Chapter 7.5, Problem 108AYU

To determine

Show that, ${\sin }^3\theta =\sin \left(\alpha -\theta \right)\sin \left(\beta -\theta \right)\sin \left(\gamma -\theta \right)$ .

Answer to Problem 108AYU

  $\boxed{\sin \left(\alpha -\theta \right)\sin \left(\beta -\theta \right)\sin \left(\gamma -\theta \right)={\sin }^3\theta }$

Explanation of Solution

Given information:

If $\alpha +\beta +\gamma =180°$ and $\cot \theta =\cot \alpha +\cot \beta +\cot \gamma ,0<\theta <90°$

Show that ${\sin }^3\theta =\sin \left(\alpha -\theta \right)\sin \left(\beta -\theta \right)\sin \left(\gamma -\theta \right)$ .

Calculation:

Now we have $\cot \theta =\cot \alpha +\cot \beta +\cot \gamma$

  $\begin{array}{l}\cot \theta -\cot \alpha =\cot \beta +\cot \gamma \\ \frac{\cos \theta }{\sin \theta }-\frac{\cos \alpha }{\sin \alpha }=\frac{\cos \beta }{\sin \beta }+\frac{\cos \gamma }{\sin \gamma }\\ \frac{\sin \alpha \cos \theta -\cos \alpha \sin \theta }{\sin \theta \sin \alpha }=\frac{\sin \gamma \cos \beta +\cos \gamma \sin \beta }{\sin \beta \sin \gamma }\end{array}$

We know that,

  $\sin \left(A-B\right)=\sin A\cos B-\cos A\sin B$

  $\sin \left(A+B\right)=\sin A\cos B+\cos A\sin B$

  $\begin{array}{l}\frac{\sin \left(\alpha -\theta \right)}{\sin \theta \sin \alpha }=\frac{\sin \left(\gamma -\beta \right)}{\sin \beta \sin \gamma }\mathrm{......}\left(1\right)\\ now,\alpha +\beta +\gamma \\ \beta +\gamma =180°-\alpha \\ \frac{\sin \left(\alpha -\theta \right)}{\sin \theta \sin \alpha }=\frac{\sin \left(180°-\alpha \right)}{\sin \beta \sin \gamma }\\ \frac{\sin \left(\alpha -\theta \right)}{\sin \theta \sin \alpha }=\frac{\sin \alpha }{\sin \beta \sin \gamma }\left[\therefore \sin \left(180°-\alpha \right)=\sin \theta \right]\\ \sin \left(\alpha -\theta \right)=\frac{{\sin }^2\alpha \sin \theta }{\sin \beta \sin \gamma }\mathrm{......}\left(2\right)\end{array}$

We have given that,

  $\begin{array}{l}\cot \theta =\cot \alpha +\cot \beta +\cot \gamma \\ \cot \theta -\cot \beta =\cot \alpha +\cot \gamma \\ \frac{\cos \theta }{\sin \theta }-\frac{\cos \beta }{\sin \beta }=\frac{\cos \alpha }{\sin \alpha }+\frac{\cos \gamma }{\sin \gamma }\\ \frac{\sin \beta \cos \theta -\cos \beta \sin \theta }{\sin \theta \sin \beta }=\frac{\sin \gamma \cos \alpha -\cos \gamma \sin \alpha }{\sin \alpha \sin \gamma }\end{array}$

  $\begin{array}{l}\frac{\sin \left(\beta -\theta \right)}{\sin \theta \sin \beta }=\frac{\sin \left(\gamma -\alpha \right)}{\sin \theta \sin \gamma }\mathrm{....}\left(3\right)\\ \alpha +\beta +\gamma =180°\\ \frac{\sin \left(\beta -\theta \right)}{\sin \theta \sin \beta }=\frac{\sin \left(180°-\beta \right)}{\sin \alpha \sin \gamma }\\ \frac{\sin \left(\beta -\theta \right)}{\sin \theta \sin \beta }=\frac{\sin \beta }{\sin \alpha \sin \gamma }\\ \sin \left(\beta -\theta \right)=\frac{{\sin }^2\beta \sin \theta }{\sin \alpha \sin \gamma }\mathrm{......}\left(4\right)\end{array}$

  $\begin{array}{l}\frac{\cos \theta }{\sin \theta }-\frac{\cos \gamma }{\sin \gamma }=\frac{\cos \alpha }{\sin \alpha }+\frac{\cos \beta }{\sin \beta }\\ \frac{\sin \gamma \cos \theta -\cos \gamma \sin \theta }{\sin \theta \sin \gamma }=\frac{\sin \beta \cos \alpha +\cos \beta \sin \alpha }{\sin \alpha \sin \beta }\end{array}$

  $\begin{array}{l}\cot \theta -\cot \gamma =\cot \alpha +\cot \beta \\ \frac{\sin \left(\gamma -\theta \right)}{\sin \theta \sin \gamma }=\frac{\sin \left(\beta -\alpha \right)}{\sin \alpha \sin \beta }\mathrm{.....}\left(5\right)\\ \alpha +\beta +\gamma =180°\\ \alpha +\beta =180°-\gamma \\ \frac{\sin \left(\gamma -\theta \right)}{\sin \theta \sin \gamma }=\frac{\sin \left(180°-\gamma \right)}{\sin \alpha \sin \beta }\\ \frac{\sin \left(\gamma -\theta \right)}{\sin \theta \sin \gamma }=\frac{\sin \gamma }{\sin \alpha \sin \beta }\\ \sin \left(\gamma -\theta \right)=\frac{{\sin }^2\gamma \sin \theta }{\sin \alpha \sin \beta }\mathrm{......}\left(6\right)\end{array}$

Multiply equation $\left(2\right)$

  $\left(4\right)$ and $\left(6\right)$

  $\sin \left(\alpha -\theta \right)\sin \left(\beta -\theta \right)\sin \left(\gamma -\theta \right)=\frac{{\sin }^2\alpha \sin \theta }{\sin \beta \sin \gamma }\times \frac{{\sin }^2\beta \sin \theta }{\sin \alpha \sin \gamma }\times \frac{{\sin }^2\gamma \sin \theta }{\sin \alpha \sin \gamma }$

  $\sin \left(\alpha -\theta \right)\sin \left(\beta -\theta \right)\sin \left(\gamma -\theta \right)={\sin }^3\theta$

Hence, $\boxed{\sin \left(\alpha -\theta \right)\sin \left(\beta -\theta \right)\sin \left(\gamma -\theta \right)={\sin }^3\theta }$ .