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

#Geometry

Easy Math Editor

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

There are various ways to solve this problem, but the most elegant way (according to me) is to make $\color{#20A900}{\text{Weierstrass Substitution}}$ , 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 $\color{#3D99F6}{\text{Viete's Theorem}}$ , $\tan\left(\dfrac{\alpha}{2}\right)+\tan\left(\dfrac{\beta}{2}\right)\,=\,\dfrac{16}{15}$ 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 :- $\sin(\alpha+\beta)\,=\,\dfrac{2 \cdot \dfrac{4}{3}}{1+\left(\dfrac{4}{3}\right)^{2}}\,=\,\boxed{\color{#D61F06}{\dfrac{24}{25}}}$ .

Aditya Sky - 5 years ago

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

Rishabh Tiwari - 5 years ago

Thanks :)

Aditya Sky - 5 years ago

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

Sharky Kesa - 5 years ago

Yes ur right!

Rishabh Tiwari - 5 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Find the value of sin⁡(α+β)\sin(\alpha + \beta) sin(α+β)

Comments

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