Сканави (А) 3.1-3.15

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№ 3.1 Теңбе-теңдікті дәлелдеңіз: $\left( {1 + {{\cos }^{ - 1}}2\alpha + {{\tg}} 2\alpha } \right)\left( {1 - {{\cos }^{ - 1}}2\alpha + {{\tg}} 2\alpha } \right) = 2{{\tg}} 2\alpha $

Шешуі: $${\left( {1 + {{\cos }^{ - 1}}2\alpha + {{\tg}} 2\alpha } \right)\left( {1 - {{\cos }^{ - 1}}2\alpha + {{\tg}} 2\alpha } \right) = 2{{\tg}} 2\alpha = }$$ $${ = \left( {1 + \dfrac{1}{{\cos 2\alpha }} + \dfrac{{\sin 2\alpha }}{{\cos 2\alpha }}} \right)\left( {1 - \dfrac{1}{{\cos 2\alpha }} + \dfrac{{\sin 2\alpha }}{{\cos 2\alpha }}} \right) = }$$ $${ = \dfrac{{\cos 2\alpha + 1 + \sin 2\alpha }}{{\cos 2\alpha }} \cdot \dfrac{{\cos 2\alpha - 1 + \sin 2\alpha }}{{\cos 2\alpha }} = }$$ $${ = \dfrac{{((\cos 2\alpha + \sin 2\alpha ) + 1)((\cos 2\alpha + \sin 2\alpha ) - 1)}}{{{{\cos }^2}2\alpha }} = \dfrac{{{{(\cos 2\alpha + \sin 2\alpha )}^2} - 1}}{{{{\cos }^2}2\alpha }} = }$$ $${ = \dfrac{{{{\cos }^2}2\alpha + 2\sin 2\alpha \cos 2\alpha + {{\sin }^2}2\alpha - 1}}{{{{\cos }^2}2\alpha }} = \dfrac{{1 + 2\sin 2\alpha {{\cos }^2}2\alpha - 1}}{{{{\cos }^2}2\alpha }} = }$$ $${ = \dfrac{{2\sin 2\alpha \cos 2\alpha }}{{{{\cos }^2}2\alpha }} = \dfrac{{2\sin 2\alpha }}{{\cos 2\alpha }} = 2{{\tg}} 2\alpha .}$$

№ 3.2 Теңбе-теңдікті дәлелдеңіз: $\left( {{{\cos }^{ - 1}}2\alpha + {{\ctg}} \left( {\dfrac{5}{2}\pi + 2\alpha } \right)} \right){{\ctg}} \left( {\dfrac{5}{4}\pi - \alpha } \right) = 1$

