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#
