Unizor – Creative Mind through Art of Mathematics: Trigonometry+ 08: UNIZOR.COM – Math+ & Problems

Notes to a video lecture on http://www.unizor.com Trigonometry+ 08Problem AProve the following inequalitycos(36°) ≥ tan(36°)Hint AFind the point where left side equals to the right side and compare it with 36°.Use the following values:π/5≅0.628 andarcsin(½(√5−1))≅0.666.Problem BSolve the equationa·sin²(x) + b·sin(x)·cos(x) ++ c·cos²(x) = dwhere a ≠ d.Hint BFor the right side of this equation use the identitysin²(x) + cos²(x) = 1Answer Bx = arctan{R/[2(a−d)]}+π·NwhereR=[−b±√b²−4·(a−d)·(c−d)]and N is any integer number.Problem CSolve the following system of equationstan(x)·tan(y) = 3sin(x)·sin(y) = 3/4Hint CConvert this system into formcos(x+y)=…cos(x−y)=…Solution CSince tan()=sin()/cos(), substitute the second equation’s value 3/4 into numerator of the first and invert the fraction(3/4)/[cos(x)·cos(y)]=3cos(x)·cos(y) = 1/4As we know, cos(x+y) =cos(x)·cos(y) − sin(x)·sin(y)and cos(x−y) =cos(x)·cos(y) + sin(x)·sin(y)Since sin(x)·sin(y) = 3/4and cos(x)·cos(y) = 1/4we can findcos(x+y) = 1/4 − 3/4 = −1/2cos(x−y) = 1/4 + 3/4 = 1Function cos() is periodical with a period of 2π.The equation cos(x+y)=−1/2 has two solutions for (x+y) within an interval [0,2π]:x + y = ±2π/3The equation cos(x−y)=1 has one solution for (x−y) within an interval [0,2π]:x − y = 0Adding periodicity, we come up with two systems of equations, each depending on some integer parametersx + y = 2π/3 + 2π·Mx − y = 2π·Nwhere M and N are any integers, andx + y = −2π/3 + 2π·Mx − y = 2π·NEach one of these systems can be easily solved by adding and subtracting the equations, which leads to the first series of solutionsx1 = π/3 + π·(M+N)y1 = π/3 + π·(M−N)and the second series of solutionsx2 = −π/3 + π·(M+N)y2 = −π/3 + π·(M−N)In both series M and N can independently take any integer value.Note CSince original system of equation contained tan(x) and tan(y), we have to make sure that by getting rid of cos() in the denominator we have not added extraneous solutions.Function cos() is zero at π/2+π·K, where K can be any integer number. If any of our solutions falls in this set, it must be excluded. Fortunately, none of our solutions coincides with this set.