Question

If$\tan \theta =\frac{1}{2},-\frac{\mathrm{\pi }}{2}<\theta <\frac{\mathrm{\pi }}{2}$then $\sin \theta =\_\_\_\_\_\_\_$.

Short answer

If$\tan \theta =\frac{1}{2},-\frac{\mathrm{\pi }}{2}<\theta <\frac{\mathrm{\pi }}{2}$then$Sin\theta =\frac{\sqrt{5}}{5}$.

Step-by-step solution

Step 1. Given information.

$\tan \theta =\frac{1}{2},-\frac{\mathrm{\pi }}{2}<\theta <\frac{\mathrm{\pi }}{2}$

. Value ofTan θ.

Here, $\theta$lies between$-\frac{\mathrm{\pi }}{2}$and $\frac{\mathrm{\pi }}{2}$.But tan positive in first and third quadrants. So, $\theta$ lies in first quadrant.

$\tan \theta =\frac{opposite}{adjacent}$

Step 3. Using Pythagorean theorem. 

$$\begin{gathered} H^2=P^2+B^2 \\ {H}^2=1^2+2^2 \\ {H}^2=1+4 \\ H=\sqrt{5} \end{gathered}$$

Step 4. Value of Sinθ.

$$\begin{gathered} Sin\theta =\frac{opposite}{hypotenuse} \\ Sin\theta =\frac{1}{\sqrt{5}} \\ Sin\theta =\frac{1\times \sqrt{5}}{\sqrt{5}\times \sqrt{5}} \\ Sin\theta =\frac{\sqrt{5}}{5} \end{gathered}$$