Trig Problem?

Question

Can anyone explain how to get tan(-pi/6) but I need to leave it in terms of pi because my professor wants it in exact form... I also need to understand how to get the answer.

Answer 1

tan(-pi/6) = -tan(pi/6) = -1/sqrt(3)
(The obtain the value of tan(pi/6), apply Pythagoras to a "half" of the equilateral triangle, whose angles are pi/3 or 60 deg.)

Answer 2

It is difficult to explain with out some kind of visual but ill do what i can.

First realize that -pi/6 is negative... so you are rotating around the origin clockwise. Since -pi/6>-pi/2 you are in the fourth quadrant so x is positive and y is negative. By definition tan(theta)=y/x you get a negative over a positive therefore your answer will be negative.

Ok start out by drawing the xy plane and drawing a line starting from the origin in the fourth quadrant, drop a line from this initial line to the x axis forming a triangle with a vertex on the origin. The angle made by this in your case will be pi/6, this is called the reference angle.

NOTE: pi/6 is 30degrees... we are dealing with a 30:60:90 triangle thier respective sides are 1:sqrt(3):2

So now you have your traingle drawn. Lable the hypoteneus 2.. becuase thats opppisite of the right angle.. the line along the x axis is sqrt(3)... this is opposite the 60degree angle...and the line going down will be -1 becuase y is negative here.

(these ratios should be remebered becuase they are commonly used: 30:60:90 (pi/6:pi/3:pi/2) = 1:sqrt(3):2
45:45:90 (pi/4:pi/4:pi/4) = 1:1:sqrt(2)
you can use pythagorean's theorem to prove them)

now you have values
x=sqrt3
y=-1
and r=2 (hypoteneus)

tan(theta)=y/x = -1/sqrt(3) or rationalize = -sqrt(3)/3

its seems like overkill but with it written out like this you can also solve for sinx cosx and all the other trig functions easily which from what i remeber is commonly asked.

did my best to not sound confusing... dunno if i succeeded

Answer 3

- pi/6 is equivalent to -30 degrees.
divide sin/cos to find tan
sin(-30) = -0.5
cos(-30) = (sqrt3)/2
-0.5/(sqrt3)/2 = -0.5 x 2/(sqrt3) = -1/(sqrt3)= -(sqrt3)/3

Answer 4

I agree with Blue

Tan Θ = sinΘ/cosΘ

Therefore

Tan - π/6 = sin -π/6 / cos - π/6

- - - - - - s-

Answer 5

tan (pi-pi/6)=tan(-pi/6)
or -tan (pi/6)=tan(-pi/6)

Answer 6

tan(-pi/6)=tan5pi/6

Answer 7

tan(Θ) = sin(Θ) / cos(Θ)
So
tan(-π/6) = sin(-π/6) / cos(-π/6)

Also:
sin(-Θ) = -sin(Θ)
and
cos(-Θ) = cos(Θ)

So
tan(-π/6) = -sin(π/6) / cos(π/6)

That's as simple as I know to make it (I assume you more or less know what sin(π/6) and cos(π/6) are) without actually solving it...
And you don't need to leave it in this form to have an exact solution:

tan(-π/6) = -sin(π/6) / cos(π/6)
= -[1/2] / [√(3)/2] = -1 / √(3) = -√(3) / 3
That's still and exact solution.

Answer 8

http://mathforum.org/dr.math/

This might help.

Good Luck.