QUESTION 4 Given that $$ \cos 42^{\circ}=k $$, determine the following in terms of $$ k $$. 4.1.1 $$ \sin 42^{\circ} $$ (3) 4.1.2 $$ \cos 21^{\circ} $$ (3) Simplify the following expression to a single trigonometric ratio. $$ \frac{\cos (-\theta) \cdot \sin \left(90^{\circ}+\theta\right)}{\cos \left(360^{\circ}-\theta\right) \cdot \sin ^{2}\left(540^{\circ}+\theta\right)} $$ March Control Test Grade 12 $$ \begin{array}{l} \text { Grade } 12 \ \text { Prove the identity : } \frac{\sin 2 x}{1+\cos 2 x}=\tan x \end{array} $$ (3) [15] TOTAL: 50

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**4.1.1 Determining $$\sin 42^\circ$$ in terms of $$k$$**

Since
$$
\cos 42^\circ = k,
$$
we use the Pythagorean identity:
$$
\sin^2 42^\circ + \cos^2 42^\circ = 1.
$$
Substitute $$ \cos^2 42^\circ = k^2 $$:
$$
\sin^2 42^\circ = 1 - k^2.
$$
Taking the positive square root (because $$42^\circ$$ is acute):
$$
\sin 42^\circ = \sqrt{1-k^2}.
$$

---

**4.1.2 Determining $$\cos 21^\circ$$ in terms of $$k$$**

Notice that $$42^\circ = 2 \times 21^\circ$$. Using the cosine double-angle formula:
$$
\cos 2\theta = 2\cos^2 \theta - 1,
$$
with $$\theta = 21^\circ$$ we have:
$$
\cos 42^\circ = 2\cos^2 21^\circ - 1.
$$
Since $$\cos 42^\circ = k$$, this becomes:
$$
k = 2\cos^2 21^\circ - 1.
$$
Solving for $$\cos^2 21^\circ$$:
$$
2\cos^2 21^\circ = k + 1 \quad \Longrightarrow \quad \cos^2 21^\circ = \frac{k+1}{2}.
$$
Taking the positive square root (because $$21^\circ$$ is acute):
$$
\cos 21^\circ = \sqrt{\frac{k+1}{2}}.
$$

---

**Simplifying the Trigonometric Expression**

Simplify
$$
\frac{\cos (-\theta) \cdot \sin \left(90^\circ+\theta\right)}{\cos \left(360^\circ-\theta\right) \cdot \sin^2\left(540^\circ+\theta\right)}.
$$

1. **Simplify $$\cos (-\theta)$$:**  
   Since cosine is even,
   $$
   \cos (-\theta) = \cos \theta.
   $$
   
2. **Simplify $$\sin (90^\circ+\theta)$$:**  
   Using the co-function identity,
   $$
   \sin (90^\circ+\theta)= \cos (-\theta)= \cos \theta.
   $$
   
3. **Simplify $$\cos (360^\circ-\theta)$$:**  
   Since cosine has a period of $$360^\circ$$,
   $$
   \cos (360^\circ-\theta)= \cos \theta.
   $$
   
4. **Simplify $$\sin^2 (540^\circ+\theta)$$:**  
   Note that
   $$
   540^\circ+\theta = 360^\circ + 180^\circ+\theta,
   $$
   and since $$\sin (\alpha+360^\circ)= \sin \alpha$$,
   $$
   \sin (540^\circ+\theta)= \sin (180^\circ+\theta).
   $$
   Using the identity
   $$
   \sin (180^\circ+\theta)= -\sin \theta,
   $$
   it follows that:
   $$
   \sin^2 (540^\circ+\theta)= (-\sin \theta)^2 = \sin^2 \theta.
   $$

Putting these together, the expression becomes:
$$
\frac{\cos \theta \cdot \cos \theta}{\cos \theta \cdot \sin^2 \theta} = \frac{\cos^2 \theta}{\cos \theta \sin^2 \theta} = \frac{\cos \theta}{\sin^2 \theta}.
$$
Alternatively, we can express it as:
$$
\frac{\cos \theta}{\sin^2 \theta} = \cot \theta \csc \theta.
$$

---

**Prove the Identity $$\frac{\sin 2x}{1+\cos 2x}=\tan x$$**

Start with the left-hand side:
$$
\frac{\sin 2x}{1+\cos 2x}.
$$
Recall the double-angle identities:
$$
\sin 2x = 2\sin x \cos x, \quad \text{and} \quad \cos 2x = 2\cos^2 x - 1.
$$
Substitute $$\cos 2x$$ into the denominator:
$$
1+\cos 2x = 1 + (2\cos^2 x - 1) = 2\cos^2 x.
$$
Thus, the expression becomes:
$$
\frac{2\sin x \cos x}{2\cos^2 x} = \frac{\sin x}{\cos x} = \tan x.
$$

This completes the proof.
**4.1.1** $$ \sin 42^\circ = \sqrt{1 - k^2} $$ **4.1.2** $$ \cos 21^\circ = \sqrt{\frac{k + 1}{2}} $$ **Simplified Expression** $$ \frac{\cos \theta}{\sin^2 \theta} = \cot \theta \csc \theta $$ **Proven Identity** $$ \frac{\sin 2x}{1 + \cos 2x} = \tan x $$

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