General Solutions of trigonometrical function

Trigonometric functions are periodic,a solution generalised by means of periodicity of the trigonometrical functions. The solution of the trigonometrical equation is called its general solution

1.We know that sin θ =0 for all integral multiples of π

$\displaystyle \sin { \theta } =0\\ \Leftrightarrow \theta =\pm \pi ,\pm 2\pi ,\pm 3\pi ,....\\ \Leftrightarrow \theta =n\pi ,\quad n\quad \epsilon \quad I\\ \therefore \sin { \theta } =0\\ \Leftrightarrow \theta =n\pi ,\quad n\quad \epsilon \quad I.$

2.We know that cos θ =0 for all odd multiples of π/2

$\displaystyle \cos { \theta } =0\\ \Leftrightarrow \theta =\pm \frac { \pi }{ 2 } ,\pm \frac { 3\pi }{ 2 } ,\pm \frac { 5\pi }{ 2 } ,....\\ \Leftrightarrow \theta =\left( 2n+1 \right) \frac { \pi }{ 2 } ,\quad n\quad \epsilon \quad I\\ \therefore \cos { \theta } =0\\ \Leftrightarrow \theta =\left( 2n+1 \right) \frac { \pi }{ 2 } ,\quad n\quad \epsilon \quad I.$

3.We know that tan θ =0 for all integral multiples of π

$\displaystyle \tan { \theta } =0\\ \Leftrightarrow \theta =n\pi ,\quad n\quad \epsilon \quad I\\ \therefore \tan { \theta } =0\\ \Leftrightarrow \theta =n\pi ,\quad n\quad \epsilon \quad I.$

$\displaystyle \sin { \theta } =\sin { \alpha } \\ \Leftrightarrow \theta =n\pi +{ \left( -1 \right) }^{ n }\alpha \quad ,where\quad n\quad \epsilon \quad I\quad and\quad \alpha \quad \epsilon \quad \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \\ \\ We\quad have,\quad \sin { \theta } =\sin { \alpha } \quad where\quad \alpha \quad \epsilon \quad \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \\ \sin { \theta } -\sin { \alpha } =0\\ 2\cos { \left( \frac { \theta +\alpha }{ 2 } \right) } \sin { \left( \frac { \theta -\alpha }{ 2 } \right) } =0\\ \cos { \left( \frac { \theta +\alpha }{ 2 } \right) } =0\quad or\quad \sin { \left( \frac { \theta -\alpha }{ 2 } \right) } =0\\ \left( \frac { \theta +\alpha }{ 2 } \right) =\left( 2m+1 \right) \frac { \pi }{ 2 } ,\quad m\quad \epsilon \quad I\quad or\left( \frac { \theta -\alpha }{ 2 } \right) =m\pi ,m\quad \epsilon \quad I\\ \left( \theta +\alpha \right) =\left( 2m+1 \right) \frac { \pi }{ 2 } ,\quad m\quad \epsilon \quad I\quad or\left( \frac { \theta -\alpha }{ 2 } \right) =m\pi ,m\quad \epsilon \quad I\\ \theta =\left( 2m+1 \right) \pi ,m\quad \epsilon \quad I\quad or\quad \left( \theta -\alpha \right) =2m\pi ,m\quad \epsilon \quad I$

θ=(any odd multiple of π)-α

θ=(any even multiple of π)+α

$\displaystyle \cos { \theta } =\cos { \alpha } \\ \Leftrightarrow \theta =2n\pi \pm \alpha ,where\quad n\quad \epsilon \quad I\quad and\quad \alpha \quad \epsilon \quad \left[ 0,\pi \right] \\ \\ We\quad have,\quad \cos { \theta } =\cos { \alpha } \quad \quad where\quad \alpha \quad \epsilon \quad\left[ 0,\pi \right] \\ \cos { \theta } -\cos { \alpha } =0\\ 2\sin { \left( \frac { \theta +\alpha }{ 2 } \right) } \sin { \left( \frac { \theta -\alpha }{ 2 } \right) } =0\\ \sin { \left( \frac { \theta +\alpha }{ 2 } \right) } =0\quad or\quad \sin { \left( \frac { \theta -\alpha }{ 2 } \right) } =0\\ \left( \frac { \theta +\alpha }{ 2 } \right) =n\pi ,\quad or\quad \left( \frac { \theta -\alpha }{ 2 } \right) =n\pi ,\quad \quad n\quad \epsilon \quad I\\ \left( \theta +\alpha \right) =2n\pi ,\quad or\quad \left( \theta -\alpha \right) =2n\pi ,\quad \quad \quad n\quad \epsilon \quad I\\ \therefore \cos { \theta } =\cos { \alpha } \\ \theta =2n\pi \pm \alpha ,\quad n\quad \epsilon \quad I,\quad where\quad \alpha \quad \epsilon \quad \left[ 0,\pi \right]$

