T-ratios of Multiple and Sub-multiple angles

T-ratios of 2A in terms of those of A

(i) sin 2A = 2 sin A cos A

(ii) cos 2 A = cos² A -sin² A = 1 -2 sin² A = 2 cos² A -1

T-ratios of 2 A in terms of tan A

(i) sin 2 A = 2 tan A / (1 + tan²A)

(ii) cos 2 A = (1 - tan² A)/(1 + tan²A)

(iii) tan 2 A = 2 tan A / (1 - tan²A)

T-ratios of A in terms of cos 2 A

(i) cos² A = (1 +cos2A)/2 i.e. cos A = ±

(ii) sin² A = (1 -cos2A)/2 i.e. sin A = ±

(iii) tan² A = (1 -cos2A)/(1 +cos2A) i.e. tan A = ±

T-ratios of sub-multiple angles

Replacing A by A/2 in the above formulae, we get

(i) sin A = 2 sin A/2 cos A/2

(ii) cos A = cos² A/2 - sin²A/2 = 1 -2 sin²A/2 2 cos²A/2 - 1

(iii) sin A = 2 tan A/2 /(1 + tan² A/2)

(iv) cos A = (1 - tan² A/2)/(1 + tan² A/2)

(v) tan A = 2 tan A/2 / (1 - tan² A/2)

(vi) cos A/2 = ±

(vii) sin A/2 = ±

(viii) tan A/2 = ±

T-ratios 3A in terms of those of A

(i) sin 3 A = 3 sin A -4 sin³ A

(ii) cos 3 A = 4 cos³ A -3 cos A

(iii) tan 3 A = (3tan A -tan³A)/(1 -3 tan² A)

T-ratios of some special angles

sin 18° = (5 -10/4, cos 18° = ([10 + 2 5]) /4

cos 36° = (5 +1)/4

sin 36° = ([10 -2 5]) /4

sin 72° = sin (90° -18°) = cos 18° = ([10 + 2 5]) /4

cos 72° = cos (90° -18°) = sin 18° = (5 -1)/ 4

sin 54° = sin (90° -36°) = cos 36° = (5 +1)/4

cos 54° = cos (90° -36°) = sin 36° = ([10 -2 5]) /4

tan = 2 -1

Illustrative Examples

Example

Prove that

  1. cos 6° cos 42° cos 66° cos 78° = 1/16
  2. sin /15 sin 2/15 sin 3/15 sin 4/15 = 5/16

Solution

  1. L.H.S. = cos 6° cos 42° cos 66° cos 78°
               = (1/4)(2 cos 66° cos 6°)(2 cos 78° cos 42°)
               = (1/4) [cos (66° +6°) +cos (66° -6°)] [cos (78° +42°) +cos (78° -42°)]
               = (1/4)(cos 72° +cos 60°)(cos 120° +cos 36°)
               = (1/4)(sin 18° +cos 60°)(cos 120° +cos 36°)
               =
               = = R.H.S.
  2. L.H.S. = sin/5 sin 2/5 sin 3/5 sin 4/5
               = sin/5 sin 2/5 sin
               = sin /5 sin 2/5 sin 2/5 sin /5
               = = (sin 36° sin 72°)² = (sin 36° cos 18°)²
               =
                = (100 - 20)/16 = 80/16 = 5 = R.H.S.

Example

Prove that = 2 cos

Solution

L.H.S. =
           = [2 + 2 cos 2 ] = [2+2(2cos² -1)]
           = [4cos² ] = 2 cos = R.H.S.

Exercise

  1. Evaluate without using tables or calculator:
    (i) 2 cos sin
    (ii) 2 cos² 15° -1
    (iii) 8 cos³20° -6 cos 20°
    (iv) 3 sin 40° -4 sin³ 40°
  2. Prove that
    (i) sin 6° sin 42° sin 66° sin 78° = 1/16
    (ii) cos 36° cos 42° cos 60° cos 78° = 1/16
    (iii) tan 6° tan 42° tan 66° tan 78° = 1
  3. Prove that
    (i) cos /7 cos 2/7 cos 4/7 = - 1/8
    (ii) (1+cos /8) (1+cos 3/8) (1+cos 5/8)(1+cos 7/8) = 1/8
  4. Prove that tan A +cot A = 2 cosec 2 A and deduce that tan 75° + cot 75° = 4.
  5. If 2 cos = x + 1/x, prove that cos 3 = (1/2)(x³ + 1/x³)
  6. Prove that
    (i) cos 4 = 1 -8 sin² cos² = 1 -8 cos² +8 cos4
    (ii) sin 4 = 4 sin cos³ -4 cos sin³
    (iii) sin 5 = 5 sin -20 sin³ +16 sin5
  7. Given that cos A/2 = 12/13, calculate without the use of tables, the values of sin A, cos A and tan A.
  8. Given that tan x = 12/5, cos y = 3/5 and the angles x and y are in the same quadrant, calculate without the use of tables the values of
    (i)sin (x +y)    (ii) cos y/2
  9. If 0 x < 2 , find sin x/2, cos x/2 and tan x/2 if
    (i) tan x = -4/3, x lies in second quadrant
    (ii) cos x = 1/3, x does not lie in second quadrant
    (iii) sin x = 1/4, x does not lie in first quadrant.
  10. Using tan (x -y) = (tan x - tan y)/(1 + tanx tan y) , evaluate tan 13/12
    [ Hint: Take x = 5/4 and y = /6]

Answers

1. (i) 1/2     (ii) 3/2    (iii) 1     (iv) 3/2
7. 120/169, 119/169, 120/169
8. (i) 56/65     (ii) -1/5
9. (i) 2/5, 1/5, -2    (ii) , -1/3, -2
(iii)
10. (3 - 1)/(3 + 1)