# Answer in Trigonometry

If t = tan(“theta” /2), show that sin(“theta” )=2t/(1-t^{2}) and cos (“theta” )=(1-t^{2})/(1+t^{2}). hence solve the equation cos(“theta” )-2sin(“theta” )=2

Let “t = tan(frac{theta}2).” Then

“sin(theta )=2sinfrac{theta}2cosfrac{theta}2=2tanfrac{theta}2cos^2frac{theta}2n=frac{2tanfrac{theta}2}{frac{1}{cos^2frac{theta}2}}n=frac{2tanfrac{theta}2}{1+tan^2frac{theta}2}=frac{2t}{1+t^2},” and

“cos (theta )=cos^2frac{theta}2-sin^2frac{theta}2n=frac{cos^2frac{theta}2-sin^2frac{theta}2}{cos^2frac{theta}2frac{1}{cos^2frac{theta}2}}n=frac{1-tan^2frac{theta}2}{1+tan^2frac{theta}2}n=frac{1-t^2}{1+t^2}.”

Let us solve the equation “cos(theta )-2sin(theta )=2.” Let “t = tan(frac{theta}2).” Then we get the equation

“frac{1-t^2}{1+t^2}-2frac{2t}{t^2+1}=2,” and hence “frac{1-t^2-4t}{1+t^2}=2.” It follows that “1-t^2-4t=2+2t^2,” and thus “3t^2+4t+1=0.” The last equation is equivalent to“(3t+1)(t+1)=0,” and hence has the roots “t_1=-1, t_2=-frac{1}3.” Then “tan(frac{theta}2)=-1” or “tan(frac{theta}2)=-frac{1}3.” It follows that “frac{theta}2=-frac{pi}4+pi n” or “frac{theta}2=-arctanfrac{1}3+pi n, ninmathbb N.”

We conclude that the solutions of the equation “cos(theta )-2sin(theta )=2” are of the form:

“theta=-frac{pi}2+2pi n” or “theta=-2arctanfrac{1}3+2pi n,” where “ninmathbb N.”

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