Question

If \(\alpha,\beta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\), \[ \cos^4\alpha=\frac1{16},\qquad \sin^4\beta=\frac1{16} \] then \[ \cos\alpha+\cos\beta= \]

• A. \(\sqrt2\cos15^\circ\)
• B. \(\sqrt2\sin15^\circ\)
• C. \(-\sqrt2\cos15^\circ\)
• D. \(-\sqrt2\sin15^\circ\)

Hint:

Always use interval restrictions to determine correct sign of trigonometric values.

Answer 1

The Correct Option is A

Concept: Since angles lie in interval \[ \left(-\frac{\pi}{2},\frac{\pi}{2}\right) \] principal positive values are taken.

Step 1: Find \(\cos\alpha\).
\[ \cos^4\alpha=\frac1{16} \] \[ \cos\alpha=\frac12 \] because positive interval chosen.

Step 2: Find \(\cos\beta\).
\[ \sin^4\beta=\frac1{16} \] \[ \sin\beta=\frac12 \] Thus \[ \cos\beta=\sqrt{1-\frac14} \] \[ =\frac{\sqrt3}{2} \]

Step 3: Add values.
\[ \cos\alpha+\cos\beta \] \[ =\frac12+\frac{\sqrt3}{2} \] \[ =\frac{1+\sqrt3}{2} \] Now \[ \sqrt2\cos15^\circ = \sqrt2\left(\frac{\sqrt6+\sqrt2}{4}\right) \] \[ =\frac{\sqrt3+1}{2} \] Hence \[ \boxed{\sqrt2\cos15^\circ} \]