Question

Помогите умоляюююю не хочу 2​

Answer 1

Объяснение:

$2)\ cos^215^0-sin^215^0=cos(2*15^0)=cos30^0=\frac{\sqrt{3} }{2}.\\ 4)\ (cos15^0+sin15^0)^2=cos^215^0+2*sin15^0*cos15^0+sin^215^0=\\=1+sin(2*15^0)=1+sin30^0=1+0,5=1,5.$

$2)\ cos^2\frac{\pi }{8} -sin^2\frac{\pi }{8} =cos(2*\frac{\pi }{8})=cos\frac{\pi }{4}=\frac{\sqrt{2} }{2}.\\ 4)\ \frac{\sqrt{2} }{2} -(cos\frac{\pi }{8}+sin\frac{\pi }{8})^2=\frac{\sqrt{2} }{2}-(cos^2\frac{\pi }{8} +2*sin\frac{\pi }{8}*cos\frac{\pi }{8} +sin^2\frac{\pi }{8} )=\\ =\frac{\sqrt{2} }{2}-(1+sin(2*\frac{\pi }{8})) =\frac{\sqrt{2} }{2}-(1+sin\frac{\pi }{4})=\frac{\sqrt{2} }{2} -1-\frac{\sqrt{2} }{2} =-1.$

$2)\ cos\alpha =-\frac{4}{5} \ \ \ \ \ \frac{\pi }{2}<\alpha <\pi \ \ \ \ sin2\alpha =?\\ sin^2\alpha +cos^2\alpha =1\\sin^2\alpha =1-cos^2\alpha =1-(-\frac{4}{5})^2=1-\frac{16}{25} =\frac{9}{25}.\\ sin\alpha =б\sqrt{\frac{9}{25} }= б\frac{3}{5} \ \ \ \ \frac{\pi }{2}<\alpha <\pi \ \ \ \ \Rightarrow\\sin\alpha =\frac{3}{5} .\\sin2\alpha =2*sin\alpha *cos\alpha =2*\frac{3}{5}*(-\frac{4}{5})=-\frac{24}{25}.$

Ответ: sin2α=-24/25.