Trigonometry Quiz

The values are: $\cos(-765^\circ) = \cos(765^\circ) = \cos(360^\circ + 405^\circ) = \cos(45^\circ) = \frac{\sqrt{2}}{2}$, $\sin(105^\circ) = \sin(90^\circ + 15^\circ) = \sin(15^\circ) = \frac{1}{2}(\sqrt{6} - \sqrt{2})$, and $\sin(100^\circ) = \sin(90^\circ + 10^\circ) = \sin(10^\circ) = \frac{1}{2}(\sqrt{5} - 1)$.

The value is: $\sin(150^\circ) + \cos(300^\circ) - \tan(315^\circ) + \sec(3660^\circ) = \frac{1}{2} + \frac{1}{2} - 1 + 1 = 1$.

The value is: $\sin(420^\circ) \cos(390^\circ) + \sin(-330^\circ) \cos(-300^\circ) = \sin(60^\circ) \cos(30^\circ) + \sin(30^\circ) \cos(60^\circ) = \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{1}{2} \cdot \frac{1}{2} = \frac{3}{4} + \frac{1}{4} = 1$.

<p>Using the identity $\cos 2\theta = 2\cos^2\theta - 1$, we have $2\cos 20^\circ \cos 40^\circ \cos 80^\circ - 1 = 0$, so $\cos 20^\circ \cos 40^\circ \cos 80^\circ = \frac{1}{2}$.</p> Signup and view all the answers

<p>Using the identity $\sin 2\theta = 2\sin\theta\cos\theta$, we have $4\sin(20^\circ) \sin(40^\circ) \sin(60^\circ) \sin(80^\circ) = \frac{3}{2}$, so $8\sin(10^\circ) \sin(30^\circ) \sin(50^\circ) \sin(70^\circ) = \frac{3}{2}$.</p> Signup and view all the answers