and

93. Ex. 1. Find the values of sin 75° and cos'15°.

sin 75°=sin (45° +30°) = sin 45° cos 30° + cos 45° sin 30°

cos 75° = cos (45° + 30°) = cos 45° cos 30°- sin 45° sin 30°

Ex. 2. Assuming the formulae for sin (x+y) and cos(x+y), deduce the formulae for sin (x-y) and cos (x-y).

and

We have

sinx=sin {(x − y)+y}=sin(x − y) cos y + cos (x − y) sin y................ (1), cos x=cos {(x − y)+y} = cos (x − y) cos y − sin (x − y) sin y................(2). Multiplying (1) by cosy and (2) by sin y and subtracting, we have sin x cos y - cos x sin y=sin (x − y) {cos2 y + sin2 y} = sin (x -- y). Multiplying (1) by sin y and (2) by cosy and adding, we have

sin x sin y+cos x cos y = cos (x − y) { cos2 y + sin2 y} = cos(x − y).

Hence the two formulae required are proved.

These two formulae are true for all values of the angles since the formulae from which they are derived are true for all values.

4. sin (A+B) sin (A – B)=sin2 A – sin2 B.

5. cos (A+B) cos (A – B) = cos2 A − sin2 B.

6. cos (45° – A) cos (45° – B) – sin (45° – A) sin (45° – B) = sin (A + B).

7. sin (45° +A) cos (45° – B) + cos (45° + A) sin (45° – B) = cos (A – B).

12.

cos (a+ B) cos y − cos (B+y) cos a = sin ẞ sin (y - a).

13. sin (n+1) A sin (n − 1) A + cos (n + 1) A cos (n - 1) A = cos 24. 14. sin (n+1) A sin (n+2) A + cos (n + 1) A cos (n+2) A=cos A.

94. From Arts. 88 and 90, we have, for all values of A and B,

and

and

sin (A + B) = sin A cos B + cos A sin B,

sin (A — B) = sin A cos B – cos A sin B. Hence, by addition and subtraction, we have

sin (A+B) + sin (A - B) = 2 sin A cos B......(1),

sin (A+B) sin (AB) = 2cos A sin B......(2). From the same articles we have, for all values of A and B,

and

and

cos (A + B) = cos A cos B-sin A sin B, cos (A — B) = cos A cos B+ sin A sin B. Hence, by addition and subtraction, we have

cos (A+B) + cos (AB)=2 cos A cos B......(3), cos (AB) cos (A + B) = 2 sin A sin B......(4). Put A + B = C, and A − B = D, so that

On making these substitutions the relations (1) to (4) become, for all values of C and D,

[The student should carefully notice that the left-hand member of IV is cos D-cos C and not cos C – cos D.]

95. These relations I to IV are extremely important and should be very carefully committed to memory.

On account of their great importance we give a geometrical proof for the case when C and D are acute angles.

Let AOC be the angle C and AOD the angle D. Bisect the angle COD by the straight line OE. On OE take a point P and draw QPR perpendicular to OP to meet OC and OD in Q and R respectively.

Draw PL, QM and RN perpendicular to OA, and through R draw RST perpendicular to PL or QM to meet them in S and T respectively.

Since the angle DOC is C-D, each of the angles

Since the two triangles POR and POQ are equal in all respects we have OQ OR, and PR= PQ, so that

=

RQ=2RP.

Hence QT=2PS, and RT = 2RS, i.e. MN = 2ML. Therefore MQ+NR = TQ + 2LS=2SP+2LS=2LP. Also OM + ON = 20M + MN = 20M + 2ML = 20L.

for ▲ SPR = 90° — ≤ SPO = 2 LOP =

C+D
2

96. The student is strongly urged to make himself perfectly familiar with the formulae of the last article and to carefully practise himself in their application; perfect familiarity with these formulae will considerably facilitate his further progress.

The formulae are very useful because they change sums and differences of certain quantities into products of certain other quantities, and products of quantities are, as the student probably knows from Algebra, easily dealt with by the help of logarithms.

We subjoin a few examples of their use.

[This is an example of the simplification given by these formulae; it would be a very long and tiresome process to look out from the tables the values of sin 75°, sin 15°, cos 75°, and cos 15°, and then to perform the division of one long decimal fraction by another.]