Question

If \(cos(θ-α) = a\) and \(sin(θ-β) = b\), then the value of \(cos^2(α-β)+2absin(α-β)+cos^2(θ-α)\) is...

• A. \(a^2-2b^2\)
• B. \(a^2+2b^2\)
• C. \(2a^2-b^2\)
• D. \(2a^2+b^2\)

Hint:

Write \(\alpha-\beta = (\theta-\beta)-(\theta-\alpha)\) and expand with the given \(a\) and \(b\).

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
Let \(u = \theta-\alpha\) and \(v = \theta-\beta\). Then \(\cos u = a\), \(\sin v = b\) and \(\alpha-\beta = v-u\).

Step 2: Expand:
\(\cos(v-u) = \cos v\cos u + \sin v\sin u = a\cos v + b\sin u\).
\(\sin(v-u) = \sin v\cos u - \cos v\sin u = ab - \cos v\sin u\).

Step 3: Combine the first two terms:
\[ \cos^2(\alpha-\beta) = a^2\cos^2v + 2ab\cos v\sin u + b^2\sin^2u \]
\[ 2ab\sin(\alpha-\beta) = 2a^2b^2 - 2ab\cos v\sin u \]
Adding, the cross terms cancel: \(a^2\cos^2v + b^2\sin^2u + 2a^2b^2\).

Step 4: Simplify:
Use \(\cos^2v = 1-b^2\) and \(\sin^2u = 1-a^2\): \(a^2(1-b^2) + b^2(1-a^2) + 2a^2b^2 = a^2 + b^2\).
The last term is \(\cos^2(\theta-\alpha) = a^2\). So the full expression is \(a^2+b^2+a^2 = 2a^2+b^2\).

Final Answer:
The value is \(2a^2+b^2\), option (D). \[ \boxed{2a^2+b^2} \]