جواب کاردرکلاس صفحه 44 درس 2 ریاضی دهم (مثلثات)

الف طرف چپ =

 \(\begin{array}{l}{\sin ^4}\theta - {\cos ^4}\theta = \left( {{{\sin }^2}\theta - {{\cos }^2}\theta } \right) \times \left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right) = \\\\\left( {{{\sin }^2}\theta - {{\cos }^2}\theta } \right) \times \left( 1 \right) = {\sin ^2}\theta - {\cos ^2}\theta \end{array}\)

ب طرف راست =

\(\begin{array}{l}\frac{{\tan \alpha + \cos \alpha }}{{\sin \alpha }} = \frac{{\tan \alpha }}{{\sin \alpha }} + \frac{{\cos \alpha }}{{\sin \alpha }} = \\\\\frac{{\frac{{\sin \alpha }}{{\cos \alpha }}}}{{\sin \alpha }} + \cot \alpha \, = \,\frac{1}{{\cos \alpha }} + \,\cot \alpha \end{array}\)

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

اتحاد نیست؛ زیرا:

α = 30°

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

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

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

اتحاد است؛ زیرا:

: طرف چپ

\(\begin{array}{*{20}{l}}\begin{array}{l}{\sin ^4}{\mkern 1mu} \alpha + {\cos ^4}{\mkern 1mu} \alpha = \\\\\left( {{{\sin }^4}{\mkern 1mu} \alpha + 2{{\sin }^2}\alpha {{\cos }^2}\alpha + {{\cos }^4}{\mkern 1mu} \alpha } \right) - 2{\sin ^2}\alpha {\cos ^2}\alpha \end{array}\\\begin{array}{l}\\ = {\left( {{{\sin }^2}{\mkern 1mu} \alpha + {{\cos }^2}{\mkern 1mu} \alpha } \right)^2} - 2{\sin ^2}\alpha {\cos ^2}\alpha \end{array}\\\begin{array}{l}\\ = 1 - 2{\sin ^2}\alpha {\cos ^2}\alpha \end{array}\end{array}\)

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

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