Question

Question #990d6

Answer 1

We know that `cos(x)=-1/2` and that `sin(x)>0`, which means that `sin(x)` is a negative number.

Based on this information, let's make a coordinate grid for this problem. That will help us draw an accurate picture.
`color(white)(----)color(white)(Sin)color(white)(-----)color(black)(|)color(white)(----)` `color(white)(All)` `color(white)(---------)`
`color(white)(----)`Sin`color(white)(-----)color(black)(|)color(white)(----)`All`color(white)(---------)`
`color(white)(----)color(white)(Sin)color(white)(-----)color(black)(|)color(white)(----)` `color(white)(All)` `color(white)(---------)`
`color(black)(----)color(white)(Sin)color(black)(-----)color(black)(|)color(black)(---------)`
`color(white)(----)color(white)(Tan)color(white)(-----)color(black)(|)color(white)(----)` `color(white)(Cos)` `color(white)(---------)`
`color(white)(----)`Tan`color(white)(-----)color(black)(|)color(white)(----)`Cos`color(white)(---------)`
`color(white)(----)color(white)(Tan)color(white)(-----)color(black)(|)color(white)(----)` `color(white)(Cos)` `color(white)(---------)`

This shows the quadrant in which the answer is positive. I like to remember it by using the mnemonic device All Students Take Calculus.
In our problem, `Sin` in negative and so is `Cos`, so the picture has to lie in the `3^(rd)` quadrant, where `Tan` is. That means all our ratios will be negative, except for those concerning tangent

Now, I like to draw a picture, but I can't do that on this program, so I'll just write it out.

`cos=("adjacent")/("hypotenuse")` or `-1/2`

If the hypotenuse is `2` and the leg is `1`, we can find the remaining side using Pythagorean's theorem `(c^2-b^2=a^2)`. That gives us `sqrt3` as the other side.

So, just to summirize:

• adjacent `=1`
• hypotenuse `=2`
• opposite `=sqrt3`

Now we can solve everything:

`sin=("opposite")/("hypotenuse")` or `-sqrt3/2`

`cos=("adjacent")/("hypotenuse")` or `-1/2`

`tan=("opposite")/("adjacent")` or `sqrt3/1` or just `sqrt3`

Now we find the inverses:

`csc=1/(sin)=-2/sqrt3` or `(-2sqrt3)/3`

`sec=1/(cos)=-2/1` or jsut `-2`

`cot=1/(tan)=1/sqrt3` or `sqrt3/3`

Answer 2

Since `cos x= -1/2` and sin x > 0, then, x is in Quadrant 2.
`sin^2 x = 1 - cos^2 x = 1 - 1/4 = 3/4`
`sin x = +- sqrt3/2`
Take the positive answer because sin x > 0
`sin x = sqrt3/2`
`tan x = sin/(cos) = (sqrt3/2)(-2/1) = - sqrt3`
`cot x = 1/(tan) = - 1/sqrt3 = - sqrt3/3`
`sec x = 1/(cos) = -2`
`csc x = 1/(sin) = 2/sqrt3 = (2sqrt3)/3`