T-ratios of Allied Angles

Aid to Memory for formulae of t-ratios of allied angles

For memorising, we can use the following table:

  - 90°- 90° + 180°- 180° + 270°- 270° + 2n - 2n +
sin -sin cos cos sin -sin -cos -cos -sin sin
cos cos sin -sin -cos -cos -sin sin cos cos
tan -tan cot -cot -tan tan cot -cot -tan tan

Illustrative Examples

Example

Find the values of
(i) cos 495°      (ii) sin 1230°   (iii) tan (-1590°).

Solution

  1. cos 495° = cos (360° +135°) = cos (135°)
                   = cos (90° +45°) = -sin 45° = -1/2
  2. sin 1230° = sin (3 x 360 +150)° = sin 150°
                   = sin (180° -30°) =  sin 30° = 1/2
  3. tan (-1590°) = -tan 1590° = -tan (4 x 360 +150)°
           = -tan 150° = -tan (180° -30°) = -(-tan 30°)
           = tan 30° =   1/3

Example

  1. If tan A = -3, find all possible values of A between 0° and 360°.
  2. Find all values of x lying between 0 and 360 such that sin 2x° = 0·6428.

Solution

  1. We know that for first quadrant, tan 60° = 3.
    Now tan A = -3, so A lies in second or fourth quadrant.
    Since required values of A are 180°-60° or 360°-60° i.e. 120° or 300°,
    we get A = 120° or 300°.
  2. From trigonometric tables, we find that 0·6428 = sin 40°.
    As sin 2x° = 0·6428, a positive value, 2x° lies in first or second quadrant,
    we get 2x° = 40° or 180°-40° i.e. 40° or 140°
    => x = 20 or 70.
  3. Since 0 < x° < 360°, so 0 < 2x° < 720°, therefore,
    360° +40° and 360° +140° should also be considered.
    So we may have 2 x = 400 or 500 =>  x = 200 or 250.
    Hence required values of x are 20, 70, 200 or 250.

Example

  1. Which is greater: sin 40° or cos 40°?
  2. If = -400°, determine the sign of (sin +cos ).

Solution

  1. In first quadrant, 1 > 2   =>  sin 1 > sin 2, as value of sin steadily increases from 0 to 1 as   increases from 0° to 90°.
    Now cos 40° = cos (90° -50°) = sin 50°.
    Therefore sin 50° > sin 40°  =>  cos 40° > sin 40°.
  2. sin (-400°) = -sin (400°) = -sin (360° +40°) = -sin (40°)
    cos (-400°) = cos (400°) = cos (360° +40°) = cos 40°
    = cos (90° -50°) = sin 50°
    Hence sin (-400°) +cos (-400°) = -sin 40° +sin 50°
    = sin 50° -sin 40° > 0.              (since  sin 50° > sin 40°)

Example

If ABCD is a cyclic quadrilateral, then show that
                cos A +cos B +cos C +cos D = 0.

Solution

Since ABCD is a cyclical quadrilateral,
A +C =, B +D = C = -A, D = -B
Hence cos A +cos B +cos C +cos D
                = cos A +cos B +cos ( -B)
                = cos A +cos B -cos A -cos B = 0.

Exercise

  1. Find all the t-ratios of (i) 120° (ii) 150° (iii) 180°
  2. Evaluate
    (i) sin 930°
    (ii) cos (-870°)
    (iii) tan (-2025°)
    (iv) cot (-315°)
    (v) sin 19/4
    (vi) cos 2/3
    (vii) tan (-/3)
  3. Express the following as functions of angles less than 45°:
    (i) sin (-1785°)   (ii) cosec (-7498°).
  4. (i) Which is bigger: sin 55° or cos 55°?
    (ii) If = 100°, determine the sign of (sin +cos ).
  5. Evaluate
    (i) 2sin 135° cos 210° tan 240° cot 300° sec 330°
    (ii) sin 690° cos 930° +tan (-765°) cosec (1170°)
  6. If 8 = , show that cos 7 +cos = 0.
  7. If cos = -(3) /2, find all possible values of between -180° and 180°.
  8. If cos = sin 200°, find all possible values of between -180° and 60°.
  9. Given that cos = -0·5150, sin is positive and that lies between 500° and 900°, find the values of .
  10. Find x from the following equation:
      cosec (90° + ) + x cos cot (90° + ) = sin (90° + ).
  11. Find all values of satisfying 0 < < and tan² +cot² = 2.
  12. If A, B, C, D are angles of a cyclical quadrilateral, prove that
    (i) cot A +cot B +cot C +cot D = 0
    (ii) sin A +sin B +sin C +sin D = 0
          [Hint. A +C = B +D = .]

Answers

1. (i) sin 120° = (3)/2, cos 120° = 1/2, tan 120° = -3,
       cot 120° = -1/3, sec 120° = -2,
       cosec 120°  = 2/3.
  (ii) sin 150° = 1/2, cos 150°  = -(3)/2,
      tan 150° = -1/ 3,  cot 150° = -3,
       sec 150° = -2/ 3,  cosec150° = 2.
(iii) sin 180° = 0, cos 180° = -1, tan 180° = 0,
     sec 180° = -1;
     cot 180° and cosec 180° are undefined.
2. (i) 1/2  (ii) -(3)/2   (iii) -1         (iv) 1/2
(v) 1/2 (vi) 1/2        (vii) -3.
3. (i) sin 15°   (ii) sec 28°
4. (i) sin 55°   (ii) +ve
5. (i) 1            (ii) (3)/4 +1
7. 120°, -120°
8. -110°, 110°, 250°
9. 841°
10. x = tan
11. /4, 3/4