Question

16cos^3x sinx=2cosx -3cosx-cos5x prove

Answer 1

Answer:

LHS=RHS

Step-by-step explanation:

2 cos(x)-cos(3 x)-cos(5 x) = ^?16×(3 cos(x)+cos(3 x))/4 sin(x)^2

(3 cos(x)+cos(3 x))/4 = (3 cos(x))/4+1/4 cos(3 x):

2 cos(x)-cos(3 x)-cos(5 x) = ^?16 (3 cos(x))/(4)+(cos(3 x))/4 sin(x)^2

sin(x)^2 = 1/2 (1-cos(2 x)):

2 cos(x)-cos(3 x)-cos(5 x) = ^?16×(1-cos(2 x))/2 ((3 cos(x))/(4)+(cos(3 x))/4)

(1-cos(2 x))/2 = 1/2-1/2 cos(2 x):

2 cos(x)-cos(3 x)-cos(5 x) = ^?16 1/2-(cos(2 x))/(2) ((3 cos(x))/(4)+(cos(3 x))/4)

16 ((3 cos(x))/(4)+(cos(3 x))/4) (1/2-(cos(2 x))/(2)) = 6 cos(x)-6 cos(x) cos(2 x)+2 cos(3 x)-2 cos(2 x) cos(3 x):

2 cos(x)-cos(3 x)-cos(5 x) = ^?6 cos(x)-6 cos(x) cos(2 x)+2 cos(3 x)-2 cos(2 x) cos(3 x)

cos(x) cos(2 x) = 1/2 (cos(x-2 x)+cos(x+2 x)) = 1/2 (cos(-x)+cos(3 x)):

2 cos(x)-cos(3 x)-cos(5 x) = ^?6 cos(x)+-3 (cos(-x)+cos(3 x))+2 cos(3 x)-2 cos(2 x) cos(3 x)

Use the identity cos(-x) = cos(x):

2 cos(x)-cos(3 x)-cos(5 x) = ^?6 cos(x)-3 (cos(x)+cos(3 x))+2 cos(3 x)-2 cos(2 x) cos(3 x)

-3 (cos(x)+cos(3 x)) = -3 cos(x)-3 cos(3 x):

2 cos(x)-cos(3 x)-cos(5 x) = ^?6 cos(x)+-3 cos(x)-3 cos(3 x)+2 ucos(3 x)-2 cos(2 x) cos(3 x)

cos(2 x) cos(3 x) = 1/2 (cos(2 x-3 x)+cos(2 x+3 x)) = 1/2 (cos(-x)+cos(5 x)):

2 cos(x)-cos(3 x)-cos(5 x) = ^?6 cos(x)-3 cos(x)-3 cos(3 x)+2 cos(3 x)+-cos(-x)-cos(5 x)

Use the identity cos(-x) = cos(x):

2 cos(x)-cos(3 x)-cos(5 x) = ^?6 cos(x)-3 cos(x)-3 cos(3 x)+2 cos(3 x)-cos(x)-cos(5 x)

6 cos(x)-3 cos(x)-3 cos(3 x)+2 cos(3 x)-cos(x)-cos(5 x) = 2 cos(x)-cos(3 x)-cos(5 x):

2 cos(x)-cos(3 x)-cos(5 x) = ^?2 cos(x)-cos(3 x)-cos(5 x)

The left hand side and right hand side are identical.