Choose the one alternative that best completes the statement
Solution
1) sin 2x sin 5x cos 2x cos 5x
let 2x = a
5x = b
sin a sin b = -1/2 [ cos (a+b) -cos (a-b) ]
-1/2 [ cos 7x - cos -3x ]
cos a cos b = 1/2 [sin (a+b) - sin (a-b) ]
cos 2x cos 5x = 1/2 [ sin 7x - sin (-3x) ]
-1/2 [ cos 7x - cos -3x ] * 1/2 [ sin 7x - sin (-3x) ] = 1/4 [ cos^2 3x - cos^2 7x ] ( option a)
2) cos 2x - cos 8x / cos 2x + cos 8x
cos 2x = cos ( 5x - 3x )
cos 8x = cos ( 5x + 3x )
plugging the values back
cos ( 5x - 3x ) - cos ( 5x + 3x ) / cos ( 5x - 3x ) + cos ( 5x + 3x )
applying cos (a-b) and cos (a+b) formula
cos (a-b) = cos a cos b + sin a sin b
cos (a+b) = cos a cos b - sin a sin b
cos ( 5x - 3x ) - cos ( 5x + 3x ) / cos ( 5x - 3x ) + cos ( 5x + 3x )
cos 5x cos 3x + sin 3x sin 5x - [ cos 5x cos 3x -sin 5x sin 3x ] / cos 5x cos 3x + sin 3x sin 5x + [ cos 5x cos 3x -sin 5x sin 3x ]
sin^2 5x sin^3x / cos^2 5x cos^2 3x
sinx / cos x= tan x
= tan 5x tan 3x
3) 2 cos x - 1 = 0
adding 1 one both sides
2 cos x = 1
dividing both sides by 2
cos x = 1/2
x = pi/3 + 2pi*n
or
x = 5pi/3 + 2 pi* n
(option a)
4) tan x sec x = - 2 tan x
adding 2 tan x on both sides
tan x sec x + 2 tan x = 0
taking tan x as gcf
tan x( sec x + 2 ) = 0
tan x = 0
sec x + 2 = 0
x = pi*n
or
x = 2pi/3 + 2pi*n , 4pi/3 + 2pi*n
option d