Шешуі: $${\left( {{{\cos }^{ - 1}}2\alpha + {{\ctg}} \left( {\dfrac{5}{2}\pi + 2\alpha } \right)} \right){{\ctg}} \left( {\dfrac{5}{4}\pi - \alpha } \right) = }$$ $${ = \left( {\dfrac{1}{{\cos 2\alpha }} + {{\ctg}} \left( {\dfrac{{4\pi + \pi }}{2} + 2\alpha } \right)} \right) \cdot {{\ctg}} \left( {\dfrac{{4\pi + \pi }}{2} - \alpha } \right) = }$$ $${ = \left( {\dfrac{1}{{\cos 2\alpha }} + {{\ctg}} \left( {2\pi + \left( {\dfrac{\pi }{2} + 2\alpha } \right)} \right)} \right) \cdot {{\ctg}} \left( {\pi + \left( {\dfrac{\pi }{4} - \alpha } \right)} \right) = }$$ $${ = \left( {\dfrac{1}{{\dfrac{{1 - {{{{\tg}} }^2}\alpha }}{{1 + {{{{\tg}} }^2}\alpha }}}} + {{\ctg}} \left( {\dfrac{\pi }{2} + 2\alpha } \right)} \right) \cdot {{\ctg}} \left( {\dfrac{\pi }{4} - \alpha } \right) = }$$ $${ = \left( {\dfrac{{1 + {{{{\tg}} }^2}\alpha }}{{1 - {{{{\tg}} }^2}\alpha }} + {{\ctg}} \left( {\dfrac{\pi }{2} + 2\alpha } \right)} \right) \cdot {{\ctg}} \left( {\dfrac{\pi }{4} - \alpha } \right) = }$$ $${ = \left( {\dfrac{{1 + {{{{\tg}} }^2}\alpha }}{{1 - {{{{\tg}} }^2}\alpha }} - {{\tg}} 2\alpha } \right) \cdot \dfrac{{1 + {{\tg}} \dfrac{\pi }{4}{{\tg}} \alpha }}{{{{\tg}} \dfrac{\pi }{4} - {{\tg}} \alpha }} = \left( {\dfrac{{1 + {{{{\tg}} }^2}\alpha }}{{1 - {{{{\tg}} }^2}\alpha }} - {{\tg}} 2\alpha } \right) \cdot \dfrac{{1 + {{\tg}} \alpha }}{{1 - {{\tg}} \alpha }} = }$$ $${ = \left( {\dfrac{{1 + {{{{\tg}} }^2}\alpha }}{{1 - {{{{\tg}} }^2}\alpha }} - \dfrac{{2{{\tg}} \alpha }}{{1 - {{\tg}} 2\alpha }}} \right) \cdot \dfrac{{1 + {{\tg}} \alpha }}{{1 - {{\tg}} \alpha }} = \dfrac{{1 + {{{{\tg}} }^2}\alpha - 2{{\tg}} \alpha }}{{1 - {{\tg}} 2\alpha }} \cdot \dfrac{{1 + {{\tg}} \alpha }}{{1 - {{\tg}} \alpha }} = }$$ $${ = \dfrac{{1 - 2{{{{\tg}} }^2}\alpha + {{\tg}} 2\alpha }}{{\left( {1 - {{{{\tg}} }^2}\alpha } \right)(1 + {{\tg}} \alpha )}} \cdot \dfrac{{1 + {{\tg}} \alpha }}{{1 - {{\tg}} \alpha }} = \dfrac{{\left( {1 - {{{{\tg}} }^2}\alpha } \right.}}{{(1 - {{\tg}} \alpha )(1 + {{\tg}} \alpha )}} \cdot \dfrac{{1 + {{\tg}} \alpha }}{{1 - {{\tg}} \alpha }} = 1}$$

№ 3.3 Теңбе-теңдікті дәлелдеңіз: $\dfrac{{\cos (3\pi - 2\alpha )}}{{2{{\sin }^2}\left( {\dfrac{{5\pi }}{4} + \alpha } \right)}} = {{\tg}} \left( {\alpha - \dfrac{{5\pi }}{4}} \right)$

Шешуі: $${\dfrac{{\cos (3\pi - 2\alpha )}}{{2{{\sin }^2}\left( {\dfrac{{5\pi }}{4} + \alpha } \right)}} = \dfrac{{ - \cos 2\alpha }}{{1 - \cos \left( {\dfrac{{5\pi }}{2} + 2\alpha } \right)}} = \dfrac{{ - \cos 2\alpha }}{{1 - \cos \left( {\dfrac{{4\pi + \pi }}{2} + 2\alpha } \right)}} = }$$ $${ = \dfrac{{ - \cos 2\alpha }}{{1 - \cos \left( {2\pi + \left( {\dfrac{\pi }{2} + 2\alpha } \right)} \right)}} = \dfrac{{ - \cos 2\alpha }}{{1 - \cos \left( {\dfrac{\pi }{2} + 2\alpha } \right)}} = \dfrac{{ - \cos 2\alpha }}{{1 + \sin 2\alpha }} = }$$ $$ = \dfrac{{ - \sin \left( {\dfrac{\pi }{2} - 2\alpha } \right)}}{{1 + \cos \left( {\dfrac{\pi }{2} - 2\alpha } \right)}} = - {{\tg}} \left( {\dfrac{\pi }{4} - \alpha } \right) = - {{\tg}} \left( {\pi + \dfrac{\pi }{4} - \alpha } \right) = $$ $$ = - {{\tg}} \left( {\dfrac{{5\pi }}{4} - \alpha } \right) = {{\tg}} \left( {\alpha - \dfrac{{5\pi }}{4}} \right)$$

