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Friday, December 7, 2018

Trigonometry and Quadratic equation

 If \displaystyle \sec \theta +\cos ec\ \theta =2 , find exact value of \displaystyle \sin \theta .

  Solution

\displaystyle \sec \theta +\cos ec\ \theta =2

\displaystyle \frac{1}{{\cos \theta }}\ +\ \frac{1}{{\sin \theta }}\ \ =\ 2

\displaystyle \frac{{\sin \theta +\cos \theta }}{{\sin \theta \cos \theta }}=2

\displaystyle \sin \theta +\cos \theta =\sin 2\theta \ \ ......(1)

\displaystyle {{(\sin \theta +\cos \theta )}^{2}}={{\sin }^{2}}2\theta

\displaystyle {{\sin }^{2}}\theta +2\sin \theta \cos \theta +{{\cos }^{2}}\theta ={{\sin }^{2}}2\theta

\displaystyle 1+\sin 2\theta ={{\sin }^{2}}2\theta

\displaystyle {{\sin }^{2}}2\theta -\sin 2\theta -1=0

\displaystyle \sin 2\theta =\frac{{1-\sqrt{5}}}{2}\ \ [\because -1\le {{\sin }^{2}}2\theta \le 1]

\displaystyle 2\sin \theta \cos \theta =\frac{{1-\sqrt{5}}}{2}

\displaystyle \cos \theta =\frac{{1-\sqrt{5}}}{{4\sin \theta }}

\displaystyle \sin \theta +\frac{{1-\sqrt{5}}}{{4\sin \theta }}=\frac{{1-\sqrt{5}}}{2}\ \ \ \ \ [\ By\ (1)\ ]

\displaystyle 4{{\sin }^{2}}\theta +1-\sqrt{5}=(2-2\sqrt{5})\sin \theta

\displaystyle 4{{\sin }^{2}}\theta -(2-2\sqrt{5})\sin \theta +1-\sqrt{5}=0

\displaystyle \sin \theta =\frac{{2-2\sqrt{5}\pm \sqrt{{{{{(2-2\sqrt{5})}}^{2}}-16(1-\sqrt{5})}}}}{8}

\displaystyle \ \ \ \ \ \ \ =\frac{{2-2\sqrt{5}\pm \sqrt{{8+8\sqrt{5}}}}}{8}

\displaystyle \ \ \ \ \ \ \ =\frac{{2-2\sqrt{5}\pm 2\sqrt{{2+2\sqrt{5}}}}}{8}

\displaystyle \ \ \ \ \ \ \ =\frac{{1-\sqrt{5}\pm \sqrt{{2+2\sqrt{5}}}}}{4}

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