شاه کلید نکات
شاه کلید سوال
1 با فرض بامعنی بودن هر کسر، درستی هر یک از تساوی های زیر را بررسی کنید:
الف \({\sin ^4}\theta - {\cos ^4}\theta = {\sin ^2}\theta - {\cos ^2}\theta \)
طرف چپ = \({\sin ^4}\theta - {\cos ^4}\theta = \left( {{{\sin }^2}\theta - {{\cos }^2}\theta } \right) \times \left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right) = ...........\)
ب \(\frac{1}{{\cos \alpha }} + \cot \alpha = \frac{{\tan \alpha + \cos \alpha }}{{\sin \alpha }}\)
طرف راست = \(\frac{{\tan \alpha + \cos \alpha }}{{\sin \alpha }} = \frac{{\tan \alpha }}{{\sin \alpha }} + \frac{{\cos \alpha }}{{\sin \alpha }} = \frac{{\frac{{\sin \alpha }}{{\cos \alpha }}}}{{\sin \alpha }} + .........\, = \,.........\, + \,.........\)
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 =
\(\begin{array}{l}\frac{{\tan \alpha + \cos \alpha }}{{\sin \alpha }} = \frac{{\tan \alpha }}{{\sin \alpha }} + \frac{{\cos \alpha }}{{\sin \alpha }} = \\\\\frac{{\frac{{\sin \alpha }}{{\cos \alpha }}}}{{\sin \alpha }} + \cot \alpha \, = \,\frac{1}{{\cos \alpha }} + \,\cot \alpha \end{array}\)
2 کدام یک از تساوی های زیر یک اتحاد است؟ چرا؟
الف \({\sin ^4}\alpha + {\cos ^4}\alpha = 1 - 2\sin \alpha \cos \alpha \)
ب \({\sin ^4}\alpha + {\cos ^4}\alpha = 1 - 2{\sin ^2}\alpha {\cos ^2}\alpha \)
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 : \(1 - 2\sin {\mkern 1mu} \alpha {\mkern 1mu} \cos {\mkern 1mu} \alpha = 1 - 2\left( {\frac{1}{2}} \right)\left( {\frac{{\sqrt 3 }}{2}} \right) = 1 - \frac{{\sqrt 3 }}{2}\)
اتحاد نیست؛ زیرا:
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 : \(1 - 2{\sin ^2}\alpha {\cos ^2}\alpha \)
3 با ضرب کردن طرفینِ اتحاد مثلثاتی \(1 + {\sin ^2}\alpha = \frac{1}{{{{\cos }^2}\alpha }}\) در cotα یک اتحاد مثلثاتی بسازید؛ سپس درستی آن را اثبات کنید.
\(\begin{array}{l}1 + {\tan ^2}\alpha = \frac{1}{{{{\cos }^2}\alpha }}\:\mathop \Rightarrow \limits^{ \times \cot \alpha } \:\\\\\cot \alpha + \cot \alpha {\tan ^2}\alpha = \cot \alpha \times \frac{1}{{{{\cos }^2}\alpha }}\\\\ \Rightarrow \cot \alpha + \left( {\cot \alpha \times \tan \alpha } \right)\tan \alpha = \\\\\frac{{\cos \alpha }}{{\sin \alpha }} \times \frac{1}{{{{\cos }^2}\alpha }} \Rightarrow \cot \alpha + \tan \alpha = \frac{1}{{\cos \alpha {\mkern 1mu} \sin \alpha }}\end{array}\)
1736019749.png)
