FIND THE VALUE OF COS 2A WHEN OTHER TRIGONOMETRIC RATIOS GIVEN

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,

In the above formula, let us choose the 3^rd 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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