We are given: \[ \cos^4 \theta - \sin^4 \theta = \frac{1}{2} \] Using the difference of squares formula: \[ \cos^4 \theta - \sin^4 \theta = (\cos^2 \theta - \sin^2 \theta)(\cos^2 \theta + \sin^2 \theta) \] Since \(\cos^2 \theta + \sin^2 \theta = 1\), the equation simplifies to: \[ \cos^2 \theta - \sin^2 \theta = \frac{1}{2} \] This is the standard identity for \(\cos(2\theta)\): \[ \cos(2\theta) = \frac{1}{2} \] From trigonometric identities, we know: \[ \cos(60^\circ) = \frac{1}{2} \] Thus, \(2\theta = 60^\circ\), which gives: \[ \theta = 30^\circ \]
The correct answer is option (A): \(30°\)
The value of \(\dfrac{\sqrt{3}\cosec 20^\circ - \sec 20^\circ}{\cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ}\) is equal to
If $\cot x=\dfrac{5}{12}$ for some $x\in(\pi,\tfrac{3\pi}{2})$, then \[ \sin 7x\left(\cos \frac{13x}{2}+\sin \frac{13x}{2}\right) +\cos 7x\left(\cos \frac{13x}{2}-\sin \frac{13x}{2}\right) \] is equal to
If \[ \frac{\cos^2 48^\circ - \sin^2 12^\circ}{\sin^2 24^\circ - \sin^2 6^\circ} = \frac{\alpha + \beta\sqrt{5}}{2}, \] where \( \alpha, \beta \in \mathbb{N} \), then the value of \( \alpha + \beta \) is ___________.