$\displaystyle \tan { \theta } =\tan { \alpha } \\ \Leftrightarrow \theta =n\pi +\alpha ,where\quad n\quad \epsilon \quad I\quad and\quad \alpha \quad \epsilon \quad \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \\ \\ We\quad have,\quad \tan { \theta } =\tan { \alpha } \quad \quad where\quad \alpha \quad \epsilon \quad \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \\ \tan { \theta } -\tan { \alpha } =0\\ \frac { \sin { \theta } }{ \cos { \theta } } =\frac { \sin { \alpha } }{ \cos { \alpha } } \\ \sin { \theta } \cos { \alpha } -\cos { \theta } \sin { \alpha } =0\\ \sin { \left( \theta -\alpha \right) } =0\\ \left( \theta -\alpha \right) =n\pi ,\quad \quad n\quad \epsilon \quad I\\ \therefore \tan { \theta } =\tan { \alpha } \\ \theta =n\pi +\alpha ,\quad n\quad \epsilon \quad I,\quad where\quad \alpha \quad \epsilon \quad \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right]$

$\displaystyle { sin }^{ 2 }\theta ={ sin }^{ 2 }\alpha ,\quad { cos }^{ 2 }\theta ={ cos }^{ 2 }\alpha ,\quad { tan }^{ 2 }\theta ={ tan }^{ 2 }\alpha \Leftrightarrow \theta =n\pi \pm \alpha \\ (i){ sin }^{ 2 }\theta ={ sin }^{ 2 }\alpha \\ \Leftrightarrow \frac { 1-\cos { 2\theta } }{ 2 } =\frac { 1-\cos { 2\alpha } }{ 2 } \\ \Leftrightarrow \cos { 2\theta } =\cos { 2\alpha } \\ \Leftrightarrow 2\theta =2n\pi \pm 2\alpha ,\quad n\quad \epsilon \quad I\\ \Leftrightarrow \theta =n\pi \pm \alpha ,\quad n\quad \epsilon \quad I$

$\displaystyle (ii){ cos }^{ 2 }\theta ={ cos }^{ 2 }\alpha \\ \Leftrightarrow \frac { 1+\cos { 2\theta } }{ 2 } =\frac { 1+\cos { 2\alpha } }{ 2 } \\ \Leftrightarrow \cos { 2\theta } =\cos { 2\alpha } \\ \Leftrightarrow 2\theta =2n\pi \pm 2\alpha ,\quad n\quad \epsilon \quad I\\ \Leftrightarrow \theta =n\pi \pm \alpha ,\quad n\quad \epsilon \quad I$

$\displaystyle (iii){ tan }^{ 2 }\theta ={ tan }^{ 2 }\alpha \\ \Leftrightarrow \frac { 1-{ tan }^{ 2 }\theta }{ 1+{ tan }^{ 2 }\theta } =\frac { 1-{ tan }^{ 2 }\alpha }{ 1+{ tan }^{ 2 }\alpha } \quad \left( applying\quad componendo\quad and\quad dividendo \right) \\ \Leftrightarrow \cos { 2\theta } =\cos { 2\alpha } \\ \Leftrightarrow 2\theta =2n\pi \pm 2\alpha ,\quad n\quad \epsilon \quad I\\ \Leftrightarrow \theta =n\pi \pm \alpha ,\quad n\quad \epsilon \quad I$

February 22, 2019

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General Solutions of trigonometrical function

Nishant Agrawal

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