Question

Given sin x=-3/7 and tan x <0 find cos 2u

Answer 1

To find the value of cos 2x we first need to find the value of cos x.

Now we know that:

`\cos ^2x+\sin ^2x=1`

Plugging the value of the sine given we have that:

`\begin{gathered} \cos ^2x+(-(3)/(7))^2=1 \\ \cos ^2x+(9)/(49)=1 \\ \cos ^2x=1-(9)/(49) \\ \cos ^2x=(40)/(49) \\ \cos x=\pm\sqrt[]{(40)/(49)} \\ \cos x=\pm\frac{2\sqrt[]{10}}{7} \end{gathered}`

Now, from finding the cosine from this identity we get two possible values; we need to determine whether the cosine is positive or negative. To do this we have to notice two things; first we notice that the sine is negative, that means that the angle x is on the third or fourth quadrant of the plane (this comes from the fact that the sine is negative for angles greater than pi and less than 2 pi). The second thing we have to notice is that the tangent is negative, this only happens if the angle is on the second or fourth quadrant. Hence, combining this two facts we conclude that the angle is on the fourth quadrant.

Now that we know where the angle is we have to remember that the cosine functions is positive if the angle is on the first or fourth quadrant, therefore we conclude that:

`\cos x=\frac{2\sqrt[]{10}}{7}`

Now that we have the value of the cosine we have to remember that we have the identity:

`\cos 2x=\cos ^2x-\sin ^2x`

Plugging the values we have for the sine and cosine we have:

`\begin{gathered} \cos 2x=(\frac{2\sqrt[]{10}}{7})^2-(-(3)/(7))^2 \\ =(40)/(49)-(9)/(49) \\ =(31)/(49) \end{gathered}`

Therefore:

`\cos 2x=(31)/(49)`

Other way to solve this problem is to remember that:

`\cos 2x=1-2\sin ^2x`

Plugging the value of the sine we have:

`\begin{gathered} \cos 2x=1-2(-(3)/(7))^2 \\ =1-2((9)/(49)) \\ =1-(18)/(49) \\ =(31)/(49) \end{gathered}`

This is the same result we got from the long way.