Question

If $sin\alpha +sin \beta = \frac{\sqrt6}2$ and $cos \alpha + cos\beta = \frac{\sqrt2}2$ then $coas(\alpha - \beta)$ is equa; to

• A. $\frac1{2}$
• B. $\frac{{3}}{2}$
• C. 0
• D. $\frac{-3}{2}$
• E. $\frac{-1}{2}$

Answer 1

Answer: (E)

Given equations:

\(\sin \alpha + \sin \beta = \frac{\sqrt{6}}{2}\) 

\(\cos \alpha + \cos \beta = \frac{\sqrt{2}}{2}\)

Using sum-to-product identities:

\(\sin \alpha + \sin \beta = 2 \sin \left(\frac{\alpha + \beta}{2} \right) \cos \left(\frac{\alpha - \beta}{2} \right)\)

\(\cos \alpha + \cos \beta = 2 \cos \left(\frac{\alpha + \beta}{2} \right) \cos \left(\frac{\alpha - \beta}{2} \right)\)

Substituting given values:

\(2 \sin \left(\frac{\alpha + \beta}{2} \right) \cos \left(\frac{\alpha - \beta}{2} \right) = \frac{\sqrt{6}}{2}\)

\(2 \cos \left(\frac{\alpha + \beta}{2} \right) \cos \left(\frac{\alpha - \beta}{2} \right) = \frac{\sqrt{2}}{2}\)

Dividing both equations:

\(\frac{\sin \left(\frac{\alpha + \beta}{2} \right)}{\cos \left(\frac{\alpha + \beta}{2} \right)} = \frac{\frac{\sqrt{6}}{2}}{\frac{\sqrt{2}}{2}} = \frac{\sqrt{6}}{\sqrt{2}} = \sqrt{3}\)

So, \(\tan \left(\frac{\alpha + \beta}{2} \right) = \sqrt{3}\), which gives:

\(\frac{\alpha + \beta}{2} = \frac{\pi}{3}\), thus \(\alpha + \beta = \frac{2\pi}{3}\).

Solving for \(\cos(\alpha - \beta)\):

\(\cos \left(\frac{\alpha - \beta}{2} \right) = \frac{\frac{\sqrt{2}}{2}}{2 \cos \left(\frac{\alpha + \beta}{2} \right)}\)

\(= \frac{\frac{\sqrt{2}}{2}}{2 \times \frac{1}{2}} = \frac{\frac{\sqrt{2}}{2}}{1} = \frac{\sqrt{2}}{2}\)

Squaring both sides and using \(\cos^2 x = \frac{1 + \cos 2x}{2}\):

\(\frac{1 + \cos(\alpha - \beta)}{2} = \frac{2}{4} = \frac{1}{2}\)

\(\cos(\alpha - \beta) = \frac{-1}{2}\)

Thus, the correct answer is:

\(\frac{-1}{2}\)