1. **Evaluate the expression:** $$8 \sin 45^\circ - \sin 60^\circ \cdot \cos 30^\circ - \frac{1}{4} \tan 45^\circ$$ 2. **Recall exact trigonometric values:** - $\sin 45^\circ = \frac{\sqrt{2}}{2}$ - $\sin 60^\circ = \frac{\sqrt{3}}{2}$ - $\cos 30^\circ = \frac{\sqrt{3}}{2}$ - $\tan 45^\circ = 1$ 3. **Substitute values:** $$8 \times \frac{\sqrt{2}}{2} - \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} - \frac{1}{4} \times 1$$ 4. **Simplify each term:** - $8 \times \frac{\sqrt{2}}{2} = 4\sqrt{2}$ - $\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} = \frac{3}{4}$ - $\frac{1}{4} \times 1 = \frac{1}{4}$ 5. **Rewrite expression:** $$4\sqrt{2} - \frac{3}{4} - \frac{1}{4}$$ 6. **Combine fractions:** $$4\sqrt{2} - \frac{3+1}{4} = 4\sqrt{2} - 1$$ --- 7. **For $f(x) = \sin x - 1$ and $g(x) = 2 \tan x$:** **3.6.1 Period of $g(x)$:** - The period of $\tan x$ is $180^\circ$. - Scaling by 2 does not change the period. - So, period of $g(x) = 180^\circ$. **3.6.2 Amplitude of $f(x)$:** - $\sin x$ has amplitude 1. - Vertical shift down by 1 does not change amplitude. - So, amplitude of $f(x)$ is 1. **3.6.3 Range of $g(x)$:** - $\tan x$ ranges over all real numbers. - Scaling by 2 stretches range but still all real numbers. - So, range of $g(x)$ is $(-\infty, \infty)$. --- **3.6.4 Sketch details for $x \in [0^\circ, 360^\circ]$:** - $f(x) = \sin x - 1$ oscillates between $-2$ and $0$. - Intercepts of $f(x)$ with x-axis where $\sin x = 1$ at $x=90^\circ$. - Turning points of $f(x)$ at $x=270^\circ$ (minimum at $-2$) and $x=90^\circ$ (maximum at $0$). - $g(x) = 2 \tan x$ has vertical asymptotes at $x=90^\circ$ and $x=270^\circ$. - Intercepts of $g(x)$ at $x=0^\circ, 180^\circ, 360^\circ$. - $g(45^\circ) = 2 \times 1 = 2$. --- **Final answers:** - Expression value: $$4\sqrt{2} - 1$$ - Period of $g(x)$: $$180^\circ$$ - Amplitude of $f(x)$: $$1$$ - Range of $g(x)$: $$(-\infty, \infty)$$