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Three Trig Problems


Date: 6/11/96 at 21:9:10
From: Anonymous
Subject: Three trig problems

I would like you to answer three trig problems that have been baffling 
me for a while.

(1). Find tan x if ((sin x)^2)/3 + ((cos x)^2)/7 = (-sin(2x)+1)/10

(2). A = 20 deg. and B = 25 deg. Find the value of (1+tanA)(1+tanB).

(3). Evaluate cos 36 - cos 72. 

                 - Thanks for your help.
                       Michael


Date: 6/12/96 at 7:55:2
From: Doctor Pete
Subject: Re: Three trig problems

Again, here are a few hints:

1.  Note that the half-angle formulas give sin^2(x) = (1-cos(2x))/2,
cos^2(x) = (1+cos(x))/2; substitute and simplify to obtain an 
expression in cos(2x) and sin(2x).  Noting that 
cos(2x) = sqrt(1-sin^2(2x)), let y = sin(2x), and appropriate solving, 
using the fact that 20^2+21^2 = 29^2, will give sin(2x) = -21/29.

2.  Since tan(A+B) = tan(45) = 1, consider tan(A) = tan((A+B)-B) and 
apply the tangent addition formula, resulting in
(1+tan(A))/(1+tan(B)) = 2.

3.  An indirect way of evaluating this is to think geometrically.  If 
you are familiar with complex numbers, consider the complex 5th roots 
of unity in the plane; these form the vertices of a regular pentagon.  
Let the solutions to z^5-1 = 0 be z(0), z(1), ..., z(4), where 
z(0) = 1 and the remaining roots are counted counterclockwise.  So we 
want to find the real part of z(1), which is cos(2*Pi/5) = cos(72).  
To do this, factor z^5-1 = (z-1)(z^4+z^3+z^2+z+1).  The latter factor 
can be broken into two quadratics by considering the product 
(z^2+a*z+1)(z^2+b*z+1) (why can we restrict the coefficients of the 
square and constant terms to be 1?), and finding a and b such that the 
expansion of the above gives z^4+z^3+z^2+z+1.  Since z(1) and z(4) are 
complex conjugates, their sum is twice the real part of z(1); but
this is precisely the negative of the coefficient of the x term of one 
of these quadratic factors (which one, and why?).  This will give you 
cos(72), and cos(36) is easily found using the half-angle formula; the 
answer is 1/2.

There should be an easier way to do (3); the above hints are just 
solutions that come immediately to mind.  If you have difficulty 
filling in the details (esp. on 3), feel free to ask.

-Doctor Pete,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
College Trigonometry
High School Trigonometry

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