Question

The value of \[ \cos^2 x + \cos^2 \left(x + \frac\pi3\right) + \cos^2 \left(x - \frac\pi3\right) \] is:

• A. zero
• B. 1
• C. $\frac12$
• D. $\frac32$

Hint:

For sums of squared trigonometric functions, use $\cos^2 \theta = \frac1 + \cos 2\theta2$ and sum-to-product identities to simplify.

Answer 1

The Correct Option is D

Step 1: Recall the trigonometric identity: \[ \cos^2 \theta = \frac1 + \cos 2\theta2 \] Step 2: Apply this to each term: \[ \cos^2 x = \frac1 + \cos 2x2 \] \[ \cos^2 \left(x + \frac\pi3\right) = \frac1 + \cos \left(2x + \frac2\pi3\right)2 \] \[ \cos^2 \left(x - \frac\pi3\right) = \frac1 + \cos \left(2x - \frac2\pi3\right)2 \] Step 3: Add them up: \[ \cos^2 x + \cos^2 \left(x + \frac\pi3\right) + \cos^2 \left(x - \frac\pi3\right) = \frac3 + \left[ \cos 2x + \cos \left(2x + \frac2\pi3\right) + \cos \left(2x - \frac2\pi3\right) \right]2 \] Step 4: Use the identity: \[ \cos(A + B) + \cos(A - B) = 2\cos A \cos B \] Here, $A = 2x$, $B = \frac2\pi3$: \[ \cos \left(2x + \frac2\pi3\right) + \cos \left(2x - \frac2\pi3\right) = 2\cos(2x)\cos\left(\frac2\pi3\right) \] Step 5: Since $\cos\left(\frac2\pi3\right) = -\frac12$: \[ = 2\cos(2x) \times \left(-\frac12\right) = -\cos(2x) \] Step 6: Therefore, \[ \cos 2x + \left[ -\cos 2x \right] = 0 \] The sum inside the bracket is zero. Step 7: The total becomes: \[ \frac3 + 02 = \frac32 \] Thus, the value is $\mathbf\frac32$.