Question

Sa se aducă la forma cea mai simpla expresia:
$e = \frac{ \sin(x + y) \cos(y ) - cos(x + y)sin(y) }{ \cos(y)cos(x + y) +sin(x + y) \sin(y) }$
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Answer 1

Răspuns:

$e= \frac{\frac{1}{2}*[sin(x+2y)+sinx]-\frac{1}{2}*[sin(x+2y)+sin(-x)]}{\frac{1}{2}*[cos(-x)+cos(x+2y)]+\frac{1}{2}*[cosx-cos(x+2y)]} \\\\e= \frac{sin(x+2y)+sinx-sin(x+2y)-(-sinx)}{cosx+cos(x+2y)+cosx-cos(x+2y)} \\e= \frac{2sinx}{2cosx} \\e= \frac{sinx}{cosx} \\e=tgx$

Explicație pas cu pas:

se folosesc formulele:

$sin a * cosb = \frac{1}{2} [sin(a+b)+sin(a-b)]\\cos a * cosb = \frac{1}{2} [cos(a-b)+cos(a+b)]\\sin a * sinb = \frac{1}{2} [cos(a-b)-cos(a+b)]$