Ответы и объяснения

2014-02-27T13:31:28+04:00
2sin^22 \alpha +2cos( \frac{ \pi }{2} - \alpha )+2cos^22 \alpha =2(sin^22 \alpha +cos^22 \alpha )+2sin \alpha = \\  \\ =2+2sin \alpha =2+2sin \frac{ \pi }{6} =2+2* \frac{1}{2} =3

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 \frac{cos^262+cos^228}{4} - \frac{1}{2} cos60= \frac{cos^2(90-28)+cos^228}{4}- \frac{1}{2}  * \frac{1}{2} = \\  \\ = \frac{sin^228+cos^228}{4} - \frac{1}{4} = \frac{1}{4} - \frac{1}{4} =0

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 \frac{6 \sqrt{3}sin15cos15 }{2cos^215-1} = \frac{3 \sqrt{3}sin30 }{cos^215+cos^215-1}= \frac{3 \sqrt{3} \frac{ 1}{2}  }{cos^215-sin^215}  = \frac{3 \sqrt{3} }{2cos^230} = \frac{3 \sqrt{3} }{2 \frac{ \sqrt{3} }{2} } =3