PLANE TRIGONOMETRY. 





6. Express * 

 similar. 1+ n. 



COT. A _ sin. (90 -A) _ 2 Bin. (45 - j A) OOB. (45 - A) 

 I -r COB. (90 -A) ' 2 ooa.* (45 - } A) 



by a single function. This IB 



7. Bring cos. 4 A - sin. 4 A to a single function. 



Cos. 4 A - sin.* A = (cos. 1 A + sin.* A) (ooa. 1 A - sin. 2 A) 

 = 1 X cos. 2 A = cos. 12 A. 



8. Bring sec. A + tan. A to a single function (remember that 

 90 + A = complement of A, and bear in mind signs of angle* 

 in second quadrant). 



1 i sin - A = ! + in. A^ I-OOB. (90^-KA) 

 "oosTA cos^A cos. A "* sin. (90+ A) 

 2. in.* J (901+ A) = ton (45 . + 4 A) 



2 sin. * (90+ A) oos. i (90+ A) 



9. Reduce sin. A tan. i A. 



Sin. A tan. 1 A = 2 sin. i A cos. } A 8m " * A = 2 sin. 2 4 A 



cos. $ A 



= I cos. A = vorsin. A. 



10. Simplify t. A + tan. A 



cot. A - tan. A 



cos. 2 A -fain.* A 



Cot. A -f tan. A _ sin. A 003. A _ cos.* A -f sis- 1 A 



Cot. A tan. A cos.' A - sin.* A oos-* A - sin.* A 



sin. A cos. A 



= s~r = sec. 2 A. 



cos. 2 A 



11. Express ain - A + sin - 3A b y a single function. By (41) 

 and (43) cos. A + cos. 3 A 



Sin. A-f sin. 3 A _ 

 COB. A -f~oos. 3 A * 



2sin. A + 3A co 8 



2 



A-3A 



A + 3A A-3A 

 .-~ -- cos. _ 



= tan. 2 A. 



12. Reduce Bm.A + sin. 3 A + sin. 5 A Apply (41)and( 4 3) 



cos. A -|- cos. 3 A + cos. 5 A 



as before, but only to first and last terms of numerator and 

 denominator ; then 

 sin. A + sin. 3 A -f sin. 5 A_sin. 3 A(l +2 cos. 2 A) _ 



cos. A~+ oos. 3 A~4-~cos. 5 A cos. 3 A (1 + 2 cos. 2 A) 



13. Show that sin. (A + B) sin. (A- B) = sin.* A - sin.* B. 

 By (33) and (34) 



Sin. (A + B) sin. (A - B) = sin. 2 A cos. 2 B - cos. 2 A sin. B 



= sin. 2 A (1 - sin. 2 B) - sin.* B (1 - sin.* A) 



= sin.* A - sin.* A sin. 2 B sin.* B + sin. 2 A sin. 1 B 



= sin. 2 A - sin. 2 B. 



14. Show that cos. (A -f B) cos. (A - B) = cos.* A - sin.* B. 

 Proceed as in last case, by (35) and (36). 



15. Solve the equation, a tan. x b cos. x. Multiply both 

 aides by cos. x ; then 



a sin. x = I cos. 2 x = b (1 - sin. 2 x) ; 



.'. sin. 2 x -f- ... sin. x - 1 = ; 

 o 



.. . _ -- 



16. Show that tan.* A - tan. 2 B = i. (A + B) sin. (A- B) 



cos.* A cos/ B 



Tan. 2 A tan. 2 B = (tan. A + tan. B) (tan. A tan. B) 

 _ /sin. A , sin. B\ /sin. A _ sin. B\ 

 \cos.A cos. B/ Voos. A .oos. B/ 



(sin. A cos. B -f cos. A sin. B\ /sin. A cos. B - oos. A sin. B\ 

 coa. A cos. I : / \ cos. A oos. B / 



= "iB-JAjfJ) sin. (A - B) _ sin. (A + B) sin. (A B) 

 COB. A cos. B cos. A cos. B oos. 2 A oos. 3 B 



17. Solve the aimultaneoux equation*, Bin. g + sin. y = . 



cot. c -f COB. y fr. 

 By (41) and (43) 



2 Bin. 4 ( + y ) OOB. I ( y) m . 

 2ooB.K + V)co.*<-y) = &. 

 Dividing the first equation by the second, we obtain 



tan. 





Again, squaring both equations, and adding together, w 

 obtain 



4 co..' 



V. 



2V 2 



But the last factor of left-hand side = 1, 



.-. cos.!(*- y) = *(< + 6). 



From these two results the unknown quantities x and y may 

 be found by addition and subtraction. 



XIX. Subsidiary Angles. Trigonometrical calculations may 

 often be simplified in form by introducing a subsidiary or imagi- 

 nary angle, by which the sum or difference of two or more 

 magnitudes may be expressed by a product or quotient often 

 a matter of importance in calculating with logarithm*. An 

 example or two will best explain our meaning : 



Solve the equation, * = a sin. A + b cos. A. 



Now assume a subsidiary angle 8, such that = tan. 9, and 

 substitute this value in above equation. 



Then z = a (sin. A + cos. A) 



= a (sin. A + tan. 8 cos. A) 

 _ sin. A oos. 8 + sic. 8 coa. A 



ooa. 9 



_ sin. (A + 0) 

 cos. 



This is a much more manageable expression to deal with, 

 being already known by the assumption. As tan. may be 

 anything from to ce, the relative values of a and 6 are im- 

 material ; but had it been desired to introduce sin. 0, it would 

 have been necessary to see that the ratio assumed to represent. 

 it did not exceed unity, as sin. cannot exceed 1. 



Solve the equation a sin. z b cos. x = c. 



a sin. x - c= b cos. x. 

 (a sin. z c)* = fc* cos.** = 6* (1 -- ain. 2 *). 



Reducing this equation, we obtain the following quadratio 

 for sin. x : 



(a 2 + b 2 ) sin. 2 * 2ac sin. * - (b* c 1 ) = 0. 



The solution of which is 



Sin. * = - 



Now assume that = tan. 0, or b = a tan. 9, and substi- 







tute this in the above equation ; then we obtain 

 a (sin. x + cos. * tan. 0) = e. 



Multiply each side by cos. 8, and 



a (sin. x + cos. 8 + cos. x sin. 9) = c cos. 9 ; 

 or, a sin. (* + 0) = c oos. 9. 



From which we obtain the value of * + 9, and ultimately of 



x, 8 being already known from tan. 8 = -. 



a 



XX. Eatios between Sides and Angles of Plane Triangles in 

 general. The solution of right-angled triangles was explained 

 in Section X., and offered little difficulty. But for the solution 

 of oblique-angled triangles more complex ratios have to be 

 established between the sides and angles, which aie contained 

 in the following propositions and formulae : 



1. Any two sides of a plane triangle are in the same ratio as 

 the sines of the opposite angles. 



For example, in anv triangle IBC, -= . '- -... (65) 



b sin. B 



