Suppose that $\sin \alpha = \frac{3}{5}$ and $90^\circ < \alpha < 180^\circ$.Find the exact values of $\sin \frac{\alpha}{2}$ and $\tan \frac{\alpha}{2}$.- $\sin \frac{\alpha}{2} = \_\_$- $\tan \frac{\alpha}{2} = \_\_$There is a small pop-up box with an icon displaying a square root symbol, followed by a symbol of $\pi$, and the text "Undefined". Below, there are three icons: an "X", a circular arrow, and a question mark.

Name: Suppose that $\sin \alpha = \frac{3}{5}$ and $90^\circ < \alpha <
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Description: Suppose that \(\sin \alpha = \frac{3}{5}$ and $90^\circ < \alpha < 180^\circ$.Find the exact values of $\sin \frac{\alpha}{2}$ and $\tan \frac{\alpha}{2}$.- $\sin \frac{\alpha}{2} = \_\_$- $\tan \frac{\alpha}{2} = \_\_$There is a small po

Given Data: Equation: sinα=35 Interval: 90°<α<180° The value of cosα is, cosα=1-sin2α…

Answered: Suppose that sin a= and 90°

Trigonometry (11th Edition)

11th Edition

ISBN:9780134217437

Author:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels

Publisher:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels

Chapter1: Trigonometric Functions

Section: Chapter Questions

Problem 1RE: 1. Give the measures of the complement and the supplement of an angle measuring 35°.

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Suppose that $\sin \alpha = \frac{3}{5}$ and $90^\circ < \alpha < 180^\circ$.

Find the exact values of $\sin \frac{\alpha}{2}$ and $\tan \frac{\alpha}{2}$.

- $\sin \frac{\alpha}{2} = \_\_$

- $\tan \frac{\alpha}{2} = \_\_$

There is a small pop-up box with an icon displaying a square root symbol, followed by a symbol of $\pi$, and the text "Undefined". Below, there are three icons: an "X", a circular arrow, and a question mark.

Expert Solution

Step 1

Given Data:

Equation: $\sin α = \frac{3}{5}$

Interval: $90 ° < α < 180 °$

The value of cosα is,

$\cos α = \sqrt{1 - \sin^{2} α}$

Substitute the value in the above equation.

$\begin{array}{rcl} \cos α & = & \sqrt{1 - {(\frac{3}{5})}^{2}} \\ = & \sqrt{1 - \frac{9}{25}} \\ = & \sqrt{\frac{16}{25}} \\ = & \pm \frac{4}{5} \\ = & - \frac{4}{5} [As cosine is negative between 90 ° < α < 180 °] \end{array}$

Step 2

The interval of $\frac{α}{2}$ is,

$\begin{array}{rcl} 90 ° & < & α < 180 ° \\ \frac{90 °}{2} & < & \frac{α}{2} < \frac{180 °}{2} \\ 45 ° & < & \frac{α}{2} < 90 ° \end{array}$

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