№ 3.4 Теңбе-теңдікті дәлелдеңіз: $\dfrac{{{{\tg}} 2\alpha + {{\ctg}} 3\beta }}{{{{\ctg}} 2\alpha + {{\tg}} 3\beta }} = \dfrac{{{{\tg}} 2\alpha }}{{{{\tg}} 3\beta }}$

Шешуі: $${\dfrac{{{{\tg}} 2\alpha + {{\ctg}} 3\beta }}{{{{\ctg}} 2\alpha + {{\tg}} 3\beta }} = \dfrac{{\dfrac{{\sin 2\alpha }}{{\cos 2\alpha }} + \dfrac{{\cos 3\beta }}{{\sin 3\beta }}}}{{\dfrac{{\cos 2\alpha }}{{\sin 2\alpha }} + \dfrac{{\sin 3\beta }}{{\cos 3\beta }}}} = \dfrac{{\dfrac{{\sin 2\alpha \sin 3\beta + \cos 2\alpha \cos 3\beta }}{{\cos 2\alpha \sin 3\beta }}}}{{\dfrac{{\cos 2\alpha \cos 3\beta + \sin 2\alpha \sin 3\beta }}{{\sin 2\alpha \cos 3\beta }}}} = }$$ $${ = \dfrac{{\sin 2\alpha \sin 3\beta + \cos 2\alpha \cos 3\beta }}{{\cos 2\alpha \sin 3\beta }} \cdot \dfrac{{\sin 2\alpha \cos 3\beta }}{{\cos 2\alpha \cos 3\beta + \sin 2\alpha \sin 3\beta }} = }$$ $${ = \dfrac{{\sin 2\alpha \cos 3\beta }}{{\cos 2\alpha \sin 3\beta }} = \dfrac{{\sin 2\alpha }}{{\cos 2\alpha }} \cdot \dfrac{{\cos 3\beta }}{{\sin 3\beta }} = {{\tg}} 2\alpha \cdot {{\ctg}} 3\beta = {{\tg}} 2\alpha \cdot \dfrac{1}{{{{\tg}} 3\beta }} = \dfrac{{{{\tg}} 2\alpha }}{{{{\tg}} 3\beta }}{\rm{. }}}$$

№ 3.5 Теңбе-теңдікті дәлелдеңіз: $\cos \alpha + \cos 2\alpha + \cos 6\alpha + \cos 7\alpha = 4\cos \dfrac{\alpha }{2}\cos \dfrac{{5\alpha }}{2}\cos 4\alpha $

Шешуі: $${\cos \alpha + \cos 2\alpha + \cos 6\alpha + \cos 7\alpha = (\cos \alpha + \cos 7\alpha ) + (\cos 2\alpha + \cos 6\alpha ) = }$$ $${ = 2\cos 4\alpha \cos 3\alpha + 2\cos 4\alpha \cos 2\alpha = 2\cos 4\alpha (\cos 3\alpha + \cos 2\alpha ) = }$$ $${ = 2\cos 4\alpha \cdot 2\cos \dfrac{{5\alpha }}{2}\cos \dfrac{\alpha }{2} = 4\cos \dfrac{\alpha }{2}\cos \dfrac{{5\alpha }}{2}\cos 4\alpha }$$

