Question

Pre-Calc I need the answers to for sin and tan.

Answer 1

Given:

`\sin (\alpha)/(2)=(15)/(17),90^(\circ)<\alpha<180^(\circ)`

Now,

`\begin{gathered} 90^(\circ)<\alpha<180^(\circ) \\ \text{replace }\alpha\text{ by }(\alpha)/(2) \\ (90^(\circ))/(2)<(\alpha)/(2)<(180^(\circ))/(2) \\ 45^(\circ)<(\alpha)/(2)<90^(\circ) \\ So,\text{ }(\alpha)/(2)\text{ lies in first quadrant} \\ \Rightarrow\sin (\alpha)/(2),\text{ cos}(\alpha)/(2)\text{ and tan}(\alpha)/(2)\text{ are positive.} \end{gathered}`

As we know that,

`\begin{gathered} \sin 2\alpha=2\sin \alpha\cos \alpha \\ \text{ replace }\alpha\text{ with }(\alpha)/(2) \\ \sin 2((\alpha)/(2))=2\sin (\alpha)/(2)\cos (\alpha)/(2) \\ \sin \alpha=2\sin (\alpha)/(2)\cos (\alpha)/(2) \\ 2\sin (\alpha)/(2)\cos (\alpha)/(2)=(15)/(17) \\ \sin (\alpha)/(2)\cos (\alpha)/(2)=(15)/(34)\ldots\ldots\ldots.\ldots(1) \end{gathered}`

Now, find the cosine function,

`\begin{gathered} \sin ^2\alpha+\cos ^2\alpha=1 \\ ((15)/(17))^2+\cos ^2\alpha=1 \\ \text{cos}^2\alpha=1-(225)/(289) \\ \cos ^2\alpha=(64)/(289) \\ \cos \alpha=\pm(8)/(17) \\ \Rightarrow\cos \alpha=-(8)/(17)\ldots.(90^(\circ)\text{<}\alpha<180^(\circ)) \end{gathered}`

Using the identity,

`\begin{gathered} \cos \alpha=2\cos ^2(\alpha)/(2)-1 \\ -(8)/(17)=2\cos ^2(\alpha)/(2)-1 \\ 2\cos ^2(\alpha)/(2)=1-(8)/(17) \\ \cos ^2(\alpha)/(2)=(9)/(17*2) \\ \cos (\alpha)/(2)=\pm\sqrt[]{(9)/(34)} \\ \Rightarrow\cos (\alpha)/(2)=\frac{3}{\sqrt[]{34}}\ldots..(45^(\circ)<(\alpha)/(2)<90^(\circ)) \end{gathered}`

From equation (1),

`\begin{gathered} \sin (\alpha)/(2)\cos (\alpha)/(2)=(15)/(34) \\ \sin (\alpha)/(2)*\frac{3}{\sqrt[]{34}}=(15)/(34) \\ \sin (\alpha)/(2)=(15)/(34)*\frac{\sqrt[]{34}}{3} \\ \sin (\alpha)/(2)=\frac{5\sqrt[]{34}}{34} \end{gathered}`

And,

`\tan (\alpha)/(2)=(\sin(\alpha)/(2))/(\cos(\alpha)/(2))=\frac{\frac{5\sqrt[]{34}}{34}}{\frac{3}{\sqrt[]{34}}}=\frac{5\sqrt[]{34}}{34}*\frac{\sqrt[]{34}}{3}=(5)/(3)`

Answer:

`\begin{gathered} \sin (\alpha)/(2)=\frac{5\sqrt[]{34}}{34} \\ \tan (\alpha)/(2)=(5)/(3) \end{gathered}`