Inverse trigonometric function جديد

ربما هناك سهو مطبعي في النقطة الاولى :

$Q2)(i)arctan(\sqrt{\frac{1-cosx}{1+cosx}})$

$=arctan(\sqrt{\frac{1-cos^{2}(\frac{x}{2})+sin^{2}(\frac{x}{2})}{1+cos^{ 2}(\frac{x}{2})-sin^{2}(\frac{x}{2})}})$

$=arctan(\sqrt{\frac{2sin^{2}\frac{x}{2}}{2cos^{2} \frac{x}{2}}})$

$=arctan(tan(\frac{x}{2}))=\frac{x}{2}$

$(ii)arctan(\frac{cosx}{1+sinx})$

$=arctan(\frac{cos^{2}\frac{x}{2}-sin^{2}\frac{x}{2}}{cos^{2}\frac{x}{2}+ sin^{2}\frac{x}{2} +2sin\frac{x}{2}cos\frac{x}{2}})$

$=arctan(\frac{(cos\frac{x}{2}+sin\frac{x}{2})(cos \frac{x}{2}-sin\frac{x}{2})}{(cos\frac{x}{2}+sin\frac{x}{2})^{ 2}})$

$=arctan(\frac{(cos\frac{x}{2}-sin\frac{x}{2})}{(cos\frac{x}{2}+sin\frac{x}{2})}) $

$=arctan(\frac{1-tan\frac{x}{2}}{1+tan\frac{x}{2}})$

$=arctan(tan(\frac{\pi }{4}-\frac{x}{2}))=\frac{\pi }{4}-\frac{x}{2}$

$(iii)arctan(\frac{cosx}{1-sinx})$

$=arctan(\frac{cos^{2}\frac{x}{2}-sin^{2}\frac{x}{2}}{cos^{2}\frac{x}{2}+ sin^{2}\frac{x}{2}-2sin\frac{x}{2}cos\frac{x}{2}})$

$=arctan(\frac{(cos\frac{x}{2}+sin\frac{x}{2})(cos \frac{x}{2}-sin\frac{x}{2})}{(cos\frac{x}{2}-sin\frac{x}{2})^{2}})$

$=arctan(\frac{(cos\frac{x}{2}+sin\frac{x}{2})}{(c os\frac{x}{2}-sin\frac{x}{2})})$

$=arctan(\frac{1+tan\frac{x}{2}}{1-tan\frac{x}{2}})$

$=arctan(tan(\frac{\pi }{4}+\frac{x}{2}))=\frac{\pi }{4}+\frac{x}{2}$