Question

Докажите равенства !!!

Answer 1

Ответ:

В каждом равенстве преобразована левая часть.

Объяснение:

Answer 2

Ответ:

Покажем, что из левой части равенства получается правая часть.

1)

$\displaystyle \tt \; cos^2 \bigg (\frac{5 \cdot \pi }{8} -\frac{\alpha }{2} \bigg ) -sin^2 \bigg (\frac{3 \cdot \pi }{8} -\frac{\alpha }{2} \bigg ) = \\\\=\dfrac{1+cos\bigg (\dfrac{5 \cdot \pi }{8} -\dfrac{\alpha }{2} \bigg )}{2} -\dfrac{1-cos\bigg (\dfrac{3 \cdot \pi }{8} -\dfrac{\alpha }{2} \bigg )}{2}=$

$\displaystyle \tt \; =\dfrac{cos\bigg (\dfrac{5 \cdot \pi }{8} -\dfrac{\alpha }{2} \bigg )+cos\bigg (\dfrac{3 \cdot \pi }{8} -\dfrac{\alpha }{2} \bigg )}{2} =\\\\=\dfrac{2 \cdot cos\bigg (\dfrac{5 \cdot \pi }{8} -\dfrac{\alpha }{2} +\dfrac{3 \cdot \pi }{8} -\dfrac{\alpha }{2} \bigg ) \cdot cos\bigg (\dfrac{5 \cdot \pi }{8} -\dfrac{\alpha }{2} -\dfrac{3 \cdot \pi }{8} +\dfrac{\alpha }{2} \bigg )}{2} =$

$\displaystyle \tt =cos\bigg (\pi-\alpha \bigg ) \cdot cos\bigg (\dfrac{2 \cdot \pi }{8} \bigg ) = -cos \frac{\pi }{4} \cdot cos\alpha =-\frac{\sqrt{2} }{2} \cdot cos\alpha.$

2)

$\displaystyle \tt \; sin^2 \bigg (\frac{7 \cdot \pi }{12} +3 \cdot \alpha \bigg ) -cos^2 \bigg (\frac{ 5\cdot \pi }{12} +3 \cdot \alpha \bigg )= \\\\=\dfrac{1-cos\bigg (\dfrac{7 \cdot \pi }{12} +3 \cdot \alpha \bigg )}{2} -\dfrac{1+cos\bigg (\dfrac{5 \cdot \pi }{12} +3 \cdot \alpha \bigg )}{2}=$

$\displaystyle \tt \; =-\dfrac{cos\bigg (\dfrac{5 \cdot \pi }{12} + 3 \cdot \alpha \bigg )+cos\bigg (\dfrac{7 \cdot \pi }{12} + 3 \cdot \alpha \bigg )}{2} =\\\\=-\dfrac{2 \cdot cos\bigg (\dfrac{5 \cdot \pi }{12} + 3 \cdot \alpha +\dfrac{7 \cdot \pi }{12} + 3 \cdot \alpha \bigg ) \cdot cos\bigg (\dfrac{5 \cdot \pi }{12} + 3 \cdot \alpha -\dfrac{7 \cdot \pi }{12} - 3 \cdot \alpha \bigg )}{2} =$

$\displaystyle \tt =-cos\bigg (\pi +6 \cdot \alpha \bigg ) \cdot cos\bigg (-\dfrac{2 \cdot \pi }{12} \bigg ) = -cos \frac{\pi }{6} \cdot (-cos (6 \cdot \alpha)) =\frac{\sqrt{3} }{2} \cdot cos(6 \cdot \alpha).$

3)

$\displaystyle \tt \; tg20^0 + tg40^0 + tg50^0 + tg70^0=(tg20^0 + tg70^0) + (tg40^0 + tg50^0)=\\\\=\frac{sin(20^0 +70^0 )}{cos 20^0 \cdot cos 70^0 } + \frac{sin(40^0 +50^0 )}{cos 40^0 \cdot cos 50^0 } =\frac{sin(90^0 )}{cos 20^0 \cdot cos 70^0 } + \frac{sin(90^0)}{cos 40^0 \cdot cos 50^0 } =$

$\displaystyle \tt \; =\frac{1}{cos 20^0 \cdot cos 70^0 } + \frac{1}{cos 40^0 \cdot cos 50^0 } =\frac{1}{\dfrac{cos 50^0 + cos 90^0}{2} } + \frac{1}{\dfrac{cos 10^0 + cos 90^0 }{2} } =$

$\displaystyle \tt \; =\frac{2}{cos 50^0} + \frac{2}{cos 10^0} =\frac{2 \cdot (cos10^0+cos50^0)}{cos 50^0 \cdot cos10^0}=\frac{2 \cdot (2 \cdot cos30^0 \cdot cos20^0)}{cos (90^0-40^0 )\cdot cos10^0}=\\\\=\frac{2 \cdot (2 \cdot \dfrac{\sqrt{3} }{2} \cdot cos20^0)}{sin 40^0 \cdot cos10^0}=\frac{2 \cdot \sqrt{3} \cdot cos20^0}{2 \cdot sin20^0 \cdot cos20^0\cdot cos10^0}=\frac{\sqrt{3} }{ sin20^0 \cdot cos10^0}=$

$\displaystyle \tt \; =\frac{ \sqrt{3} }{ \dfrac{sin(20^0-10^0) +sin(20^0+10^0)}{2} }=\frac{ 2 \cdot \sqrt{3} }{ sin10^0 +sin30^0}=\\\\=\frac{ 2 \cdot \sqrt{3} }{ sin10^0+\dfrac{1}{2} }= \frac{ 4 \cdot \sqrt{3} }{ 1+2 \cdot sin10^0 }.$