№ 3.6 Теңбе-теңдікті дәлелдеңіз: $\sin 9\alpha + \sin 10\alpha + \sin 11\alpha + \sin 12\alpha = 4\cos \dfrac{\alpha }{2}\cos \alpha \sin \dfrac{{21\alpha }}{2}$

Шешуі: $${\sin 9\alpha + \sin 10\alpha + \sin 11\alpha + \sin 12\alpha = }$$ $${ = (\sin 9\alpha + \sin 12\alpha ) + (\sin 10\alpha + \sin 11\alpha ) = }$$ $${ = 2\sin \dfrac{{21\alpha }}{2}\cos \dfrac{{3\alpha }}{2} + 2\sin \dfrac{{21\alpha }}{2}\cos \dfrac{\alpha }{2} = 2\sin \dfrac{{21\alpha }}{2}\left( {\cos \dfrac{{3\alpha }}{2} + \cos \dfrac{\alpha }{2}} \right) = }$$ $${ = 2\sin \dfrac{{21\alpha }}{2} \cdot 2\cos \alpha \cos \dfrac{\alpha }{2} = 4\cos \dfrac{\alpha }{2}\cos \alpha \sin \dfrac{{21\alpha }}{2}.}$$

№ 3.7 Теңбе-теңдікті дәлелдеңіз: $\cos 2\alpha - \cos 3\alpha - \cos 4\alpha + \cos 5\alpha = - 4\sin \dfrac{\alpha }{2}\sin \alpha \cos \dfrac{{7\alpha }}{2}$

Шешуі: $${(\cos 2\alpha + \cos 5\alpha ) - (\cos 3\alpha + \cos 4\alpha ) = }$$ $${ = 2\cos \dfrac{{7\alpha }}{2}\cos \dfrac{{3\alpha }}{2} - 2\cos \dfrac{{7\alpha }}{2}\cos \dfrac{\alpha }{2} = 2\cos \dfrac{{7\alpha }}{2}\left( {\cos \dfrac{{3\alpha }}{2} - \cos \dfrac{\alpha }{2}} \right) = }$$ $${ = 2\cos \dfrac{{7\alpha }}{2} \cdot \left( { - 2\sin \alpha \sin \dfrac{\alpha }{2}} \right) = - 4\sin \dfrac{\alpha }{2}\sin \alpha \cos \dfrac{{7\alpha }}{2}}$$

№ 3.8 Теңбе-теңдікті дәлелдеңіз: $\sin 4\alpha - \sin 5\alpha - \sin 6\alpha + \sin 7\alpha = - 4\sin \dfrac{\alpha }{2}\sin \alpha \sin \dfrac{{11\alpha }}{2}$

Шешуі: $${\sin 4\alpha + \sin 7\alpha - (\sin 5\alpha + \sin 6\alpha ) = }$$ $${ = 2\sin \dfrac{{11\alpha }}{2}\cos \dfrac{{3\alpha }}{2} - 2\sin \dfrac{{11\alpha }}{2}\cos \dfrac{\alpha }{2} = 2\sin \dfrac{{11\alpha }}{2}\left( {\cos \dfrac{{3\alpha }}{2} - \cos \dfrac{\alpha }{2}} \right) = }$$ $${ = 2\sin \dfrac{{11\alpha }}{2} \cdot \left( { - 2\sin \alpha \sin \dfrac{\alpha }{2}} \right) = - 4\sin \dfrac{\alpha }{2}\sin \alpha \sin \dfrac{{11\alpha }}{2}.}$$

№ 3.9 Теңбе-теңдікті дәлелдеңіз: $\cos \alpha + \sin \alpha + \cos 3\alpha + \sin 3\alpha = 2\sqrt 2 \cos \alpha \sin \left( {\dfrac{\pi }{4} + 2\alpha } \right)$

