Question

вычислите значение выражения:
${ \cos }^{2} ( \frac{\pi}{8} ) + \sin( \frac{\pi}{8} ) \cos( \frac{3\pi}{8} )$
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Answer 1

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

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

Answer 2

сos²π/8=(1+cos(π/4))/2=(1+√2/2)/2=(2+√2)/4

sin(π/8)*cos(3π/8)=sin(π/8)*cos(π/2-π/8)=sin²(π/8)=1-((2+√2)/4)

сos²π/8+sin(π/8)*cos(3π/8)=

(2+√2)/4+1-((2+√2)/4)=(2+√2+4-2-√2)/4=1