please help with this problem, I solved through but got it wrong, my answers were: sin = 7root58/58, cos= -3root58/58, and tan= -7/3

ANSWER
[tex]\begin{gathered} \text{Sin }\frac{\alpha}{2}\text{ = 0.3939} \\ \cos \text{ }\frac{\alpha}{2}\text{ = }0.9191 \\ \tan \text{ }\frac{\alpha}{2}\text{ = 0.4286} \end{gathered}[/tex]STEP-BY-STEP EXPLANATION:
Given information
[tex]\tan \text{ }\alpha\text{ = }\frac{21}{20}[/tex]Recall that,
[tex]\tan \text{ }\alpha\text{ = }\frac{opposite\text{ }}{\text{adjacent}}[/tex]This implies that,
The opposite side of the triangle = 21
Adjacent side of the triangle = 20
This can be represented pictorially below as
[tex]The\text{ next step is to find }\alpha[/tex][tex]\begin{gathered} \tan \text{ }\alpha\text{ = }\frac{21}{20} \\ \tan \text{ }\alpha\text{ = 1.05} \\ \alpha=tan^{-1}\text{ (1.05)} \\ \alpha\text{ = 46.40}\degree \end{gathered}[/tex][tex]\text{ since }\alpha\text{ = 46.40, we can now find the following}[/tex][tex]\begin{gathered} \sin \text{ }\frac{\alpha}{2} \\ \text{where }\alpha\text{ = 46.40} \\ \sin \text{ }\frac{46.40}{2} \\ \sin \text{ }\frac{\alpha}{2}\text{ = 0.3939} \end{gathered}[/tex][tex]\begin{gathered} \cos \text{ }\frac{\alpha}{2}\text{ = cos }\frac{46.40}{2} \\ \cos \text{ }\frac{\alpha}{2}\text{ = cos }\frac{46.40}{2} \\ \cos \text{ }\frac{\alpha}{2}\text{ = cos 23.20} \\ \cos \text{ }\frac{\alpha}{2}\text{ = 0.9191} \end{gathered}[/tex][tex]\begin{gathered} \tan \text{ }\frac{\alpha}{2}\text{ = tan }\frac{46.40}{2} \\ \tan \text{ }\frac{\alpha}{2}\text{ = tan }23.20 \\ \tan \text{ }\frac{\alpha}{2}\text{ = 0.4286} \end{gathered}[/tex]