(g) $$ \frac{\sin 2 \alpha}{1+\cos 2 \alpha}= an \alpha $$ TRIGONOMETRIC GRAPHS In Grade 10 and 11 you studied amplitude shifts, vertical horizontal shifts. In the following exercise, you will revise these and exercises that follow, you will link trigonometric graphs angles. EXERCISE $$ 4.11 \quad $$ Revision of Grade 10-11 Trigonometric Sketch the graphs of the following. Write down the maxim minimum value, amplitude and period. (1) $$ y=2 \sin x-1 $$ for $$ x \in\left[0^{\circ} ; 360^{\circ} ight] $$ (3) $$ y=\cos 2 x $$ for $$ x \in\left[-45^{\circ} ; 315^{\circ} ight] $$ (5) $$ y=-2 \sin 2 x $$ for $$ x \in\left[-90^{\circ} ; 180^{\circ} ight] $$ (7) $$ y=\cos \left(x+60^{\circ} ight) $$ for $$ x \in\left[-150^{\circ} ; 300^{\circ} ight] $$

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(g) $$ \frac{\sin 2 \alpha}{1+\cos 2 \alpha}=	an \alpha $$ TRIGONOMETRIC GRAPHS In Grade 10 and 11 you studied amplitude shifts, vertical horizontal shifts. In the following exercise, you will revise these and exercises that follow, you will link trigonometric graphs angles. EXERCISE $$ 4.11 \quad $$ Revision of Grade 10-11 Trigonometric Sketch the graphs of the following. Write down the maxim minimum value, amplitude and period. (1) $$ y=2 \sin x-1 $$ for $$ x \in\left[0^{\circ} ; 360^{\circ}ight] $$ (3) $$ y=\cos 2 x $$ for $$ x \in\left[-45^{\circ} ; 315^{\circ}ight] $$ (5) $$ y=-2 \sin 2 x $$ for $$ x \in\left[-90^{\circ} ; 180^{\circ}ight] $$ (7) $$ y=\cos \left(x+60^{\circ}ight) $$ for $$ x \in\left[-150^{\circ} ; 300^{\circ}ight] $$

UpStudy AI · Accepted Answer

**(g) Verify that**  
$$
\frac{\sin 2\alpha}{1+\cos 2\alpha}=\tan \alpha
$$

1. **Express the double‑angle formulas:**  
   $$
   \sin 2\alpha = 2 \sin \alpha \cos \alpha
   $$
   $$
   \cos 2\alpha = 2\cos^2 \alpha - 1
   $$

2. **Rewrite the denominator using a double‑angle identity:**  
   Notice that:
   $$
   1+\cos 2\alpha = 1 + (2\cos^2\alpha - 1) = 2\cos^2\alpha
   $$

3. **Substitute these into the fraction:**  
   $$
   \frac{\sin 2\alpha}{1+\cos 2\alpha} = \frac{2\sin\alpha\cos\alpha}{2\cos^2\alpha} = \frac{\sin\alpha}{\cos\alpha} = \tan\alpha.
   $$

---

**EXERCISE 4.11 – Trigonometric Graphs**

---

**(1) $$y=2\sin x-1$$ for $$x \in [0^\circ,360^\circ]$$**

1. **Amplitude:**  
   $$
   \text{Amplitude} = |2| = 2.
   $$

2. **Vertical shift:**  
   The graph is shifted downward by 1 unit.

