Chapter 12.3, Problem 48HP

To determine

To find: Whether $3\sin \left({60}^0\right)=\sin \left({180}^0\right)$ is True of false.

Answer to Problem 48HP

  $3\sin \left({60}^0\right)=\sin \left({180}^0\right)$ is FALSE.

Explanation of Solution

Given information:

  $3\sin \left({60}^0\right)=\sin \left({180}^0\right)$

Concept:

If $\theta$ is the non-quadrantal angle in standard position, its reference angle ${\theta }^{\text{'}}$ is the acute angle formed by the terminal side of $\theta$ and the x-axis.

The ruled for finding the measures of reference angles for $0^0<\theta <{360}^0$ or $0^0<\theta <2\pi$ are shown

For Quadrant − I

  ${\theta }^{\text{'}}=\theta$

For Quadrant − II

  ${\theta }^{\text{'}}={180}^0-\theta$ or ${\theta }^{\text{'}}=\pi -\theta$

For Quadrant − III

  ${\theta }^{\text{'}}=\theta -{180}^0$ or ${\theta }^{\text{'}}=\theta -\pi$

For Quadrant − IV

  ${\theta }^{\text{'}}={360}^0-\theta$ or ${\theta }^{\text{'}}=2\pi -\theta$

Given

  $3\sin \left({60}^0\right)=\sin \left({180}^0\right)$

Taking Left hand side

Left hand side $=3\sin \left({60}^0\right)$

  $\Rightarrow$ Left hand side $=3\times \frac{\sqrt{3}}{2}$

  $\Rightarrow$ Left hand side $=\frac{3\sqrt{3}}{2}$

Taking Right hand side

Right hand side $=\sin \left({180}^0\right)$

Here ${180}^0$ lies in Quadrant-III

  $\begin{array}{l}{\theta }^{\text{'}}=\theta -{180}^0\\ \Rightarrow {\theta }^{\text{'}}={180}^0-{180}^0\\ \Rightarrow {\theta }^{\text{'}}=0^0\end{array}$

Therefore,

  $\sin \left({180}^0\right)=-\sin \left(0^0\right)$

Because sine function is negative in Quadrant-III

  $\begin{array}{l}\Rightarrow \sin \left({180}^0\right)=-0\\ \Rightarrow \sin \left({180}^0\right)=0\end{array}$

Therefore,

Right hand side $=0$

We got

Left hand side $=\frac{3\sqrt{3}}{2}$

Right hand side $=0$

So,

  $3\sin \left({60}^0\right)\ne \sin \left({180}^0\right)$

Therefore,

  $3\sin \left({60}^0\right)=\sin \left({180}^0\right)$ is FALSE.