T-ratios of Standard Angles

	0°
sin	0	1/2	1/2	(3)/2	1
cos	1	(3)/2	1/2	1/2	0
tan	0	1/3	1	3	not defined
cot	not defined	3	1	1/3	0
sec	1	2/3	2	2	not defined
cosec	not defined	2	2	2/3	1

For memorising, we can use the following table:

	0°
sin
cos

Illustrative Examples

Example

Solve for a and c in the given triangle. Also find the area of the ABC.

Solution

sin A = height / hypotenuse = BC/AC
=> sin 30° = a/12
=> a = 12 sin 30° = 12.(1/2)= 6
Similarly cos A = AB/AC => cos 30° = c/12
=> c = 12 cos 30° = 12.(3)/2 = 63
Area of ABC = (1/2) x base x height = (1/2) x c x a = (1/2) x 63 x 6
= 183 sq. units.

Example

If A, B, A +B, A -B are positive acute angles, find the values of A and B from the equations:
sin (A -B) = 1/2, cos (A +B) = 1/2

Solution

The given equations are
sin (A -B) = 1/2 = sin 30° => A -B = 30° ...(i)
cos (A +B) = 1/2 = cos 60° => A +B = 60° ...(ii)
Solving (i) and (ii) simultaneously, we get
A = 45°, B = 15°

Exercise

Show that
(i) sin 30° cos 0° +sin 45° cos 45° +sin 60° cos 30° = 7/4
(ii) 4 sin (/6) sin²(/3) +3 cos(/3) tan (/4) +cosec²(/2) = 2 sec² (/4)
Evaluate sec 30° tan 60° +sin 45° cosec 45° +cos 30° cot 60°
Taking A = 30°, verify that
(i) sin2 A +cos 2 A = 1 (ii) sin 3 A = 3 sin A -4 sin³ A
(iii) sin 2 A = (2 tan A)/(1 -tan² A)
(iv) cos 2 A = (1 -tan² A)/(1 +tan²A)
Taking A = 60°, B = 30°, verify that
(i) sin (A +B) sin A +sin B
(ii) cos (A +B) cos A +cos B
Find the value of (0° < < 90°) satisfying
(i) cos / (cosec +1) + cos / (cosec -1) = 2
(ii) (cos² -3cos +2)/ sin² = 1
(iii) 2 sin² = 1/2
(iv) 3 cos = 2 sin²
(v) 2 cosec = 3 sec²
(vi) tan + cot = 2
(vii) sec² = 1 + tan
Assuming A, B, A +B, A -B to be positive acute angles, find A and B when
(i) sin (A +B) = (3)/2, cos (A -B) = (3)/2
(ii) tan (A +B) = 3, tan (A -B) = 1

Answers

2. 7/2
5. (i) 45° (ii) 60° (iii) 30° (iv) 60°
(v) 30° (vi) 45° (vii) 45° (discarding

= 0° as 0° <

< 90°)
6. (i) A = 45°, B = 15° (ii) A = 52·5°, B = 7·5°