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

1. 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)
2. Evaluate sec 30° tan 60° +sin 45° cosec 45° +cos 30° cot 60°
3. 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)
4. Taking A = 60°, B = 30°, verify that
   (i) sin (A +B) sin A +sin B
   (ii) cos (A +B) cos A +cos B
5. 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
6. 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