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:
We are given an expression involving the sum of squares of sines and cosines. Expanding the squares will trigger Pythagorean identities.
Step 2: Key Formula or Approach:
Expand \((x + y)^2 = x^2 + 2xy + y^2\).
Use \(\sin^2 \theta + \cos^2 \theta = 1\).
Use the cosine difference formula: \(\cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta\).
Step 3: Detailed Explanation:
Let \(E = (\sin \alpha + \sin \beta)^2 + (\cos \alpha + \cos \beta)^2\).
Expand the squares:
\[ E = (\sin^2 \alpha + 2\sin \alpha \sin \beta + \sin^2 \beta) + (\cos^2 \alpha + 2\cos \alpha \cos \beta + \cos^2 \beta) \] Regroup the terms:
\[ E = (\sin^2 \alpha + \cos^2 \alpha) + (\sin^2 \beta + \cos^2 \beta) + 2(\cos \alpha \cos \beta + \sin \alpha \sin \beta) \] Apply Pythagorean identities:
\[ E = 1 + 1 + 2\cos(\alpha - \beta) \] \[ E = 2 + 2\cos(\alpha - \beta) \] We are given \(\alpha - \beta = \frac{\pi}{4}\). Substitute this in:
\[ E = 2 + 2\cos\left(\frac{\pi}{4}\right) \] Since \(\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}\):
\[ E = 2 + 2\left(\frac{1}{\sqrt{2}}\right) = 2 + \sqrt{2} \] Step 4: Final Answer:
The value is \(2 + \sqrt{2}\).