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দেওয়া আছে:

\( \sin x+\sin y=a \quad \ldots (i) \)
\( \cos x+\cos y=b \quad \ldots (ii) \)

\((i)\) নং সমীকরণকে বর্গ করে পাই:

\( (\sin x+\sin y)^2=a^2 \)
\( \sin^2 x+2\sin x \sin y+\sin^2 y=a^2 \quad \ldots (iii) \)

\((ii)\) নং সমীকরণকে বর্গ করে পাই:

\( (\cos x+\cos y)^2=b^2 \)
\( \cos^2 x+2\cos x \cos y+\cos^2 y=b^2 \quad \ldots (iv) \)

এখন, \((iii)\) ও \((iv)\) নং সমীকরণ যোগ করে পাই:

\( (\sin^2 x+\cos^2 x)+(\sin^2 y+\cos^2 y)+2(\sin x \sin y+\cos x \cos y)=a^2+b^2 \)
\( 1+1+2\cos(x-y)=a^2+b^2 \)
\( 2+2\cos(x-y)=a^2+b^2 \)

আমরা জানি, \( \cos \theta = 1-2\sin^2 \frac{\theta}{2} \)। এখানে \( \theta = (x-y) \)।

সুতরাং,

\( 2+2\left(1-2\sin^2 \frac{x-y}{2}\right)=a^2+b^2 \)
\( 2+2-4\sin^2 \frac{x-y}{2}=a^2+b^2 \)
\( 4-4\sin^2 \frac{x-y}{2}=a^2+b^2 \)
\( 4\sin^2 \frac{x-y}{2}=4-a^2-b^2 \)
\( \sin^2 \frac{x-y}{2}=\frac{4-a^2-b^2}{4} \)
\( \sin \frac{x-y}{2}=\pm \sqrt{\frac{4-a^2-b^2}{4}} \)
\( \sin \frac{x-y}{2}=\pm \frac{1}{2}\sqrt{4-a^2-b^2} \)

প্রমাণিত।

যদি ( sin x+ sin y=a ) এবং ( cos x+ cos y=b ) হয় তবে দেখাও যে ( sin 1 2 (x-y)= pm 1 2 sqrt 4-a^ 2 -b^ 2 )।