Question

If \( \alpha \) and \( \beta \) are two acute angles of a right triangle, then \[ (\sin \alpha + \sin \beta)^2 + (\cos \alpha + \cos \beta)^2 = \] 

• A. 1+sin2α
• B. 2(1+sin2α)
• C. 1+cos2α
• D. 2(1+2cos2α)
• E. 2+sin2α

Answer 1

The Correct Option is B

We are given the expression: \[ (\sin \alpha + \sin \beta)^2 + (\cos \alpha + \cos \beta)^2 \] We can expand each square term: \[ (\sin \alpha + \sin \beta)^2 = \sin^2 \alpha + \sin^2 \beta + 2 \sin \alpha \sin \beta \] \[ (\cos \alpha + \cos \beta)^2 = \cos^2 \alpha + \cos^2 \beta + 2 \cos \alpha \cos \beta \] Now, adding the two expanded expressions: \[ \sin^2 \alpha + \sin^2 \beta + 2 \sin \alpha \sin \beta + \cos^2 \alpha + \cos^2 \beta + 2 \cos \alpha \cos \beta \] Using the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we simplify: \[ 1 + 1 + 2 (\sin \alpha \sin \beta + \cos \alpha \cos \beta) \] Using the sum formula for cosines, \( \cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta \), we get: \[ 2 + 2 \cos(\alpha - \beta) \] Since \( \alpha + \beta = 90^\circ \) in a right triangle, \( \cos(\alpha - \beta) = \cos 2\alpha \). Therefore, the final expression becomes: \[ 2(1 + \sin 2\alpha) \]

The correct option is (B) : \({2(1 + \sin 2\alpha)}\)

Answer 2

Since \(\alpha\) and \(\beta\) are two acute angles of a right triangle, we have \(\alpha + \beta = 90^\circ\), so \(\beta = 90^\circ - \alpha\).

We are asked to find the value of \((\sin\alpha+\sin\beta)^2+(\cos\alpha+\cos\beta)^2\).

Expanding the expression, we get:

\(\sin^2\alpha + 2\sin\alpha\sin\beta + \sin^2\beta + \cos^2\alpha + 2\cos\alpha\cos\beta + \cos^2\beta\)

Rearranging the terms, we have:

\((\sin^2\alpha + \cos^2\alpha) + (\sin^2\beta + \cos^2\beta) + 2(\sin\alpha\sin\beta + \cos\alpha\cos\beta)\)

Using the identity \(\sin^2\theta + \cos^2\theta = 1\), we get:

\(1 + 1 + 2(\cos\alpha\cos\beta + \sin\alpha\sin\beta)\)

Using the identity \(\cos(\alpha - \beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta\), we get:

\(2 + 2\cos(\alpha - \beta)\)

Since \(\beta = 90^\circ - \alpha\), we have \(\alpha - \beta = \alpha - (90^\circ - \alpha) = 2\alpha - 90^\circ\).

Then, \(\cos(\alpha - \beta) = \cos(2\alpha - 90^\circ) = \cos(-(90^\circ - 2\alpha)) = \cos(90^\circ - 2\alpha) = \sin(2\alpha)\).

Substituting this back into our expression, we get:

\(2 + 2\sin(2\alpha) = 2(1 + \sin(2\alpha))\)

Therefore, the expression simplifies to \(2(1 + \sin2\alpha)\).