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
- cos 6° cos 42° cos 66° cos 78° = 1/16
- sin /15 sin 2/15 sin
3/15 sin 4/15 = 5/16
Solution
- 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.
- 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
- 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°
- 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
- 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
- Prove that tan A +cot A = 2 cosec 2 A and deduce that tan 75° + cot 75° = 4.
- If 2 cos = x + 1/x, prove
that cos 3 = (1/2)(x³ + 1/x³)
- 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
- Given that cos A/2 = 12/13, calculate without the use of tables, the
values of sin A, cos A and tan A.
- 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
- 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.
- 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)