ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 1293

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\[1)\ 1 + \cos a + \sin a =\]

\[= \left( 1 + \cos a \right) + \sin a =\]

\[= 2\cos^{2}\frac{a}{2} + 2\sin\frac{a}{2} \bullet \cos\frac{a}{2} =\]

\[= 2\cos\frac{a}{2} \bullet \left( \cos\frac{a}{2} + \sin\frac{a}{2} \right) =\]

\[= 2\cos\frac{a}{2} \bullet \left( \sin\left( \frac{\pi}{2} + \frac{a}{2} \right) + \sin\frac{a}{2} \right) =\]

\[= 4\cos\frac{a}{2} \bullet \sin\left( \frac{\pi}{4} + \frac{a}{2} \right) \bullet \cos\frac{\pi}{4} =\]

\[= 2\sqrt{2} \bullet \cos\frac{a}{2} \bullet \sin\left( \frac{a}{2} + \frac{\pi}{4} \right);\]

\[2)\ 1 - \cos a - \sin a =\]

\[= \left( 1 - \cos a \right) - \sin a =\]

\[= 2\sin^{2}\frac{a}{2} - 2\sin\frac{a}{2} \bullet \cos\frac{a}{2} =\]

\[= 2\sin\frac{a}{2} \bullet \left( \sin\frac{a}{2} - \cos\frac{a}{2} \right) =\]

\[= 2\sin\frac{a}{2} \bullet \left( \sin\frac{a}{2} + \sin\left( \frac{\pi}{2} - \frac{a}{2} \right) \right) =\]

\[= 4\sin\frac{a}{2} \bullet \sin\left( \frac{a}{2} - \frac{\pi}{4} \right) \bullet \cos\frac{\pi}{4} =\]

\[= 2\sqrt{2} \bullet \sin\frac{a}{2} \bullet \sin\left( \frac{a}{2} - \frac{\pi}{4} \right);\]

\[3)\ 3 - 4\sin^{2}a = 4 - 4\sin^{2}a - 1 =\]

\[= 1 + \left( 2\cos^{2}a - 2\sin^{2}a \right) =\]

\[= 1 + 2\cos{2a} = 2 \bullet \frac{1}{2} + 2\cos{2a} =\]

\[= 2 \bullet \left( \cos{2a} - \left( - \frac{1}{2} \right) \right) =\]

\[= 2 \bullet \left( \cos{2a} - \cos\frac{2\pi}{3} \right) =\]

\[= 2 \bullet ( - 2) \bullet \sin\frac{2a + \frac{2\pi}{3}}{2} \bullet \sin\frac{2a - \frac{2\pi}{3}}{2} =\]

\[= - 4 \bullet \sin\left( a + \frac{\pi}{3} \right) \bullet \sin\left( a - \frac{\pi}{3} \right);\]

\[4)\ 1 - 4\cos^{2}a = 2 - 1 - 4\cos^{2}a =\]

\[= 2\sin^{2}a + 2\cos^{2}a - 1 - 4\cos^{2}a =\]

\[= - 1 - \left( 2\cos^{2}a - 2\sin^{2}a \right) =\]

\[= - 1 - 2\cos{2a} =\]

\[= - 2 \bullet \frac{1}{2} - 2\cos{2a} =\]

\[= - 2 \bullet \left( \cos{2a} - \left( - \frac{1}{2} \right) \right) =\]

\[= - 2 \bullet \left( \cos{2a} - \cos\frac{2\pi}{3} \right) =\]

\[= - 2 \bullet ( - 2) \bullet \sin\frac{2a + \frac{2\pi}{3}}{2} \bullet \sin\frac{2a - \frac{2\pi}{3}}{2} =\]

\[= 4 \bullet \sin\left( a + \frac{\pi}{3} \right) \bullet \sin\left( a - \frac{\pi}{3} \right).\]