Question

Prove that cos2x + cos2(x+π/3)+〖cos〗^2 (x-π/3)=3/2

Answer 1

we know,
cos(90 + ø) = sinø
cos²x = (1 + sin2x)/2

we have to $\underline{prove\: :-}$
$cos^{2} x + cos {}^{2} ({x}{} + \frac{\pi}{3}) + cos {}^{2} ( x - \frac{\pi}{3} ) = \frac{3}{2}$

now, (1 + cos2x)/2 + [1 + cos(2x + $\frac{2\pi}{3}$)]/2 + [ 1 + cos(2x - $\frac{2\pi}{3}$)]/2 $\to\to\to\to\to\to\to\to\to\therefore\boxed{1 + cos2X = 2cos^2x}$

1/2 [ 1 + cos2x + 1 + cos(2x + $\frac{2\pi}{3}$) + 1 + cos(2x - $\frac{2\pi}{3}$)]

1/2[3 + cos2x + cos(2x + $\frac{2\pi}{3}$) + cos(2x - $\frac{2\pi}{3}$)] $\to\to\to\to\to\to\to\to\to\therefore\boxed{cos(x + y) + cos(x - y) = 2cosx * cosy}$

1/2[ 3 + cos2x + (2cos(2x) × cos$\frac{2\pi}{3}$ ]

1/2[3 + cos2x + 2cos2X × cos(120) ]
1/2[3 + cos2X + 2cos2X × cos(90 + 30) ]

1/2[3 + cos2X + 2cos2X × (-sin30)] $\to\to\to\to\to\to\to\to\to\therefore\boxed{cos(\frac{\pi}{2} + \theta) = -sin\theta}$

1/2[3 + cos2X + 2cos2X × (-1/2)] $\to\to\to\to\to\to\to\to\to\therefore\boxed{sin30\degree= 1/2}$ 1/2[ 3 + cos2X - cos2x] 3/2

Answer 2

this solution may help you