Question 4 4.1. Given $$ \sin \alpha=-\frac{8}{17} $$, where $$ 0^{\circ}

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Question 4 4.1. Given $$ \sin \alpha=-\frac{8}{17} $$, where $$ 0^{\circ}<\alpha<270^{\circ} $$, determine with the aid of a sketch, the value of the following: 4.1.1. $$ 	an \alpha $$ 4.1.2. $$ \cos 2 \alpha $$ 4.1.3. $$ \sin \frac{\alpha}{2} $$ 4.2. Simplify the following trigonometric expression, leaving your answer to only one trigonometric ratio: $$ \left[\sin \left(x-1980^{\circ}ight)-\cos (-x)ight]^{2} $$

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**4.1. Given**  
$$
\sin \alpha = -\frac{8}{17}\quad \text{with } 0^\circ < \alpha < 270^\circ.
$$

Since $$\sin \alpha$$ is negative and $$\alpha$$ is between $$0^\circ$$ and $$270^\circ$$, the angle must lie in the third quadrant. In the third quadrant, both sine and cosine are negative, and tangent is positive.

Using the Pythagorean identity:  
$$
\cos^2 \alpha = 1 - \sin^2 \alpha = 1 - \left(\frac{64}{289}\right) = \frac{225}{289}.
$$
Since $$\alpha$$ is in the third quadrant,  
$$
\cos \alpha = -\frac{15}{17}.
$$

---

**4.1.1. Find $$\tan \alpha$$**

$$
\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{-\frac{8}{17}}{-\frac{15}{17}} = \frac{8}{15}.
$$

---

**4.1.2. Find $$\cos 2\alpha$$**

Using the double-angle formula:  
$$
\cos 2\alpha = 1 - 2\sin^2 \alpha.
$$
Substitute $$\sin^2 \alpha = \frac{64}{289}$$:
$$
\cos 2\alpha = 1 - 2\left(\frac{64}{289}\right) = 1 - \frac{128}{289} = \frac{289 - 128}{289} = \frac{161}{289}.
$$

---

**4.1.3. Find $$\sin \frac{\alpha}{2}$$**

Since $$\alpha$$ is in the third quadrant ($$180^\circ < \alpha < 270^\circ$$), dividing by 2 gives  
$$
90^\circ < \frac{\alpha}{2} < 135^\circ,
$$
so $$\frac{\alpha}{2}$$ lies in the second quadrant where sine is positive.

Using the half-angle formula:  
$$
\sin\frac{\alpha}{2} = \sqrt{\frac{1 - \cos\alpha}{2}},
$$
substitute $$\cos \alpha = -\frac{15}{17}$$:
$$
\sin\frac{\alpha}{2} = \sqrt{\frac{1 - \left(-\frac{15}{17}\right)}{2}} = \sqrt{\frac{1 + \frac{15}{17}}{2}} = \sqrt{\frac{\frac{17}{17} + \frac{15}{17}}{2}} = \sqrt{\frac{\frac{32}{17}}{2}}.
$$
Simplify the fraction:
$$
\sin\frac{\alpha}{2} = \sqrt{\frac{32}{34}} = \sqrt{\frac{16}{17}} = \frac{4}{\sqrt{17}}.
$$
Optionally, rationalize the denominator:
$$
\sin\frac{\alpha}{2} = \frac{4\sqrt{17}}{17}.
$$

---

**4.2. Simplify $$\left[ \sin(x-1980^\circ) - \cos(-x) \right]^2$$**

First, note that cosine is an even function, so:
$$
\cos(-x) = \cos x.
$$

Next, simplify the sine term. Since sine is periodic with period $$360^\circ$$:
$$
\sin(x-1980^\circ) = \sin\Bigl(x - \bigl(5\times360^\circ + 180^\circ\bigr)\Bigr) = \sin\bigl((x-180^\circ) - 5\times360^\circ\bigr) = \sin(x-180^\circ).
$$

Using the identity:
$$
\sin(x-180^\circ) = -\sin x,
$$
the expression becomes:
$$
\left[-\sin x - \cos x\right]^2.
$$
Since squaring removes the minus sign:
$$
\left[-\left(\sin x+\cos x\right)\right]^2 = \left(\sin x+\cos x\right)^2.
$$

Expanding the square:
$$
\left(\sin x+\cos x\right)^2 = \sin^2 x + \cos^2 x + 2\sin x\cos x.
$$
Recall that:
$$
\sin^2 x + \cos^2 x = 1 \quad \text{and} \quad 2\sin x\cos x = \sin 2x.
$$
Thus:
$$
\left(\sin x+\cos x\right)^2 = 1 + \sin 2x.
$$

---

**Final Answers**

$$
\tan \alpha = \frac{8}{15}, \quad \cos 2\alpha = \frac{161}{289}, \quad \sin\frac{\alpha}{2} = \frac{4\sqrt{17}}{17},
$$
$$
\left[ \sin(x-1980^\circ) - \cos(-x) \right]^2 = 1 + \sin 2x.
$$
$$ \tan \alpha = \frac{8}{15}, \quad \cos 2\alpha = \frac{161}{289}, \quad \sin\frac{\alpha}{2} = \frac{4\sqrt{17}}{17}, $$ $$ \left[ \sin(x-1980^\circ) - \cos(-x) \right]^2 = 1 + \sin 2x. $$

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