Шешуі: $${\cos \alpha + \cos \left( {\dfrac{\pi }{2} - \alpha } \right) + \cos 3\alpha + \cos \left( {\dfrac{\pi }{2} - 3\alpha } \right) = }$$ $${ = 2\cos \dfrac{\pi }{4}\cos \left( {\dfrac{\pi }{4} - \alpha } \right) + 2\cos \dfrac{\pi }{4}\cos \left( {\dfrac{\pi }{4} - 3\alpha } \right) = }$$ $${ = 2\cos \dfrac{\pi }{4}\left( {\cos \left( {\dfrac{\pi }{4} - \alpha } \right) + \cos \left( {\dfrac{\pi }{4} - 3\alpha } \right)} \right) = 2 \cdot \dfrac{{\sqrt 2 }}{2} \cdot 2\cos \left( {\dfrac{\pi }{4} - 2\alpha } \right)\cos \alpha = }$$ $${ = 2\sqrt 2 \cos \alpha \cos \left( {\dfrac{\pi }{4} - 2\alpha } \right) = 2\sqrt 2 \cos \alpha \cos \left( {\dfrac{\pi }{2} - \dfrac{\pi }{4} - 2\alpha } \right) = }$$ $${ = 2\sqrt 2 \cos \alpha \cos \left( {\dfrac{\pi }{2} - \left( {\dfrac{\pi }{4} + 2\alpha } \right)} \right) = 2\sqrt 2 \cos \alpha \sin \left( {\dfrac{\pi }{4} + 2\alpha } \right)}$$

№ 3.10 Теңбе-теңдікті дәлелдеңіз: ${{\tg}} \alpha + {{\ctg}} \alpha + {{\tg}} 3\alpha + {{\ctg}} 3\alpha = \dfrac{{8{{\cos }^2}2\alpha }}{{\sin 6\alpha }}$

Шешуі: $${\dfrac{{\sin \alpha }}{{\cos \alpha }} + \dfrac{{\cos \alpha }}{{\sin \alpha }} + \dfrac{{\sin 3\alpha }}{{\cos 3\alpha }} + \dfrac{{\cos 3\alpha }}{{\sin 3\alpha }} = \dfrac{{{{\sin }^2}\alpha + {{\cos }^2}\alpha }}{{\sin \alpha \cos \alpha }} + \dfrac{{{{\sin }^2}3\alpha + {{\cos }^2}3\alpha }}{{\sin 3\alpha \cos 3\alpha }} = }$$ $${ = \dfrac{1}{{\sin \alpha \cos \alpha }} + \dfrac{1}{{\sin 3\alpha \cos 3\alpha }} = \dfrac{2}{{2\sin \alpha \cos \alpha }} + \dfrac{2}{{2\sin 3\alpha \cos 3\alpha }} = }$$ $${ = \dfrac{2}{{\sin 2\alpha }} + \dfrac{2}{{\sin 6\alpha }} = \dfrac{{2\sin 6\alpha + 2\sin 2\alpha }}{{\sin 2\alpha \sin 6\alpha }} = \dfrac{{2(\sin 6\alpha + \sin 2\alpha )}}{{\sin 2\alpha \sin 6\alpha }} = }$$ $${ = \dfrac{{2 \cdot 2\sin 4\alpha \cos 2\alpha }}{{\sin 2\alpha \sin 6\alpha }} = \dfrac{{4\sin 4\alpha \cos 2\alpha }}{{\sin 2\alpha \sin 6\alpha }} = \dfrac{{4 \cdot 2\sin 2\alpha \cos 2\alpha \cos 2\alpha }}{{\sin 2\alpha \sin 6\alpha }} = }$$ $${ = \dfrac{{8\sin 2\alpha {{\cos }^2}2\alpha }}{{\sin 2\alpha \sin 6\alpha }} = \dfrac{{8{{\cos }^2}2\alpha }}{{\sin 6\alpha }}}$$

№ 3.11 Теңбе-теңдікті дәлелдеңіз: ${(\sin \alpha )^{ - 1}} + {({{\tg}} \alpha )^{ - 1}} = {{\ctg}} \dfrac{\alpha }{2}$

