute angle, then find
(i) sin (π/4 - θ/2), when sin θ = 1/25
Solution :
sin (A - B) = sin A cos B - cos A sin B
sin (π/4 - θ/2) = sin π/4 cos θ/2 - cos π/4 sin θ/2
= (1/√2)cos θ/2 - (1/√2) sin θ/2
sin (π/4 - θ/2) = (1/√2)[cos θ/2 - sin θ/2]
Taking squares on both sides
sin2 (π/4 - θ/2) = (1/√2)2[cos θ/2 - sin θ/2]2
= (1/2)[cos2 θ/2 + sin2 θ/2 - 2 sin θ/2 cos θ/2]
= (1/2) [1 - sin θ]
= (1/2) [1 - (1/25)]
= (1/2) (24/25)
sin2 (π/4 - θ/2) = 12/25
sin (π/4 - θ/2) = √12/√25 = 2√3/5
(ii) cos (π/4 + θ/2), when sin θ = 8/9
Solution :
sin (A - B) = sin A cos B - cos A sin B
cos (π/4 + θ/2) = cos π/4 cos θ/2 - sin π/4 sin θ/2
= (1/√2)cos θ/2 - (1/√2) sin θ/2
cos (π/4 + θ/2) = (1/√2)[cos θ/2 - sin θ/2]
Taking squares on both sides
cos2 (π/4 - θ/2) = (1/√2)2[cos θ/2 - sin θ/2]2
= (1/2)[cos2 θ/2 + sin2 θ/2 - 2 sin θ/2 cos θ/2]
= (1/2) [1 - sin θ]
= (1/2) [1 - (8/9)]
= (1/2) (1/9)
cos2 (π/4 + θ/2) = 1/18
cos(π/4 + θ/2) = 1/√18 = 1/3√2
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