Question

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Answer 1

$42)\ \ cos^4a=\dfrac{1}{8}\,cos4a+\dfrac{1}{2}\, cos2a+\dfrac{3}{8}\ \ \ \Rightarrow \ \ \ 8cos^4a=cos4a+4cos2a+3\\\\\\cos4a+4cos2a+3=(2cos^22a-1)+4cos2a+3=2(cos^22a+2cos2a+1)=\\\\=2(1+cos2a)^2=2\cdot (2cos^2a)^2=2\cdot 4\, cos^4a=8cos^4a\\\\8\, cos^4a=8\, cos^4a$

$\star \ \ \ cos^2\alpha =\dfrac{1+cos2a}{2}\ \ \ \to \ \ \ \ 1+cos2a=2cos^2a\ \ \star\\\\\star \ \ \ cos2a=cos^2a-sin^2a=\underline {2cos^2a-1}=1-2sin^2a\ \ \star$

$43)\ \ sin^4\beta +cos^4\beta =(\underbrace{sin^2\beta +cos^2\beta }_{1})^2-2sin^2\beta \cdot cos^2\beta =1-2\cdot \Big(\dfrac{1}{2}sin2\beta \Big)^2=\\\\=1-\dfrac{1}{2}sin^2\beta =1-\dfrac{1}{2}\cdot (1-cos^2\beta )=\dfrac{1}{2}+\dfrac{1}{2}\, cos^2\beta =\dfrac{1}{2} \cdot (1+cos^2\beta )\ ;\\\\\\\dfrac{1}{2} \cdot (1+cos^2\beta )=\dfrac{1}{2} \cdot (1+cos^2\beta )\\\\\\\star \ \ \ sin^2\beta +cos^2\beta =1\ \ \to \ \ \ sin^2\beta =1-cos^2\beta \ \ \star$

$\star \ \ \ 2\, sin\beta \cdot cos\beta =sin2\beta \ \ \star$

$44)\ \ 4\cdot (sin^6a+cos^6a)=1+3cos^22a\\\\\\sin^6a+cos^6a=(sin^2a)^3+(cos^2a)^3=\\\\=(\underbrace{sin^2a+cos^2a}_{1})(sin^4a-sin^2a\cdot cos^2a+cos^4a)=\\\\=(sin^4a+cos^4a)-sin^2a\cdot cos^2a=\Big[\ \#43\ \Big]=\dfrac{1}{2}\ccdot (1+cos^22a)-(sina\cdot cosa)^2=\\\\=\dfrac{1}{2}+\dfrac{1}{2}\cdot cos^22a-\Big(\dfrac{1}{2}\cdot sin2a\Big)^2=\dfrac{1}{2}+\dfrac{1}{2}\cdot cos^22a-\dfrac{1}{4}\cdot sin^22a=$

$=\dfrac{1}{2}+\dfrac{1}{2}\cdot cos^22a-\dfrac{1}{4}\cdot (1-cos^22a)=\dfrac{1}{2}+\dfrac{1}{2}\cdot cos^22a-\dfrac{1}{4}+\dfrac{1}{4}\cdot cos^22a=\\\\=\dfrac{1}{4}+\dfrac{3}{4}\cdot cos^22a=\dfrac{1}{4}\cdot (1+3cos^22a)\ ;\\\\\\4\cdot (sin^6a+cos^6a)=4\cdot \dfrac{1}{4}\cdot (1+3cos^22a)=1+3\, cos^22a\ ;\\\\1+3\, cos^22a=1+3\, cos^22a$

$45)\ \ 8\cdot \Big(sin^8\dfrac{\beta }{2}+cos^8\dfrac{\beta }{2}\Big)=1+6cos^2\beta +cos^4\beta \\\\\\sin^8\dfrac{\beta }{2}+cos^8\dfrac{\beta }{2}=\Big(sin^4\dfrac{\beta }{2}\Big)^2+\Big(cos^4\dfrac{\beta }{2}\Big)^2=\\\\=\Big(sin^4\dfrac{\beta }{2}+cos^4\dfrac{\beta }{2}\Big)^2-2sin^4\dfrac{\beta }{2}\cdot cos^4\dfrac{\beta }{2}=\Big[\ \#43\ \Big]=\\\\=\Big(\dfrac{1}{2}\cdot (1+cos^2\beta )\Big)^2-2\cdot \Big(sin\dfrac{\beta }{2}\cdot cos\dfrac{\beta }{2}\Big)^4=$

$=\dfrac{1}{4}\cdot (1+2cos^2\beta +cos^4\beta )-2\cdot \Big(\dfrac{1}{2}\cdot sin\beta \Big)^4=\\\\=\dfrac{1}{4}+\dfrac{1}{2}\cdot cos^2\beta +\dfrac{1}{4}\cdot cos^4\beta -2\cdot \dfrac{1}{16}\cdot sin^4\beta =\\\\=\dfrac{1}{4}+\dfrac{1}{2}\cdot cos^2\beta +\dfrac{1}{4}\cdot cos^4\beta -\dfrac{1}{8}\cdot (sin^2\beta )^2=\\\\=\dfrac{1}{4}+\dfrac{1}{2}\cdot cos^2\beta +\dfrac{1}{4}\cdot cos^4\beta -\dfrac{1}{8}\cdot (1-cos^2\beta )^2=$

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

$8\cdot \Big(sin^8\dfrac{\beta }{2}+cos^8\dfrac{\beta }{2}\Big)=1+6\, cos^2\beta +cos^4\beta$

P.S.   Значок #43  отсылает к выкладкам в номере 43 .