cos^2 x + cos^ 2( x + π/3) + cos^2( x - π/3) = 3/2 According to Ch 3 Trigonometric Functions std 11th
Answers
Answer:
Step-by-step explanation:
let us first calculate all three terms separately:
we know that
cos 2x = 2 cos^2 x -1
cos 2x + 1 = 2 cos^2 x \
cos 2x + 1 /2 = cos^2 x
so cos^2 x = cos 2x + 1/2.
replacing x with (x + pi/3) is about
cos^2(x + pi/3) = cos 2(x + pi/3) + 1/2
= cos(2x + 2pi/3) + 1/2
similarly,
replacing x with (x - pi/3) in cos^2 x = cos + 2x + 1/2
cos^2 (x - pi/3) = cos 2(x - pi/3) + 1/2
= cos(2x - 2pi/3) + 1/2
Solving LHS:
cos^2 x + cos^2(x + pi/3) + cos^2( x- pi/3)
= 1 + cos 2x/2 + 1 + cos(2x + 2pi/3)/2 + 1 + cos( 2x -2pi/3)/2
= 1/2{( 1 + cos 2x + 1 + cos(2x + 2pi/3) + 1 + cos (2x - 2pi/3)}
= 1/2{( 3 + cos 2x + cos( 2x + 2pi/3) + cos (2x - 2pi/3)}
Using cos x + cos y = 2 cos (x + y/2). cos (x - y/2)
replace x by ( 2x + 2pi/3) and y by (2x - 2pi/3)
= 1/2{ 3 + cos 2x + 2 cos( 2x + 2pi/3 + 2x - 2pi/3/2) .cos (2x + 2pi/3 -( 2x - 2pi/3/2)
= 1/2{( 3 + cos 2x + 2cos (4x + 0/2 ). cos( 0+ 4pi/3/2)
= 1/2{( 3 + cos 2x + 2cos (4x/2).cos (4pi/3/2)
= 1/2( 3 + cos 2x + 2cos 2x cos 2pi/3)
= 1/2{( 3 + cos 2x + 2cos 2x cos (pi-pi/3)}
= 1/2{( 3 + cos 2x + 2cos 2x (-cos(pi/3))} ( as cos (pi - tita) = -cos tita)
= 1/2{( 3 + cos 2x + 2cos 2x ( -1/2)
= 1/2 ( 3 + cos 2x - 2 *1/2*cos 2x)
= 1/2 ( 3 + cos 2x -cos 2x
= 1/2 (3 + 0)
= 3/2 which is RHS
HENCE LHS = RHS
HENCE PROVED