QUESTION 4 1.1 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. $$ \frac{\cos (-\theta) \cdot \sin \left(90^{\circ}+\theta\right)}{\cos \left(360^{\circ}-\theta\right) \cdot \sin ^{2}\left(540^{\circ}+\theta\right)} $$

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1.1.1  
We are given  
$$
\cos 42^\circ = k.
$$  
Using the Pythagorean identity  
$$
\sin^2 42^\circ + \cos^2 42^\circ = 1,
$$  
we substitute $$\cos 42^\circ = k$$ to obtain  
$$
\sin^2 42^\circ + k^2 = 1,
$$  
so that  
$$
\sin^2 42^\circ = 1 - k^2.
$$  
Since $$42^\circ$$ is in the first quadrant, $$\sin 42^\circ$$ is positive. Therefore,  
$$
\sin 42^\circ = \sqrt{1 - k^2}.
$$

1.1.2  
To express $$\cos 21^\circ$$ in terms of $$k$$, we use the double angle formula for cosine. Since $$42^\circ = 2 \cdot 21^\circ$$, we have  
$$
\cos 42^\circ = 2\cos^2 21^\circ - 1.
$$  
Substitute $$\cos 42^\circ = k$$ to get  
$$
k = 2\cos^2 21^\circ - 1.
$$  
Solving for $$\cos^2 21^\circ$$, we add 1 to both sides and then divide by 2:  
$$
2\cos^2 21^\circ = k + 1 \quad\Longrightarrow\quad \cos^2 21^\circ = \frac{k+1}{2}.
$$  
Since $$21^\circ$$ is in the first quadrant and cosine is positive there,  
$$
\cos 21^\circ = \sqrt{\frac{k+1}{2}}.
$$

2.  
Simplify the following expression to a single trigonometric ratio:
$$
\frac{\cos (-\theta) \cdot \sin \left(90^\circ+\theta\right)}{\cos \left(360^\circ-\theta\right) \cdot \sin ^{2}\left(540^\circ+\theta\right)}.
$$

Step 1: Simplify each trigonometric function individually.

- Since cosine is an even function,  
  $$
  \cos (-\theta)=\cos \theta.
  $$

- Express $$\sin (90^\circ+\theta)$$ using the co-function identity. Recall that for any angle $$x$$,  
  $$
  \sin (90^\circ + x) = \cos (90^\circ - (90^\circ+x)) = \cos(-x)=\cos x,
  $$  
  hence,  
  $$
  \sin (90^\circ+\theta)=\cos \theta.
  $$

- Since cosine is periodic and even,  
  $$
  \cos(360^\circ-\theta)=\cos \theta.
  $$

- For the sine term, note that  
  $$
  540^\circ+\theta = 360^\circ+180^\circ+\theta.
  $$  
  Using periodicity,  
  $$
  \sin(540^\circ+\theta)=\sin(180^\circ+\theta).
  $$  
  Recall that  
  $$
  \sin (180^\circ+\theta)=-\sin\theta,
  $$  
  so  
  $$
  \sin(540^\circ+\theta)=-\sin\theta.
  $$  
  Then, squaring gives  
  $$
  \sin^2(540^\circ+\theta)= (-\sin\theta)^2 = \sin^2\theta.
  $$

Step 2: Substitute the simplified expressions back into the original expression.

The numerator becomes:  
$$
\cos\theta\cdot \cos\theta = \cos^2\theta.
$$

The denominator becomes:  
$$
\cos\theta \cdot \sin^2\theta.
$$

Thus the expression is  
$$
\frac{\cos^2\theta}{\cos\theta \cdot \sin^2\theta}.
$$

Step 3: Cancel the common factor $$\cos\theta$$ (assuming $$\cos \theta\neq 0$$) to obtain:  
$$
\frac{\cos\theta}{\sin^2\theta}.
$$

This is the simplified single trigonometric ratio.

Final Answers:
$$
\sin 42^\circ = \sqrt{1-k^2},\quad \cos 21^\circ = \sqrt{\frac{k+1}{2}},
$$
$$
\frac{\cos (-\theta) \cdot \sin \left(90^\circ+\theta\right)}{\cos \left(360^\circ-\theta\right) \cdot \sin ^{2}\left(540^\circ+\theta\right)} = \frac{\cos \theta}{\sin^2 \theta}.
$$
$$ \sin 42^\circ = \sqrt{1 - k^2}, \quad \cos 21^\circ = \sqrt{\frac{k + 1}{2}}, \quad \frac{\cos (-\theta) \cdot \sin \left(90^\circ + \theta\right)}{\cos \left(360^\circ - \theta\right) \cdot \sin^2 \left(540^\circ + \theta\right)} = \frac{\cos \theta}{\sin^2 \theta} $$

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