вычислите значение выражения:[tex] { \cos }^{2} ( \frac{\pi}{8} ) + \sin( \frac{\pi}{8} ) \cos( \frac{3\pi}{8} ) [/tex]. Created by tanya15t1. matematika-ru

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вычислите значение выражения:
[tex] { \cos }^{2} ( \frac{\pi}{8} ) + \sin( \frac{\pi}{8} ) \cos( \frac{3\pi}{8} ) [/tex]

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$\boxed{\; cos^2a=\dfrac{1+cos2a}{2}\; }\; \; \; \Rightarrow \; \; \; cos^2\frac{\pi}{8}=\dfrac{1+cos\frac{\pi}{4}}{2}=\dfrac{1+\frac{\sqrt2}{2}}{2}=\dfrac{2+\sqrt2}{4}\\\\\\\boxed {\; sin^2a+cos^2a=1\; }\; \; \; \Rightarrow \; \; \; sin^2\dfrac{\pi}{8}=1-cos^2\dfrac{\pi}{8}=1-\dfrac{2+\sqrt2}{4}=\dfrac{2-\sqrt2}{4}\\\\\\sin\dfrac{\pi}{8}=\dfrac{\sqrt{2-\sqrt2}}{2}\\\\cos\dfrac{3\pi}{8}=cos\Big(\dfrac{\pi}{2}-\dfrac{\pi}{8}\Big)=sin\dfrac{\pi}{8}=\dfrac{\sqrt{2-\sqrt2}}{2}$

$cos^2\dfrac{\pi}{8}+sin\dfrac{\pi}{8}\cdot cos\dfrac{3\pi }{8}=\dfrac{2+\sqrt2}{4}+\Big(\dfrac{\sqrt{2-\sqrt2}}{2}\Big)^2=\dfrac{2+\sqrt2}{4}+\dfrac{2-\sqrt2}{4}=\\\\\\=\dfrac{2+\sqrt2+2-\sqrt2}{4}=\dfrac{4}{4}=1\\\\\\\boxed {\; cos^2\dfrac{\pi}{8}+sin\dfrac{\pi}{8}\cdot cos\dfrac{3\pi }{8}=1\; }$

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