161. It is important that the student should be thoroughly familiar with the second set of formulæ on p. 126.

Written as follows, they may be regarded as the inverse of the 'S, T' formulæ.

2 sin A. cos B = sin (A + B) + sin (A – B),

2 cos A. sin B = sin (4 + B) − sin (A – B),
2 cos A. cos B = cos (A + B) + cos (A – B),
2 sin A. sin B = cos (A – B) — cos (A + B)..

iv.

EXAMPLES. XXXVIII,

Express as the sum or as the difference of two trigonometrical ratios the ten following expressions:

(11) Simplify 2 cos 20. cos 0 - 2 sin 40. sin 0.

** MISCELLANEOUS EXAMPLES. XXXIX.

(1) If tan a=

and tan ß=}, prove that tan (a+B)=1.

and tan ß=, prove that one of the values

COS a- cos 5a
sin a+ sin 5a
sin 5x - sin 3x
cos 5x+cos 3x
cos A+ cos 34
cos 34+ cos 5A
sin 3x - sin x
cos 3x + cos x

(6) Prove that

(7) Simplify

(8) Simplify

(9) Prove that

+

COS 24

=

COS 44 sin 3x + sin x

cos 3x

(sin 44 - sin 24) (cos A
(cos 44 + cos 24) (sin A + sin 34)*

2 sin 2a. cos a +2 cos 4a. sin a=sin 5a+sin a.

(10) Prove that

cos 2a. cos a - sin 4a. sin a=cos 3a. cos 2a. (11) tan 24. tan 34. tan 5A=tan 54 - tan 34 - tan 24. (12) Solve 4 sin (0+). cos (0-4)=3)

4 cos (8+). sin (0-6)=1*

(13) Prove that

sin A. sin 2A + sin 2A. sin 5A + sin 34. sin 104 cos A. sin 24+ sin 24. cos 5A-cos 3A. sin 104

CHAPTER XII.

ON THE TRIGONOMETRICAL RATIOS OF MULTIPLE

ANGLES.

162. To express the Trigonometrical Ratios of the angle 24 in terms of those of the angle 4.

Since sin (A+B) = sin A. cos B + cos A . sin B ;

... sin (4 + 4) = sin A. cos A + cos A. sin A ;

.: sin 24 2 sin A. cos A

=

Also, since cos (A + B) = cos A. cos B - sin A. sin B;

(1).

.. cos (A + A) = cos A. cos Asin A. sin A;

..cos 24 cos2 A - sin3 A

=

1 = cos3 A + sin3 A ;

But

.. 1+ cos 24 = 2 cos3 A,

(2).

and 1-cos 24 = 2 sin3 4.

The last two results are usually written

cos 242 cos3 A-1

(3),

163.** To prove the '2A' formula geometrically.

M

02A M

R

Let ROP be the angle 24. describe the semicircle RPL. Join RP. PL.

With centre O and any radius Draw PM perpendicular to OR.

Then the angle RPL in a semicircle is a right angle. The angle ROP=OLP+OPL=20LP [since OL=0P]. .. OLP= a half of ROP=A. Also MPR and OLP are each the comple

ment of MPL. :. MPR=OLP=A.

Then

=

2LM. LP

OP
OP

LP. LR OP

=2 cos2 A - 1.

=2 cos MLP. cos PLR-1

20M=M'M=LM-LM'=LM-MR. Hence,

166. The student must notice that A is any angle, and therefore these formulæ will be true whatever we put for A.

A

Example. Write 4 instead of A, and we get