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About Abe Mantell
Expertise
Hello, I am a college professor of mathematics and regularly teach all levels from elementary mathematics through differential equations, and would be happy to assist anyone with such questions!

Experience
Over 15 years teaching at the college level.

Organizations
NCTM, NYSMATYC, AMATYC, MAA, NYSUT, AFT.

Education/Credentials
B.S. in Mathematics from Rensselaer Polytechnic Institute
M.S. (and A.B.D.) in Applied Mathematics from SUNY @ Stony Brook


 
   

You are here:  Experts > Teens > Homework/Study Tips > Calculus > Trigonometry show part

Calculus - Trigonometry show part


Expert: Abe Mantell - 8/14/2009

Question
Show that the equation tan(30degree+a) =2tan(60degree-a) can be written in the form (tan^2) a+ (6*(3^.5))tan a- 5=0. Many thanks

Answer
Hello from the US!  :-)

Use the sum and difference formulas for tangent:
tan(30+a)=[tan(30)+tan(a)]/[1-tan(30)tan(a)]=[sqrt(3)+3tan(a)]/[3-sqrt(3)tan(a)]
tan(60-a)=[tan(60)-tan(a)]/[1+tan(60)tan(a)]=[sqrt(3)-tan(a)]/[1+sqrt(3)tan(a)]

Substitute those into tan(30+a)=2tan(60-a) and "cross multiply" to get:
sqrt(3)+6tan(a)+3sqrt(3)tan(a)^2=6sqrt(3)-12tan(a)+2sqrt(3)tan(a)^2
Combining terms gives: sqrt(3)tan(a)^2+18tan(a)-5sqrt(3)=0
Now divide through by sqrt(3) ==> tan(a)^2+6sqrt(3)tan(a)-5=0

OK?  I hope you can follow this and fill in the details yourself.

In case you are curious to know the solutions for a, they are approximately:
.4317178425, -1.478915394, 1.662677260, -2.709874813 (plus or minus multiples of 2Pi).

Abe


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