T-ratios of Standard Angles

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:

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

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°