Demonstrati identitatea :
sin x + sin 3x + sin 5x + sin 7x = 4 × cos x × cos 2x × sin 4x
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Salut,
![sina+sinb=2\cdot sin\left(\dfrac{a+b}{2}\right)\cdot cos\left(\dfrac{a-b}{2}\right).\\\\cosa+cosb=2\cdot cos\left(\dfrac{a+b}{2}\right)\cdot cos\left(\dfrac{a-b}{2}\right).\\\\sin(7x)+sinx+sin(5x)+sin(3x)=\\\\=2\cdot sin\left(\dfrac{7x+x}{2}\right)\cdot cos\left(\dfrac{7x-x}{2}\right)+2\cdot sin\left(\dfrac{5x+3x}{2}\right)\cdot cos\left(\dfrac{5x-3x}{2}\right)=\\\\=2\cdot sin(4x)\cdot cos(3x)+2\cdot sin(4x)\cdot cosx=2\cdot sin(4x)[cos(3x)+cosx]=\\\\=2\cdot sin(4x)\cdot 2\cdot cos\left(\dfrac{3x+x}{2}\right)\cdot cos\left(\dfrac{3x-x}{2}\right)=\\\\=4\cdot sin(4x)\cdot cos(2x)\cdot cosx,\ ceea\ ce\ trebuia\ demonstrat.](https://tex.z-dn.net/?f=sina%2Bsinb%3D2%5Ccdot+sin%5Cleft%28%5Cdfrac%7Ba%2Bb%7D%7B2%7D%5Cright%29%5Ccdot+cos%5Cleft%28%5Cdfrac%7Ba-b%7D%7B2%7D%5Cright%29.%5C%5C%5C%5Ccosa%2Bcosb%3D2%5Ccdot+cos%5Cleft%28%5Cdfrac%7Ba%2Bb%7D%7B2%7D%5Cright%29%5Ccdot+cos%5Cleft%28%5Cdfrac%7Ba-b%7D%7B2%7D%5Cright%29.%5C%5C%5C%5Csin%287x%29%2Bsinx%2Bsin%285x%29%2Bsin%283x%29%3D%5C%5C%5C%5C%3D2%5Ccdot+sin%5Cleft%28%5Cdfrac%7B7x%2Bx%7D%7B2%7D%5Cright%29%5Ccdot+cos%5Cleft%28%5Cdfrac%7B7x-x%7D%7B2%7D%5Cright%29%2B2%5Ccdot+sin%5Cleft%28%5Cdfrac%7B5x%2B3x%7D%7B2%7D%5Cright%29%5Ccdot+cos%5Cleft%28%5Cdfrac%7B5x-3x%7D%7B2%7D%5Cright%29%3D%5C%5C%5C%5C%3D2%5Ccdot+sin%284x%29%5Ccdot+cos%283x%29%2B2%5Ccdot+sin%284x%29%5Ccdot+cosx%3D2%5Ccdot+sin%284x%29%5Bcos%283x%29%2Bcosx%5D%3D%5C%5C%5C%5C%3D2%5Ccdot+sin%284x%29%5Ccdot+2%5Ccdot+cos%5Cleft%28%5Cdfrac%7B3x%2Bx%7D%7B2%7D%5Cright%29%5Ccdot+cos%5Cleft%28%5Cdfrac%7B3x-x%7D%7B2%7D%5Cright%29%3D%5C%5C%5C%5C%3D4%5Ccdot+sin%284x%29%5Ccdot+cos%282x%29%5Ccdot+cosx%2C%5C+ceea%5C+ce%5C+trebuia%5C+demonstrat. "sina+sinb=2\cdot sin\left(\dfrac{a+b}{2}\right)\cdot cos\left(\dfrac{a-b}{2}\right).\\\\cosa+cosb=2\cdot cos\left(\dfrac{a+b}{2}\right)\cdot cos\left(\dfrac{a-b}{2}\right).\\\\sin(7x)+sinx+sin(5x)+sin(3x)=\\\\=2\cdot sin\left(\dfrac{7x+x}{2}\right)\cdot cos\left(\dfrac{7x-x}{2}\right)+2\cdot sin\left(\dfrac{5x+3x}{2}\right)\cdot cos\left(\dfrac{5x-3x}{2}\right)=\\\\=2\cdot sin(4x)\cdot cos(3x)+2\cdot sin(4x)\cdot cosx=2\cdot sin(4x)[cos(3x)+cosx]=\\\\=2\cdot sin(4x)\cdot 2\cdot cos\left(\dfrac{3x+x}{2}\right)\cdot cos\left(\dfrac{3x-x}{2}\right)=\\\\=4\cdot sin(4x)\cdot cos(2x)\cdot cosx,\ ceea\ ce\ trebuia\ demonstrat.")
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