If sinα + sinβ = a and cosα + cosβ = b, Show that sin(α+β) and cos(α+β)
Question
If
\[ \sin\alpha+\sin\beta=a \]
and
\[ \cos\alpha+\cos\beta=b \]
show that:
\[ \text{(i)}\quad \sin(\alpha+\beta) = \frac{2ab}{a^2+b^2} \]
\[ \text{(ii)}\quad \cos(\alpha+\beta) = \frac{b^2-a^2}{b^2+a^2} \]
Proof
Given,
\[ \sin\alpha+\sin\beta=a \]
\[ \cos\alpha+\cos\beta=b \]
Squaring and adding,
\[ (\sin\alpha+\sin\beta)^2 + (\cos\alpha+\cos\beta)^2 = a^2+b^2 \]
\[ \sin^2\alpha+\sin^2\beta+2\sin\alpha\sin\beta \]
\[ +\cos^2\alpha+\cos^2\beta+2\cos\alpha\cos\beta = a^2+b^2 \]
\[ 2+2(\sin\alpha\sin\beta+\cos\alpha\cos\beta) = a^2+b^2 \]
\[ 2+2\cos(\alpha-\beta) = a^2+b^2 \]
\[ a^2+b^2 = 4\cos^2\frac{\alpha-\beta}{2} \]
Also,
\[ a = 2\sin\frac{\alpha+\beta}{2} \cos\frac{\alpha-\beta}{2} \]
\[ b = 2\cos\frac{\alpha+\beta}{2} \cos\frac{\alpha-\beta}{2} \]
Therefore,
\[ ab = 4\sin\frac{\alpha+\beta}{2} \cos\frac{\alpha+\beta}{2} \cos^2\frac{\alpha-\beta}{2} \]
\[ = 2\sin(\alpha+\beta) \cos^2\frac{\alpha-\beta}{2} \]
Since
\[ a^2+b^2 = 4\cos^2\frac{\alpha-\beta}{2} \]
\[ \sin(\alpha+\beta) = \frac{2ab}{a^2+b^2} \]
Now,
\[ b^2-a^2 = 4\cos^2\frac{\alpha-\beta}{2} \]
\[ \times \left( \cos^2\frac{\alpha+\beta}{2} – \sin^2\frac{\alpha+\beta}{2} \right) \]
\[ = 4\cos^2\frac{\alpha-\beta}{2} \cos(\alpha+\beta) \]
Therefore,
\[ \cos(\alpha+\beta) = \frac{b^2-a^2}{b^2+a^2} \]
Hence proved.