Question

If $\alpha$ and $\beta$ are real constants such that $\alpha - \beta = \frac{\pi}{4}$, then the value of $(\sin \alpha + \sin \beta)^2 + (\cos \alpha + \cos \beta)^2$ is equal to

• A. $2$
• B. $\sqrt{2}$
• C. $2\sqrt{2}$
• D. $2 + \sqrt{2}$
• E. $2 - \sqrt{2}$

Hint:

The expression $( \dots \alpha + \dots \beta )^2 + ( \dots \alpha + \dots \beta )^2$ almost always simplifies to $2 + 2 \cos(\alpha - \beta)$. Memorizing this pattern saves time.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Expand the squares and use the fundamental identity $\sin^2 \theta + \cos^2 \theta = 1$ and the cosine difference formula.

Step 2: Detailed Explanation:
1. Let $E = (\sin \alpha + \sin \beta)^2 + (\cos \alpha + \cos \beta)^2$.
2. Expand both terms:
\[ E = (\sin^2 \alpha + \sin^2 \beta + 2\sin \alpha \sin \beta) + (\cos^2 \alpha + \cos^2 \beta + 2\cos \alpha \cos \beta) \]
3. Rearrange and group the squared terms:
\[ E = (\sin^2 \alpha + \cos^2 \alpha) + (\sin^2 \beta + \cos^2 \beta) + 2(\cos \alpha \cos \beta + \sin \alpha \sin \beta) \]
4. Apply the identity $\sin^2 \theta + \cos^2 \theta = 1$:
\[ E = 1 + 1 + 2(\cos \alpha \cos \beta + \sin \alpha \sin \beta) \]
5. Use the identity $\cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$:
\[ E = 2 + 2\cos(\alpha - \beta) \]
6. Substitute the given $\alpha - \beta = \pi/4$:
\[ E = 2 + 2\cos\left(\frac{\pi}{4}\right) \]
\[ E = 2 + 2\left(\frac{1}{\sqrt{2}}\right) \]
\[ E = 2 + \sqrt{2} \]

Step 3: Final Answer:
The value is $2 + \sqrt{2}$.