T-ratios of Allied Angles

• Two angles are said to be allied if their sum or difference is a multiple of 90°. For example, 30° and 60° are allied, 150° and -30° are allied.

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

• Any function of an angle (n± ) treating as acute is numerically equal to same function of , with sign depending upon the quadrant in which the revolving line terminates. The proper sign can be ascertained by "All-Sin-Tan-Cos" formula. For example sin (180°+ ) = -sin ; -ve sign was chosen because 180° + lies in third quadrant and sin is -ve in third quadrant.
• Any function of an angle, treating as acute, is numerically equal to co-function of , with sign depending upon the quadrant in which the revolving line terminates. Note that sin and cos are co-functions of each other; tan and cot are co-functions of each other; sec and cosec are co-functions of each other.

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