Шешуі: $${{{(\sin \alpha )}^{ - 1}} + {{({{\tg}} \alpha )}^{ - 1}} = \dfrac{1}{{\sin \alpha }} + \dfrac{1}{{{{\tg}} \alpha }} = \dfrac{1}{{\sin \alpha }} + \dfrac{1}{{\dfrac{{\sin \alpha }}{{\cos \alpha }}}} = \dfrac{1}{{\sin \alpha }} + \dfrac{{\cos \alpha }}{{\sin \alpha }} = }$$ $${ = \dfrac{{1 + \cos \alpha }}{{\sin \alpha }} = \dfrac{{1 + 2{{\cos }^2}\dfrac{\alpha }{2} - 1}}{{2\sin \dfrac{\alpha }{2}\cos \dfrac{\alpha }{2}}} = \dfrac{{2{{\cos }^2}\dfrac{\alpha }{2}}}{{2\sin \dfrac{\alpha }{2}\cos \dfrac{\alpha }{2}}} = \dfrac{{\cos \dfrac{\alpha }{2}}}{{\sin \dfrac{\alpha }{2}}} = {{\ctg}} \dfrac{\alpha }{2}}$$

№ 3.12 Теңбе-теңдікті дәлелдеңіз: $\dfrac{{\sin \left( {\dfrac{\pi }{2} + 3\alpha } \right)}}{{1 - \sin (3\alpha - \pi )}} = {{\ctg}} \left( {\dfrac{5}{4}\pi + \dfrac{3}{2}\alpha } \right)$

Шешуі: $$\dfrac{{\sin \left( {\dfrac{\pi }{2} + 3\alpha } \right)}}{{1 - \sin (3\alpha - \pi )}} = \dfrac{{\cos 3\alpha }}{{1 + \sin 3\alpha }} = \dfrac{{\cos \left( {2 \cdot \dfrac{3}{2}\alpha } \right)}}{{1 + \sin \left( {2 \cdot \dfrac{3}{2}\alpha } \right)}} = \dfrac{{{{\cos }^2}\dfrac{{3\alpha }}{2} - {{\sin }^2}\dfrac{{3\alpha }}{2}}}{{1 + 2\sin \dfrac{{3\alpha }}{2}\cos \dfrac{{3\alpha }}{2}}} = $$ $${ = \dfrac{{\left( {\cos \dfrac{{3\alpha }}{2} - \sin \dfrac{{3\alpha }}{2}} \right)\left( {\cos \dfrac{{3\alpha }}{2} + \sin \dfrac{{3\alpha }}{2}} \right)}}{{{{\cos }^2}\dfrac{{3\alpha }}{2} + {{\sin }^2}\dfrac{{3\alpha }}{2} + 2\sin \dfrac{{3\alpha }}{2}\cos \dfrac{{3\alpha }}{2}}} = \dfrac{{\left( {\cos \dfrac{{3\alpha }}{2} - \sin \dfrac{{3\alpha }}{2}} \right)\left( {\cos \dfrac{{3\alpha }}{2} + \sin \dfrac{{3\alpha }}{2}} \right)}}{{{{\left( {\cos \dfrac{{3\alpha }}{2} + \sin \dfrac{{3\alpha }}{2}} \right)}^2}}} = }$$ $${ = \dfrac{{\cos \dfrac{{3\alpha }}{2} - \sin \dfrac{{3\alpha }}{2}}}{{\cos \dfrac{{3\alpha }}{2} + \sin \dfrac{{3\alpha }}{2}}} = \dfrac{{1 - {{\tg}} \dfrac{{3\alpha }}{2}}}{{1 + {{\tg}} \dfrac{{3\alpha }}{2}}} = \dfrac{{{{\tg}} \dfrac{\pi }{4} - {{\tg}} \dfrac{\pi }{4}{{\tg}} \dfrac{{3\alpha }}{2}}}{{{{\tg}} \dfrac{\pi }{4} + {{\tg}} \dfrac{{3\alpha }}{2}}} = \dfrac{1}{{{{\tg}} \left( {\dfrac{\pi }{4} + \dfrac{{3\alpha }}{2}} \right)}} = }$$ $${ = {{\ctg}} \left( {\dfrac{\pi }{4} + \dfrac{{3\alpha }}{2}} \right) = {{\ctg}} \left( {\pi + \dfrac{\pi }{4} + \dfrac{{3\alpha }}{2}} \right) = {{\ctg}} \left( {\dfrac{{5\pi }}{4} + \dfrac{{3\alpha }}{2}} \right)}$$

