Ucale

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 sinÂ Î¸ =0 for all integral multiples ofÂ Ï€

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

2.We know thatÂ Â cos Î¸ =0 for all odd multiples ofÂ Ï€/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 thatÂ  tan Î¸ =0 for all integral multiples ofÂ Ï€

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

4.

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

Î¸=(any odd multiple of Ï€)-Î±

Î¸=(any even multiple of Ï€)+Î±

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

5.

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

6.

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

7.

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

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

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

February 22, 2019
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