1. Problem: Calculate the values for parts a) and f) of each given trigonometric task. 2. We will solve the first problem from each set (a) and the last problem (f) as requested. --- **Task 2. Find sin t, cos t for:** a) $t=27\pi$ - Since $\sin(t)$ and $\cos(t)$ are periodic with period $2\pi$, reduce $27\pi$ modulo $2\pi$: $$27\pi \equiv 27\pi - 13 \cdot 2\pi = 27\pi - 26\pi = \pi$$ - So $\sin(27\pi) = \sin(\pi) = 0$, $\cos(27\pi) = \cos(\pi) = -1$. f) $t=\frac{33\pi}{4}$ - Reduce modulo $2\pi$: $$\frac{33\pi}{4} = 8\pi + \frac{1\pi}{4}$$ - Since $8\pi$ is $4$ full rotations, effectively $t = \frac{\pi}{4}$. - $\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$, $\cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$. --- **Task 6. Calculate:** a) $\cos \pi - \cos 4\pi + \sin \frac{\pi}{2} \cdot \cos(-\pi)$ - $\cos \pi = -1$ - $\cos 4\pi = 1$ - $\sin \frac{\pi}{2} = 1$ - $\cos(-\pi) = \cos \pi = -1$ - Substitute: $$-1 - 1 + 1 \cdot (-1) = -1 - 1 - 1 = -3$$ f) $\frac{\cos^2 7\pi - 2 \sin^2 7\pi}{\cos^2 \frac{17\pi}{2} + 2 \sin^2 \frac{17\pi}{2}}$ - Reduce angles modulo $2\pi$: $7\pi = 3\pi + \pi = \pi$ (since $6\pi$ is $3$ full rotations), so: $\cos 7\pi = \cos \pi = -1$, $\sin 7\pi = \sin \pi = 0$ - Numerator: $$(-1)^2 - 2 \cdot 0^2 = 1 - 0 = 1$$ - For denominator: $$\frac{17\pi}{2} = 8\pi + \frac{\pi}{2}$$ - So effectively angle is $\frac{\pi}{2}$. - $\cos \frac{\pi}{2} = 0$, $\sin \frac{\pi}{2} = 1$ - Denominator: $$0^2 + 2 \cdot 1^2 = 0 + 2 = 2$$ - Final value: $$\frac{1}{2} = 0.5$$ --- **Summary of answers:** - Task 2a: $\sin 27\pi = 0$, $\cos 27\pi = -1$ - Task 2f: $\sin \frac{33\pi}{4} = \frac{\sqrt{2}}{2}$, $\cos \frac{33\pi}{4} = \frac{\sqrt{2}}{2}$ - Task 6a: $-3$ - Task 6f: $0.5$