Expert: Josh - 3/18/2008

Question
QUESTION: find the value of tan300*+tan210*/ 1 - tan300*tan210* in surd form.

ANSWER: Emmy,

A helpful starting point is to divide one complete revolution of a circle (i.e., 360 degrees) into four equal quadrants. Basically, the trigonometric values of sine, cosine and tangent repeat themselves every 90 degrees (up to a sign change).

The quadrants are defined by
Quadrant 1:    0 <= x <90
Quadrant 2:    90 <= x <180
Quadrant 3:    180 <= x <270
Quadrant 4:    270 <= x <360

For example, we can work out the following knowing only the value of sine in the range from 0 to 90 degrees.

sin(180-30) = sin(150) = sin(30) = 0.5  [relating 2nd to 1st quadrant]
sin(180+30) = sin(210) = -sin(30) = -0.5 [relating 3rd to 1st quadrant]
sin(360-30) = sin(330) = -sin(30) = -0.5 [relating 4th to 1st quadrant]

Similarly, for cosine(x)

cos(180-30) = cos(150) = cos(30) = sqrt(3)/2 = approx. 0.8660
[relating 2nd to 1st quadrant]
cos(180+30) = cos(210) = -cos(30) = -0.8660 [relating 3rd to 1st quadrant]
cos(360-30) = cos(330) = -cos(30) = 0.8660 [relating 4th to 1st quadrant]

Observe the magnitude is the same in each case (if we consider the absolute value). Key question is, how do we know when there is a sign change?

Remember this:

A S T C (All Stations To Central) if you prefer.

"A" means that ALL (sin, cos and tan) are positive when the angle is in the first quadrant, between 0 and 90 degrees.

"S" means only SINE is positive (cos and tan are negative) when the angle is in the second quadrant, between 90 and 180.

"T" means only TANGENT is positive (sin and cos are negative) when the angle is in the third quadrant, between 180 and 270.

"C" means only COSINE is positive (sin and tan are negative) when the angle is in the fourth quadrant, between 270 and 360.

In summary: assuming that 0<= x < 90 deg.
sin(180-x) = +sin(x)  2nd quad (S) <= POSITIVE
sin(180+x) = -sin(x)  3rd quad (T)
sin(360-x) = -sin(x)  4th quad (C)

cos(180-x) = -cos(x)  2nd quad (S)
cos(180+x) = -cos(x)  3rd quad (T)
cos(360-x) = +cos(x)  4th quad (C) <= POSITIVE

tan(180-x) = -tan(x)  2nd quad (S)
tan(180+x) = +tan(x)  3rd quad (T) <= POSITIVE
tan(360-x) = -tan(x)  4th quad (C)

This is the most important thing for you to learn.

Once you understand this, you just have to put tan(300) as
tan(360-60) which is either +tan(60) or -tan(60). So, which is correct? Since the angle 300 is in the 4th quadrant, from "ASTC" we see that only cosine is positive in this quadrant. Thus, tan(300)= -tan(60).

I don't like using calculators. Most of the time we only need to know 6 numbers.

sin(30)=0.5, sin(45)=1/sqrt(2) and sin(60)=sqrt(3)/2.
cos(30)=sqrt(3)/2, cos(45)=1/sqrt(2) and cos(60)=1/2.

Often, the other values can be derived from these.
e.g., tan(60) = sin(60)/cos(60) = [sqrt(3)/2]*[2/1] = sqrt(3)
So, tan(300) = -tan(60)= -sqrt(3)

You should now be able to attempt the question yourself.

Observe that tan(210) = tan(180+30). Identify angle as in the 3rd quadrant. So, out of sin, cos and tan, which entity is positive?

Ans: From "ASTC", tangent is positive in the 3rd quadrant. So, tan(210)=tan(30)

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---------- FOLLOW-UP ----------

QUESTION: Thanks for the answer. I followed the principles you outline carefully, but the answer i got i.e {-2/sqrt(3)} is different from the answer in my text book which says {-1/sqrt(3)}.
Who do you think is wrong 'The Book' or 'Myself'.

Answer
Hi Emmy,

[tan(300)+tan(210)]/[1-tan(300)*tan(210)]
=[-tan(60)+tan(30)]/[1+tan(60)*tan(30)]
=[-sqrt(3)+1/sqrt(3)]/[1+sqrt(3)*(1/sqrt(3))]
....multiply first term in numerator by sqrt(3)/sqrt(3)
=[(-3+1)/sqrt(3)]/2
=-1/sqrt(3)

It's good to compare with the answer in the book. Typographical errors are not unheard of.