Question

$f\left(x\right)=\sin x,g\left(x\right)=\cos x$use the figures to evaluate each function given that and $h\left(x\right)=\tan x$.

$h\left(\frac{\alpha }{2}\right)$

Short answer

Thus

$$\begin{gathered} h\left(\frac{\alpha }{2}\right)=\tan \left(\frac{\alpha }{2}\right) \\ =-\frac{\sqrt{15}}{3} \end{gathered}$$

Step-by-step solution

Step 1. Given

$$\begin{gathered} h\left(x\right)=\tan x \\ thus \\ h\left(\frac{\alpha }{2}\right)=\tan \left(\frac{\alpha }{2}\right) \end{gathered}$$

Step 2. Concept

we will use standard trigonometric formula's for solving the given expression.

Using half-angle formula

$\tan \left(\frac{\alpha }{2}\right)=\frac{\sin \alpha }{1+\cos \alpha }$

Step 3. Calculation

We have

$$\begin{gathered} \cos \alpha =-\frac{1}{4},\sin \alpha =basthepoint\left(-\frac{1}{4},b\right)liesonthecircle{x}^2+y^2=1 \\ Now \\ {\left(-\frac{1}{4}\right)}^2+b^2=1 \\ {b}^2=1-\frac{1}{16} \\ {b}^2=\frac{15}{16} \\ b=\pm \frac{\sqrt{15}}{4} \\ Sincepointlieinthe{3}^{rd}quadrantthereforebwillbenegative \\ thus, \\ \sin \alpha =-\frac{\sqrt{15}}{4} \\ Substitutingthevalueof\sin \alpha \&\cos \alpha inhalf-angleformulaweget \\ \tan \left(\frac{\alpha }{2}\right)=\frac{-{\displaystyle \frac{\sqrt{15}}{4}}}{1+\left(-{\displaystyle \frac{1}{4}}\right)} \\ =-\frac{\sqrt{15}}{3} \end{gathered}$$