THE EXPRESSION a sin (9+6 cos 



31 



This converts the quantities a and b into fractions which are 

 trigonometrical ratios of the angle a, the base angle of a right- 

 angled triangle whose perpendicular is a and whose base is b. 

 (Fig. 19.) 



The expression thus becomes 



Va 2 + b 2 {cos cos a sin . sin a} 



a 



and finally, Va 2 + b 2 cos (6 + a) where tan a = r 



If b is negative and a is positive, then 

 a sin b cos = (b cos a sin 0) 



= Va 2 + b 2 cos (0 + a) where tan a = -7 



oc 



a 



FIG. 12. 



b 

 FIG. 13. 



a 



These results may be summarised as follows : 

 (b cos + a sin 0) = Va 2 + b 2 sin (0 + (3) where tan (3 = - 







(6 cos a sin 0) = Va 2 + b 2 cos (0 H a) where tan a = v 



18. By taking the relation y = a sin + b cos and putting it 







in the form y = Va 2 + 6 2 sin (0 + (3), where tan (3 = -, we are en- 







abled to very easily ascertain some of the impor ant features of 

 the function. Since the sine of an angle is never greater than 1 

 and never less than 1, y will be greatest when sin (0 + (3) = 1 



that is, when + (3 has the values -, , , etc., or when has 

 the values - - (3, - - (3, - - p, etc. 



a a a 



Thus the greatest value of y is Va 2 + b 2 , and this occurs when 



has the values - p, - ' - p, etc., at intervals of 2?r. 



' 