3. **Period:**  
   $$
   \text{Period} = 360^\circ \quad (\text{since the coefficient of } x \text{ is } 1).
   $$

4. **Maximum and minimum values:**  
   $$
   \text{Maximum value} = -1 + 2 = 1,
   $$
   $$
   \text{Minimum value} = -1 - 2 = -3.
   $$

---

**(3) $$y=\cos 2x$$ for $$x \in [-45^\circ,315^\circ]$$**

1. **Amplitude:**  
   $$
   \text{Amplitude} = 1.
   $$

2. **Period:**  
   $$
   \text{Period} = \frac{360^\circ}{2} = 180^\circ.
   $$

3. **Maximum and minimum values:**  
   $$
   \text{Maximum value} = 1, \quad \text{Minimum value} = -1.
   $$

---

**(5) $$y=-2\sin 2x$$ for $$x \in [-90^\circ,180^\circ]$$**

1. **Amplitude:**  
   $$
   \text{Amplitude} = |-2| = 2.
   $$

2. **Period:**  
   $$
   \text{Period} = \frac{360^\circ}{2} = 180^\circ.
   $$

3. **Reflection:**  
   The negative sign indicates a reflection about the $$x$$-axis.

4. **Maximum and minimum values:**  
   Since the standard sine function varies between $$-1$$ and $$1$$:
   $$
   y = -2\sin 2x \quad \text{varies between } -2(1) = -2 \text{ and } -2(-1) = 2.
   $$
   Thus,
   $$
   \text{Maximum value} = 2, \quad \text{Minimum value} = -2.
   $$

---

**(7) $$y=\cos (x+60^\circ)$$ for $$x \in [-150^\circ,300^\circ]$$**

1. **Amplitude:**  
   $$
   \text{Amplitude} = 1.
   $$

2. **Period:**  
   $$
   \text{Period} = 360^\circ \quad (\text{since the coefficient of } x \text{ is } 1).
   $$

3. **Phase shift:**  
   The graph is shifted to the left by $$60^\circ$$.

4. **Maximum and minimum values:**  
   $$
   \text{Maximum value} = 1, \quad \text{Minimum value} = -1.
   $$
**(g) Verify that** $$ \frac{\sin 2\alpha}{1+\cos 2\alpha}=\tan \alpha $$ 1. **Express double-angle formulas:** $$ \sin 2\alpha = 2 \sin \alpha \cos \alpha $$ $$ \cos 2\alpha = 2\cos^2 \alpha - 1 $$ 2. **Rewrite the denominator:** $$ 1 + \cos 2\alpha = 1 + (2\cos^2 \alpha - 1) = 2\cos^2 \alpha $$ 3. **Substitute into the fraction:** $$ \frac{\sin 2\alpha}{1+\cos 2\alpha} = \frac{2\sin\alpha\cos\alpha}{2\cos^2\alpha} = \frac{\sin\alpha}{\cos\alpha} = \tan\alpha $$ --- **EXERCISE 4.11 – Trigonometric Graphs** --- **(1) $$y=2\sin x-1$$ for $$x \in [0^\circ,360^\circ]$$** 1. **Amplitude:** $$ 2 $$ 2. **Vertical shift:** Downward by 1 unit. 3. **Period:** $$ 360^\circ $$ 4. **Maximum and minimum values:** $$ 1 \quad \text{and} \quad -3 $$ --- **(3) $$y=\cos 2x$$ for $$x \in [-45^\circ,315^\circ]$$** 1. **Amplitude:** $$ 1 $$ 2. **Period:** $$ 180^\circ $$ 3. **Maximum and minimum values:** $$ 1 \quad \text{and} \quad -1 $$ --- **(5) $$y=-2\sin 2x$$ for $$x \in [-90^\circ,180^\circ]$$** 1. **Amplitude:** $$ 2 $$ 2. **Period:** $$ 180^\circ $$ 3. **Reflection:** About the $$x$$-axis. 4. **Maximum and minimum values:** $$ 2 \quad \text{and} \quad -2 $$ --- **(7) $$y=\cos (x+60^\circ)$$ for $$x \in [-150^\circ,300^\circ]$$** 1. **Amplitude:** $$ 1 $$ 2. **Period:** $$ 360^\circ $$ 3. **Phase shift:** Left by $$60^\circ$$. 4. **Maximum and minimum values:** $$ 1 \quad \text{and} \quad -1 $$

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