Amath Trigo

1)sinx+sin2x+sin 3x = 1+cosx+cos2x
2 sin2x cosx + sin2x = 1+cosx+2cos2x﹣1
sin2x(2 cosx + 1) = cosx(1 + 2 cosx)
(2cosx+1)(sin2x﹣cosx) = 0
(2cosx+1)(2 sinx cosx﹣cosx) = 0
(2cosx+1)(2 sinx﹣1)cosx = 0

cosx = -1/2 or 0 OR sinx = 1/2
x = 360no ± 120oOR180no ± 90oOR180no+(-1)n 30o

2a)sin4 x + cos4 x

= (sin2 x + cos2 x)2﹣2sin2 xcos2 x
= 1﹣(1/2)×(2 sin x cos x)2
= 1﹣(1/2)×(sin 2x)2
= 1﹣(1/2)×sin2 2x
= (2﹣sin2 2x)/2
= [2﹣(1/2) × (1﹣cos 4x)]/2
= [4﹣(1﹣cos 4x)]/4
= (3 + cos 4x)/4
b)sin4 x + cos4 x = 7/8
(3 + cos 4x)/4 = 7/8
3 + cos 4x = 7/2
cos 4x = 1/2
4x = 360no ± 60o
x = 90no ± 15o

sinx + sin 3x = 2 sin2x cos x
cos2x + 1 = 2cos2 x
sinx+sin2x+sin 3x=1+cosx+cos2x
4 sinx cos2 x + 2 sinx cosx = 2cos2 x + cosx
2 sinx cosx(2cosx + 1) = cosx(2cosx + 1)
2cosx + 1=0　or　cosx(2sinx - 1) = 0



2cosx + 1=0　
or
cosx(2sinx - 1) = 0



cosx = -1/2
or
cosx = 0
or
sinx = 1/2

x = 2nPi + 2Pi/3 or 2nPi - 2Pi/3

x = 2nPi + Pi/2 or 2nPi - Pi/2
or
x = nPi + (-1)n Pi/6

2a) LHS
= (1 - cos2 x)2 + cos4 x
= 1 - 2cos2 x + 2cos4 x
RHS
= (3 + cos4x) / 4
= (3 + 2cos2 2x - 1)/4
= [2 + 2(2cos2 x - 1)2 ] / 4
= (2 + 8cos4 x - 8cos2 x + 2)/ 4
= 1 - 2cos2 x + 2cos4 x
= LHS
b) sin4 x + cos4 x = 7/8
(3 + cos4x)/4 = 7/8 (by (a))
6 + 2cos4x = 7
cos4x = 1/2
4x = 2nPi + Pi/3 or 2nPi - Pi/3
x = nPi/2 + Pi/12 or nPi/2 - Pi/12