Question

Sin80/cos40
АЛГЕБРА СПРОСТІТЬ ВИРАЗ
13.10 с объяснением

Answer 1

ответ:в файле

Объяснение:

Answer 2

Задание 13.10

1) $\frac{sin80}{cos40} = \frac{2sin40cos40}{cos40} = 2sin40$

2) $cos4\beta +sin2\beta *2=cos4\beta +2sin2\beta$

3) $cos6\alpha +2sin3\alpha *2=cos6\alpha +4sin3\alpha$

4) $\frac{1+sin2\alpha }{(sin\alpha +cos\alpha )^2} =\frac{1+sin2\alpha }{sin\alpha ^2+2sin\alpha cos\alpha +cos\alpha ^2} = \frac{1+sin2\alpha }{1+sin2\alpha } = 1$

5) $sin\alpha cos\alpha *(cos\alpha ^2-sina^2)=sin\alpha cos\alpha cos2\alpha$

6) $\frac{sin4\alpha }{cos\alpha ^4-sin\alpha ^4} = \frac{2sin2\alpha cos2\alpha }{(cos\alpha ^2-sin\alpha ^2)*(cos\alpha ^2+sin\alpha ^2)} = \frac{2sin2\alpha cos2\alpha }{cos2\alpha *1} =\frac{2sin2\alpha cos2\alpha }{cos2\alpha } = 2sin2\alpha$

7) $sin(\frac{\pi }{4} -\alpha )cos(\frac{\pi }{4} -\alpha )=(sin(\frac{\pi }{4} )cos\alpha -cos(\frac{\pi }{4} )sin\alpha )*(cos(\frac{\pi }{4} )cos\alpha +sib(\frac{\pi }{4} )sin\alpha )= (\frac{\sqrt{2} }{2} *cos\alpha -\frac{\sqrt{2} }{2} * sin\alpha )*(\frac{\sqrt{2} }{2} *cos\alpha *\frac{\sqrt{2} }{2} *sin\alpha )=(\frac{\sqrt{2} *cos\alpha }{2} - \frac{\sqrt{2} *sin\alpha }{2} )*(\frac{\sqrt{2}*cos\alpha }{2} + \frac{\sqrt{2} *sin\alpha }{2} )=\frac{\sqrt{2} *cos\alpha -\sqrt{2}*sin\alpha }{2} *$

$* \frac{\sqrt{2}*cos\alpha +\sqrt{2}*sin\alpha }{2} =\frac{(\sqrt{2}cos\alpha -\sqrt{2} *sin\alpha )*(\sqrt{2}*cos\alpha +\sqrt{2} *sin\alpha ) }{4} =\frac{2cos\alpha ^2-2sin\alpha ^2}{4} = \frac{2(cos\alpha ^2-sin\alpha ^2)}{4} = \frac{cos2\alpha }{2}$