Addition and Subtraction Formulae
(i) sin (A +B) = sin A cos B +cos A sin B
(ii) cos (A +B) = cos A cos B -sin A sin B
(iii) tan (A +B) = (tan A + tan B)/(1 - tan A tan B)
(iv) sin (A -B) = sin A cos B - cos A sin B
(v) cos (A -B) = cos A cos B + sin A sin B
(vi) tan (A -B) = (tan A - tan B)/(1 + tan A tan B)
Corollaries
(i) tan(/4 +A) = tan(45° + A) = (1 + tan A)/(1 -tan A)
(ii) tan(/4 -A) = tan(45° -A) = (1 - tan A)/(1 + tan A)
(iii) cot (A +B) = (cot A cot B - 1)/(cot B + cot A)
(iv) cot (A -B) = (cot A cot B + 1)/(cot B -cot A)
(v) tan (A +B +C) = (tan A +tan B +tan C -tan A tan B tan C)/(1 - tan A tan B
- tan B tan C - tan C tan A)
(vi) sin (A +B) sin (A -B) = sin²A -sin²B
(vii) cos (A +B) cos (A -B) = cos²A -sin²B
Illustrative Examples
Example
Using t-ratios of 45° and 60°, evaluate
(i) sin 105° (ii) tan (13 /12).
Solution
- sin 105° = sin (60° +45°) = sin 60° cos 45° + cos 60° sin 45°
=((3)/2)(1/2) + (1/2)(1/2)
= (3 + 1)/22
=
= (-1 + 3)(1 - (-1)3) = (3
- 1)/(3 + 1)
Example
- Prove that cos - sin
= 2 cos( +
/4)
- Find the maximum and minimum values of 7 cos + 24
sin .
Solution
- cos - sin =
2 [(cos ) (1/2) -
(sin )(1/2)]=
2(cos cos/4
-sin sin /4)= 2
cos ( + /4)
- Since 7² +24² = 49 +576 = 625 = 25²,
7 cos + 24 sin
= 25 [(cos )(7/25) + (sin
)(24/25)]
Now, we can find an angle such that cos
= 7/25 and sin = 24/25
7 cos + 24 sin = 25 (cos
cos + sin
sin ) = 25 cos (-)
Now -1 cos (-)
1
=> -25 25 cos (-)
25
=> -25 (7 cos + 24 sin
) 25
Hence the given expression varies between -25 and 25.
Therefore, the maximum and minimum values of 7 cos + 24
sin are 25 and -25 respectively.
In general, a cos + b sin =
(a² +b²). cos (-),
where is given by
cos = a/(a² +b²), sin
= b/((a² +b²)
Thus a cos +b sin varies between -(a²
+b²) and (a² +b²).
Exercise
- Prove that sin 75° = [6 + 2]/4
- Find tan 15° and hence show that tan 15° +cot 15° = 4
- Evaluate (i) cos 195° (ii) sin(3 /4)
- Simplify by reducing to a single term:
(i) sin 38° cos 22° +cos 38° sin 22°
(ii) cos 70° cos 10° +sin 70° sin 10°
(iii) sin (x -y) cos x -cos (x -y) sin x
(iv) (tan 69° +tan 66°)/(1 -tan 69° tan 66°)
- Evaluate
(i) cos 105° +sin 105° (ii) cos 15° -sin 15°
(iii) cot 105° -tan 105°
- (i) A positive acute angle is divided into two parts whose tangents are
1/2 and 1/3. Show that the angle is /4.
(ii) Prove that tan 22° +tan 23° +tan 22° tan 23° = 1.
(iii) If A +B = 45°, show that (1 +tan A)(1 +tan B) = 2 and hence find the value of tan 22½°.
(iv) If A +B = 225°, prove that tan A +tan B = 1 -tan A tan B.
- Prove that
(i) tan 70° = tan 20° + 2 tan 50°
(ii) tan 13A = tan 4A + tan 9A + tan 4A tan 9A tan 13A
(iii) (1 + tan A)(1 + tan B) = 2 tan A, if A -B = 45°
(iv) tan (x - y) + tan (y - z) + tan (z - x) = tan (x -y) tan (y - z) tan (z - x)
(v) tan 56° = (cos 11° +sin 11°)/(cos 11° -sin 11°)
- Prove that
sin (x + y) sin (x - y) + sin (y + z) sin (y - z) + sin (z + x) sin (z - x) = 0.
- Prove that sin (A + B + C) = cos A cos B cos C (tan A + tan B + tan C-tan A tan B tan C).
- Prove that
sin A sin (B - C) + sin B sin (C - A) + sin C sin (A -B) = 0.
- In any quadrilateral ABCD, show that
cos A cos B -cos C cos D = sin A sin B -sin C sin D.
Answers
2. (3 -1)/(3 +1)
3. (i) -(3 + 1)/22
(ii) 1/2
4. (i) (3)/2
(ii) 1/2 (iii) - sin y (iv) -1
5. (i) 1/2 (ii) 1/2
(iii) 23
6. (iii)2 -1