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# Find the value of $$\sin(\alpha + \beta)$$

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 $$\sin(\alpha + \beta)$$.

Note by Pritthijit Nath
1 year, 10 months ago

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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 :- $$\sin(\alpha+\beta)\,=\,\dfrac{2 \cdot \dfrac{4}{3}}{1+\left(\dfrac{4}{3}\right)^{2}}\,=\,\boxed{\color{red}{\dfrac{24}{25}}}$$.

- 1 year, 10 months ago

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

- 1 year, 10 months ago

Thanks :)

- 1 year, 10 months ago

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

- 1 year, 10 months ago

Yes ur right!

- 1 year, 10 months ago