Question

помогите решить
cos^2(a+2b)+sin^2(a-b)-1

Answer 1

$\cos^2{ ( a + 2b ) } + \sin^2{ ( a - b ) } - 1 = \cos^2{ ( a + 2b ) } - ( 1 - \sin^2{ ( a - b ) } ) = \\\\ = \cos^2{ ( a + 2b ) } - \cos^2{ ( a - b ) } = \\\\ = ( \cos{ ( a + 2b ) } - \cos{ ( a - b ) } ) ( \cos{ ( a + 2b ) } + \cos{ ( a - b ) } ) = \\\\ = 2 \sin{ \frac{ [ a - b ] - [ a + 2b ] }{2} } \sin{ \frac{ [ a - b ] + [ a + 2b ] }{2} } \cdot 2 \cos{ \frac{ [ a + 2b ] + [ a - b ] }{2} } \cos{ \frac{ [ a + 2b ] - [ a - b ] }{2} } = \\\\ = 2 \sin{ \frac{ - b - 2b }{2} } \sin{ \frac{ a - b + a + 2b }{2} } \cdot 2 \cos{ \frac{ a + 2b + a - b }{2} } \cos{ \frac{ 2b + b }{2} } = \\\\ = 2 \sin{ \frac{-3b}{2} } \sin{ \frac{ 2a + b }{2} } \cdot 2 \cos{ \frac{ 2a + b }{2} } \cos{ \frac{3b}{2} } = \\\\ = -2 \sin{ \frac{3b}{2} } \cos{ \frac{3b}{2} } \cdot 2 \sin{ \frac{ 2a + b }{2} } \cos{ \frac{ 2a + b }{2} } = -\sin{ ( 2 \cdot \frac{3b}{2} ) } \sin{ ( 2 \cdot \frac{ 2a + b }{2} ) } \ ;$



О т в е т :    $\cos^2{ ( a + 2b ) } + \sin^2{ ( a - b ) } - 1 = -\sin{3b} \cdot \sin{ ( 2a + b ) } \ .$

Answer 2

cos²x=(1+cos2x)/2;sin²x=(1-cos2x)/2;cosx-cosy=-2sin(x-y)/2*sin(x+y)/2. 
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cos²(α+2β) +sin²(α -β) -1= (1+cos2(α+2β)) /2 +(1-cos2(α -β)) /2 -1 =
(cos2(α+2β)-cos2(α -β))/2 = -sin3β*sin(2α+β).