№ 3.13 Теңбе-теңдікті дәлелдеңіз: $\dfrac{{\sin 2\alpha - \sin 3\alpha + \sin 4\alpha }}{{\cos 2\alpha - \cos 3\alpha + \cos 4\alpha }} = {{\tg}} 3\alpha $

Шешуі: $${\dfrac{{(\sin 2\alpha + \sin 4\alpha ) - \sin 3\alpha }}{{(\cos 2\alpha + \cos 4\alpha ) - \cos 3\alpha }} = \dfrac{{2\sin 3\alpha \cos \alpha - \sin 3\alpha }}{{2\cos 3\alpha \cos \alpha - \cos 3\alpha }} = \dfrac{{\sin 3\alpha (2\cos \alpha - 1)}}{{\cos 3\alpha (2\cos \alpha - 1)}} = }$$ $${ = \dfrac{{\sin 3\alpha }}{{\cos 3\alpha }} = {{\tg}} 3\alpha }$$

№ 3.14 Теңбе-теңдікті дәлелдеңіз: $2{\sin ^2}(3\pi - 2\alpha ){\cos ^2}(5\pi + 2\alpha ) = \dfrac{1}{4} - \dfrac{1}{4}\sin \left( {\dfrac{5}{2}\pi - 8\alpha } \right)$

Шешуі: ${\sin ^2}\dfrac{x}{2} = \dfrac{{1 - \cos x}}{2}$ және ${\cos ^2}\dfrac{x}{2} = \dfrac{{1 + \cos x}}{2}$ дәрежесін төмендету формулаларын қолданамыз: $${2 \cdot \dfrac{{(1 - \cos (6\pi - 4\alpha ))(1 + \cos (10\pi + 4\alpha ))}}{4} = \dfrac{1}{2}(1 - \cos (6\pi - 4\alpha ))(1 + \cos (10\pi + 4\alpha )) = }$$ $${ = \dfrac{1}{2}(1 - \cos 4\alpha )(1 + \cos 4\alpha ) = \dfrac{1}{2}\left( {1 - {{\cos }^2}4\alpha } \right) = \dfrac{1}{2}\left( {1 - \dfrac{{1 + \cos 8\alpha }}{2}} \right) = }$$ $${ = \dfrac{1}{2}\left( {1 - \dfrac{1}{2} - \dfrac{1}{2}\cos 8\alpha } \right) = \dfrac{1}{2}\left( {\dfrac{1}{2} - \dfrac{1}{2}\cos 8\alpha } \right) = \dfrac{1}{4} - \dfrac{1}{4}\sin \left( {\dfrac{{5\pi }}{2} - 8\alpha } \right).}$$

№ 3.15 Теңбе-теңдікті дәлелдеңіз: $\sin 2\alpha (1 + {{\tg}} 2\alpha {{\tg}} \alpha ) + \dfrac{{1 + \sin \alpha }}{{1 - \sin \alpha }} = {{\tg}} 2\alpha + {{{\tg}} ^2}\left( {\dfrac{\pi }{4} + \dfrac{\alpha }{2}} \right)$

