ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 1102

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\[= 2\cos a \bullet\]

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

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

\[= - 4\cos a \bullet \sin^{2}\frac{a}{2}\]