If \(\alpha\) and \(\beta\) are two distinct values of \(\theta\) lying between 0 and \(2\pi\), and they satisfy the equation \(6\cos \theta + 8\sin\theta = 9 \), find \(\sin(\alpha + \beta) \).

There are various ways to solve this problem, but the most elegant way (according to me) is to make \(\color{green}{\text{Weierstrass Substitution}}\), i.e, replacing \(\sin(\theta)\) and \(\cos(\theta)\) by \(\dfrac{2x^{2}}{1+x^{2}}\) and \(\dfrac{1-x^{2}}{1+x^{2}}\) respectively, where \(x\,=\,\tan\left(\dfrac{\theta}{2}\right)\).

After making this substitution, you would get :- \(6 \cdot \dfrac{1-x^{2}}{1+x^{2}} + 16 \cdot \dfrac{2x}{1+x^{2}}\,=\,9\). Manipulating this gives a quadratic equation in \(x\), i.e, \(\tan\left(\dfrac{\theta}{2}\right)\) which is \(15x^{2}-16x+3=0\).
By \(\color{blue}{\text{Viete's Theorem}}\), \(\tan\left(\dfrac{\alpha}{2}\right)+\tan\left(\dfrac{\beta}{2}\right)\,=\,\dfrac{16}{15}\) and \(\tan\left(\dfrac{\alpha}{2}\right) \cdot \tan\left(\dfrac{\beta}{2}\right)\,=\,\dfrac{1}{5}\).

Using this, we get :- \(\tan\left(\dfrac{\alpha+\beta}{2}\right)\,=\,\dfrac{\dfrac{16}{15}}{1-\dfrac{1}{5}}\,=\,\dfrac{4}{3}\).

Now again using the identity, \(\sin(\alpha+\beta)\,=\,\dfrac{2 \cdot \tan\left(\dfrac{\alpha+\beta}{2}\right)}{1+\tan^{2}\left(\dfrac{\alpha+\beta}{2}\right)}\), we get :- \(\sin(\alpha+\beta)\,=\,\dfrac{2 \cdot \dfrac{4}{3}}{1+\left(\dfrac{4}{3}\right)^{2}}\,=\,\boxed{\color{red}{\dfrac{24}{25}}}\).

Have you attempted to do the substitution, \(x=\tan \frac{\theta}{2}\)? It would give \(\sin \theta = \dfrac {2x}{x^2+1}\) and \(\cos \theta = \dfrac {1-x^2}{x^2+1}\).

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## Comments

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TopNewestThere are various ways to solve this problem, but the most elegant way (according to me) is to make \(\color{green}{\text{Weierstrass Substitution}}\), i.e, replacing \(\sin(\theta)\) and \(\cos(\theta)\) by \(\dfrac{2x^{2}}{1+x^{2}}\) and \(\dfrac{1-x^{2}}{1+x^{2}}\) respectively, where \(x\,=\,\tan\left(\dfrac{\theta}{2}\right)\).

After making this substitution, you would get :- \(6 \cdot \dfrac{1-x^{2}}{1+x^{2}} + 16 \cdot \dfrac{2x}{1+x^{2}}\,=\,9\). Manipulating this gives a quadratic equation in \(x\), i.e, \(\tan\left(\dfrac{\theta}{2}\right)\) which is \(15x^{2}-16x+3=0\). By \(\color{blue}{\text{Viete's Theorem}}\), \(\tan\left(\dfrac{\alpha}{2}\right)+\tan\left(\dfrac{\beta}{2}\right)\,=\,\dfrac{16}{15}\) and \(\tan\left(\dfrac{\alpha}{2}\right) \cdot \tan\left(\dfrac{\beta}{2}\right)\,=\,\dfrac{1}{5}\).

Using this, we get :- \(\tan\left(\dfrac{\alpha+\beta}{2}\right)\,=\,\dfrac{\dfrac{16}{15}}{1-\dfrac{1}{5}}\,=\,\dfrac{4}{3}\).

Now again using the identity, \(\sin(\alpha+\beta)\,=\,\dfrac{2 \cdot \tan\left(\dfrac{\alpha+\beta}{2}\right)}{1+\tan^{2}\left(\dfrac{\alpha+\beta}{2}\right)}\), we get :- \(\sin(\alpha+\beta)\,=\,\dfrac{2 \cdot \dfrac{4}{3}}{1+\left(\dfrac{4}{3}\right)^{2}}\,=\,\boxed{\color{red}{\dfrac{24}{25}}}\).

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Nice solution ! I really liked the method..+1 !

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Thanks :)

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Have you attempted to do the substitution, \(x=\tan \frac{\theta}{2}\)? It would give \(\sin \theta = \dfrac {2x}{x^2+1}\) and \(\cos \theta = \dfrac {1-x^2}{x^2+1}\).

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Yes ur right!

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