Question

If \(cos43^{\circ}+sin43^{\circ} = k^3\), then \(cos2^{\circ} = \ldots\)

• A. \(-\frac{k^3}{\sqrt{2}}\)
• B. \(\frac{k^3}{\sqrt{2}}\)
• C. \(-\frac{k^3}{\sqrt{3}}\)
• D. \(\frac{k^3}{\sqrt{3}}\)

Hint:

Write \(\cos43^{\circ}+\sin43^{\circ}=\sqrt2\cos(43^{\circ}-45^{\circ})\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
A sum \(\cos\theta+\sin\theta\) can be written as \(\sqrt2\cos(\theta-45^{\circ})\).

Step 2: Key Formula or Approach
\[ \cos\theta+\sin\theta = \sqrt2\left(\cos\theta\cdot\tfrac{1}{\sqrt2}+\sin\theta\cdot\tfrac{1}{\sqrt2}\right)=\sqrt2\cos(\theta-45^{\circ}) \]

Step 3: Detailed Explanation
With \(\theta=43^{\circ}\): \(\cos43^{\circ}+\sin43^{\circ}=\sqrt2\cos(-2^{\circ})=\sqrt2\cos2^{\circ}\).
So \(\sqrt2\cos2^{\circ}=k^3\), which gives
\[ \cos2^{\circ}=\frac{k^3}{\sqrt2} \]

Final Answer:
\(\cos2^{\circ}=\dfrac{k^3}{\sqrt2}\), option (B). \[ \boxed{\dfrac{k^3}{\sqrt2}\ \text{(B)}} \]