Question

помогите пожалуйста с алгеброй 78 баллов!​

Answer 1

$1)\ \ sina=-\dfrac{1}{5}\ \ ,\\\\\dfrac{3\pi}{2}<a<2\pi \ \Rightarrow \ \ cosa>0\ \ \Rightarrow \ \ cosa=+\sqrt{1-sin^2a}\\\\cosa=\sqrt{1-\dfrac{1}{25}}=\sqrt{\dfrac{24}{25}}=\dfrac{\sqrt{24}}{5}=\dfrac{2\sqrt6}{5}\\\\\\\dfrac{3\pi}{4}<\dfrac{a}{2}<\pi \ \Rightarrow \ sin\dfrac{a}{2}>0\ ,\ cos\dfrac{a}{2}<0\ ,\ tg\dfrac{a}{2}<0\ ,\ ctg\dfrac{a}{2}<0\\\\\\sin^2\dfrac{a}{2}=\dfrac{1-cosa}{2}=\dfrac{1-\frac{\sqrt{24}}{5}}{2}=\dfrac{5-2\sqrt6}{10}=\dfrac{1}2}-\dfrac{\sqrt6}{5}=0,5-0,2\cdot \sqrt6$

$sin\dfrac{a}{2}=+\sqrt{\dfrac{5-2\sqrt6}{10}}\\\\\\cos^2\dfrac{a}{2}=\dfrac{1+cosa}{2}=\dfrac{1+\frac{\sqrt{24}}{5}}{2}=\dfrac{5+2\sqrt6}{10}\ \ \Rightarrow \ \ \ cos\dfrac{a}{2}=-\sqrt{\dfrac{5+2\sqrt6}{10}}\\\\\\tg\dfrac{a}{2}=\dfrac{sin\dfrac{a}{2}}{cos\dfrac{a}{2}}=-\sqrt{\dfrac{5-2\sqrt6}{5+2\sqrt6}}=-\sqrt{\dfrac{(5-2\sqrt6)^2}{(5-2\sqrt6)(5+2\sqrt6)}}=-\sqrt{\dfrac{49-20\sqrt6}{25-24}}=\\\\\\=-\sqrt{49-20\sqrt6}$

$ctg\dfrac{a}{2}=\dfrac{1}{tg\dfrac{a}{2}}=-\sqrt{\dfrac{5+2\sqrt6}{5-2\sqrt6}}=-\sqrt{\dfrac{(5+2\sqrt6)^2}{(5-2\sqrt6)(5+2\sqrt6)}}=-\sqrt{\dfrac{49+20\sqrt6}{25-24}}=\\\\\\=-\sqrt{49+20\sqrt6}$

$2)\ \ \ \ \boxed {\ \ sin^2\alpha =\dfrac{1-cos2\alpha }{2}\ \ \ \to \ \ \ 2sin^2\alpha =1-cos2\alpha \ \ }\\\\\\{}\ \ \ \ \ \ \boxed{\ \ cos^2\alpha =\dfrac{1+cos2\alpha }{2}\ \ \ \to \ \ \ 2cos^2\alpha =1+cos2\alpha\ \ }\\\\\\2sin^2\Big(\dfrac{\pi}{4}-\dfrac{x}{2}\Big)=1-cos\Big(\dfrac{\pi}{2}-x\Big)=1-sinx\\\\\\4cos^2(\pi +x)=4\cdot (-cosx)^2=4\, cos^2x\\\\\\tg^2\Big(\dfrac{3\pi}{4}-\dfrac{3x}{2}\Big)=\dfrac{1-cos\Big(\dfrac{3\pi}{2}-3x\Big)}{1+cos\Big(\dfrac{3\pi}{2}+x\Big)}=\dfrac{1+sin3x}{1-sin3x}$