Шешуі: $${X = \sin 2\alpha (1 + {{\tg}} 2\alpha {{\tg}} \alpha ) = \sin 2\alpha \left( {1 + \dfrac{{\sin 2\alpha }}{{\cos 2\alpha }} \cdot \dfrac{{\sin \alpha }}{{\cos \alpha }}} \right) = \dfrac{{\sin 2\alpha (\cos 2\alpha \cos \alpha + \sin 2\alpha \sin \alpha )}}{{\cos 2\alpha \cos \alpha }}}$$ деп алайық. $\cos x\cos y + \sin x\sin y = \cos (x - y)$ формуласын қолданып, өрнекті мына түрге келтіреміз: $$X = \dfrac{{\sin 2\alpha \cos \alpha }}{{\cos 2\alpha \cos \alpha }} = \dfrac{{\sin 2\alpha }}{{\cos 2\alpha }} = {{\tg}} 2\alpha $$ Енді $$Y = \dfrac{{1 + \sin \alpha }}{{1 - \sin \alpha }} = \dfrac{{1 + \sin \left( {2 \cdot \dfrac{\alpha }{2}} \right)}}{{1 - \sin \left( {2 \cdot \dfrac{\alpha }{2}} \right)}}$$ деп алайық. Онда $\sin 2x = 2\sin x\cos x$ формуласынан $$Y = \dfrac{{1 + 2\sin \dfrac{\alpha }{2}\cos \dfrac{\alpha }{2}}}{{1 - 2\sin \dfrac{\alpha }{2}\cos \dfrac{\alpha }{2}}} = \dfrac{{{{\cos }^2}\dfrac{\alpha }{2} + {{\sin }^2}\dfrac{\alpha }{2} + 2\sin \dfrac{\alpha }{2}\cos \dfrac{\alpha }{2}}}{{{{\cos }^2}\dfrac{\alpha }{2} + {{\sin }^2}\dfrac{\alpha }{2} - 2\sin \dfrac{\alpha }{2}\cos \dfrac{\alpha }{2}}} = $$ $$ = \dfrac{{{{\left( {\cos \dfrac{\alpha }{2} + \sin \dfrac{\alpha }{2}} \right)}^2}}}{{{{\left( {\cos \dfrac{\alpha }{2} - \sin \dfrac{\alpha }{2}} \right)}^2}}} = {\left( {\dfrac{{\cos \dfrac{\alpha }{2} + \sin \dfrac{\alpha }{2}}}{{\cos \dfrac{\alpha }{2} - \sin \dfrac{\alpha }{2}}}} \right)^2}$$ Жақша ішіндегі өрнектердің барлығын $\cos \dfrac{\alpha }{2} \ne 0$ өрнегіне бөліп және $\dfrac{{{{\tg}} x + {{\tg}} y}}{{1 - {{\tg}} x{{\tg}} y}} = {{\tg}} (x + y),{\rm{ }}x,y,x + y \ne \dfrac{\pi }{2} + \pi n,n \in Z$ формуласын пайдаланып, $$Y = {\left( {\dfrac{{1 + {{\tg}} \dfrac{\alpha }{2}}}{{1 - {{\tg}} \dfrac{\alpha }{2}}}} \right)^2} = {\left( {\dfrac{{{{\tg}} \dfrac{\pi }{4} + {{\tg}} \dfrac{\alpha }{2}}}{{1 - {{\tg}} \dfrac{\pi }{4}{{\tg}} \dfrac{\alpha }{2}}}} \right)^2} = {\left( {{{\tg}} \left( {\dfrac{\pi }{4} + \dfrac{\alpha }{2}} \right)} \right)^2} = {{{\tg}} ^2}\left( {\dfrac{\pi }{4} + \dfrac{\alpha }{2}} \right)$$ аламыз. Онда $$X + Y = {{\tg}} 2\alpha + {{{\tg}} ^2}\left( {\dfrac{\pi }{4} + \dfrac{\alpha }{2}} \right)$$

 
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