Question

Given that cosα=−2√2 and tanα<0, draw and use a diagram to find the value of all six trig functions.

Answer 1

Given:

`\begin{gathered} \cos \alpha=-\frac{\sqrt[]{2}}{2} \\ \tan \alpha<0 \end{gathered}`

To find the six trigonometric functions:

Since,

`\cos \alpha<0\text{ and }\tan \alpha<0`

So, the angle lies in the second quadrant.

Let us draw the diagram.

Using the Pythagoras theorem,

`\begin{gathered} AB^2=AC^2+BC^2 \\ 2^2=AC^2+(\sqrt[]{2})^2 \\ 4=AC^2+2 \\ AC^2=2 \\ AC=\sqrt[]{2}\ldots\ldots\ldots(1) \end{gathered}`

Next, find the six trig functions.

`\begin{gathered} \sin \alpha=(AC)/(AB) \\ =\frac{\sqrt[]{2}}{2} \\ co\sec \alpha=(AB)/(AC) \\ =\frac{2}{\sqrt[]{2}}*\frac{\sqrt[]{2}}{\sqrt[]{2}} \\ =\frac{2\sqrt[]{2}}{2} \\ =\sqrt[]{2} \end{gathered}`

And,

`\begin{gathered} \cos \alpha=-(BC)/(AB)(Since,II\text{ quadrant)} \\ =-\frac{\sqrt[]{2}}{2} \\ \sec \alpha=-(AB)/(BC)(Since,II\text{ quadrant)} \\ =-\frac{2}{\sqrt[]{2}}*\frac{\sqrt[]{2}}{\sqrt[]{2}} \\ =-\frac{2\sqrt[]{2}}{2} \\ =-\sqrt[]{2} \end{gathered}`

Similarly,

`\begin{gathered} \tan \alpha=-(AC)/(BC)(Since,II\text{ quadrant)} \\ =-\frac{\sqrt[]{2}}{\sqrt[]{2}} \\ =-1 \\ \cot \alpha=-(BC)/(AC)(Since,II\text{ quadrant)} \\ =-\frac{\sqrt[]{2}}{\sqrt[]{2}} \\ =-1 \end{gathered}`

Therefore, the answers are,

`\begin{gathered} \sin \alpha=\frac{\sqrt[]{2}}{2} \\ co\sec \alpha=\sqrt[]{2} \\ \cos \alpha=-\frac{\sqrt[]{2}}{2} \\ \sec \alpha=-\sqrt[]{2} \\ \tan \alpha=-1 \\ \cot \alpha=-1 \end{gathered}`