گام به گام کاردرکلاس صفحه 44 درس 2 ریاضی (1) (مثلثات)

1)

(الف \({\sin ^4}\theta  - {\cos ^4}\theta  = {\sin ^2}\theta  - {\cos ^2}\theta\)

طرف چپ = \({\sin ^4}\theta  - {\cos ^4}\theta  = \left( {{{\sin }^2}\theta  - {{\cos }^2}\theta } \right) \times \left( {{{\sin }^2}\theta  + {{\cos }^2}\theta } \right) =\) \({\sin ^2}\theta  - {\cos ^2}\theta\)

نکته: \({\sin ^2}\theta  + {\cos ^2}\theta  = 1\)

(ب \(\frac{1}{{\cos \alpha }} + \cot \,\alpha  = \frac{{\tan \alpha  + \cos \alpha }}{{\sin \alpha }}\)

طرف راست  = \(\frac{{\tan \alpha  + \cos \alpha }}{{\sin \alpha }} = \frac{{\tan \alpha }}{{\sin \alpha }} + \frac{{\cos \alpha }}{{\sin \alpha }} = \frac{{\frac{{\sin \alpha }}{{\cos \alpha }}}}{{\sin \alpha }} +\) \(\cot \alpha  = \frac{1}{{\cos \alpha }} + \cot \alpha\)

2)

(الف \({\sin ^4}\,\alpha  + {\cos ^4}\,\alpha = 1 - 2\sin \,\alpha \,\cos \,\alpha\)

اتحاد نیست؛ زیرا :

\(\alpha  = {30^ \circ }\)

طرف چپ : \({\sin ^4}\,\alpha  + {\cos ^4}\,\alpha  = {\left( {\frac{1}{2}} \right)^4} + {\left( {\frac{{\sqrt 3 }}{2}} \right)^4} = \frac{1}{{16}} + \frac{9}{{16}} = \frac{{10}}{{16}}\)

طرف راست : \(1 - 2\sin \,\alpha \,\cos \,\alpha  = 1 - 2\left( {\frac{1}{2}} \right)\left( {\frac{{\sqrt 3 }}{2}} \right) = 1 - \frac{{\sqrt 3 }}{2}\)

(ب \({\sin ^4}\,\alpha  + {\cos ^4}\,\alpha = 1 - 2{\sin ^2}\alpha {\cos ^2}\alpha\)

اتحاد است؛ زیرا :

طرف چپ : \({\sin ^4}\,\alpha + {\cos ^4}\,\alpha =\)

\(\begin{array}{l}\left( {{{\sin }^4}\,\alpha + 2{{\sin }^2}\alpha {{\cos }^2}\alpha + {{\cos }^4}\,\alpha } \right) - 2{\sin ^2}\alpha {\cos ^2}\alpha \\ = {\left( {{{\sin }^2}\,\alpha + {{\cos }^2}\,\alpha } \right)^2} - 2{\sin ^2}\alpha {\cos ^2}\alpha \end{array}\)

\(= 1 - 2{\sin ^2}\alpha {\cos ^2}\alpha\)  : طرف راست

3)

\(\begin{array}{l}1 + {\tan ^2}\alpha  = \frac{1}{{{{\cos }^2}\alpha }}\\\\\mathop  \Rightarrow \limits^{ \times \cot \alpha } \cot \alpha  + \cot \alpha {\tan ^2}\alpha  = \cot \alpha  \times \frac{1}{{{{\cos }^2}\alpha }}\\\\ \Rightarrow \cot \alpha  + \left( {\cot \alpha  \times \tan \alpha } \right)\tan \alpha \\\\ = \frac{{\cos \alpha }}{{\sin \alpha }} \times \frac{1}{{{{\cos }^2}\alpha }}\\\\ \Rightarrow \cot \alpha  + \tan \alpha  = \frac{1}{{\cos \alpha \,\sin \alpha }}\end{array}\)