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
- cos 495° = cos (360° +135°) = cos (135°)
= cos (90° +45°) = -sin 45° = -1/2
- sin 1230° = sin (3 x 360 +150)° = sin 150°
= sin (180° -30°) = sin 30° = 1/2
- tan (-1590°) = -tan 1590° = -tan (4 x 360 +150)°
= -tan 150° = -tan (180° -30°) = -(-tan 30°)
= tan 30° = 1/3
Example
- If tan A = -3, find all possible values of A between
0° and 360°.
- Find all values of x lying between 0 and 360 such that sin 2x° = 0·6428.
Solution
- 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°.
- 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.
- 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
- Which is greater: sin 40° or cos 40°?
- If = -400°, determine the sign of (sin
+cos ).
Solution
- 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°.
- 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
- Find all the t-ratios of (i) 120° (ii) 150° (iii) 180°
- Evaluate
(i) sin 930°
(ii) cos (-870°)
(iii) tan (-2025°)
(iv) cot (-315°)
(v) sin 19/4
(vi) cos 2/3
(vii) tan (-/3)
- Express the following as functions of angles less than 45°:
(i) sin (-1785°) (ii) cosec (-7498°).
- (i) Which is bigger: sin 55° or cos 55°?
(ii) If = 100°, determine the sign of (sin
+cos ).
- Evaluate
(i) 2sin 135° cos 210° tan 240° cot 300° sec 330°
(ii) sin 690° cos 930° +tan (-765°) cosec (1170°)
- If 8 = , show that cos 7
+cos = 0.
- If cos = -(3) /2, find all
possible values of between -180° and 180°.
- If cos = sin 200°, find all possible values of
between -180° and 60°.
- Given that cos = -0·5150, sin
is positive and that lies between 500° and 900°, find
the values of .
- Find x from the following equation:
cosec (90° + ) + x cos cot (90° + )
= sin (90° + ).
- Find all values of satisfying 0 <
< and tan²
+cot² = 2.
- 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