Prove that:
Sin3x-cos3x=(sin2x-cos2x)
X(1-2 sin2x cos2x)
Answers
ANSWER: The Answer is 2π/3 , 4π/3 , π/8 and 5π/8 .
Given that
sin x + sin 2x + sin 3x = cos x + cos 2x + cos 3x
(sin 3x + sin x) + sin 2x = ( cos 3x + cos x ) + cos 2x
2[sin(3x+x)/2].[cos(3x-x)/2] + sin 2x = 2[cos(3x+x)/2].[cos(3x-x)/2] + cos 2x
2[sin(4x)/2].[cos(2x)/2] + sin 2x = 2[cos(4x)/2].[cos(2x)/2] + cos 2x
2 sin 2x . cos x + sin 2x = 2 cos 2x . cos x + cos 2x
2 sin 2x . cos x + sin 2x - 2 cos 2x . cos x - cos 2x = 0
sin 2x ( 2 cos x + 1) - cos 2x (2 cos x + 1) = 0
( 2 cos x + 1) ( sin 2x - cos 2x ) = 0
⇒
2 cos x + 1 = 0 And sin 2x - cos 2x = 0
2 cos x = - 1 And sin 2x = cos 2x
cos x = - 1/2 And sin 2x/cos 2x = cos 2x/cos 2x
cos x = - 1/2 And tan 2x = 1
For cos x = - 1/2 :-
The Reference Angle is π/3
Cos is -ve in Quadrant II & III
For Quad II :- x = π - π/3 = 2π/3
For Quad III :- x = π + π/3 = 4π/3
For tan 2x = 1 :-
The Reference Angle is π/4
tan is +ve in Quadrant I & III
For Quad I :- 2x = π/4 ⇒ x = π/8
For Quad III :- 2x = π + π/4 = 5π/4 ⇒ x = 5π/8
The Answer is 2π/3 , 4π/3 , π/8 and 5π/8 .