Question

Решите и 10
желательно с решением.
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Answer 1

$1)\ \ tga=\dfrac{\sqrt5}{5}\\\\2tg^22a=2\cdot \Big(\dfrac{2tga}{1-tg^2a}\Big)^2=2\cdot \dfrac{4tg^2a}{(1-tg^2a)^2}=2\cdot \dfrac{4\cdot \frac{1}{5}}{(1-\frac{1}{5})^2}=\dfrac{8\cdot 5^2}{5\cdot 4^2}=\dfrac{5}{2}=2,5$

$2)\ \ sin10^\circ +sin50^\circ =2\cdot sin\dfrac{10^\circ +50^\circ }{2}\cdot cos\dfrac{10^\circ -50^\circ }{2}=\\\\=2\cdot sin30^\circ \cdot cos(-20^\circ )=2\cdot \dfrac{1}{2}\cdot cos20^\circ =cos20^\circ \\\\1\ -\ d\\\\\\sin70^\circ -sin10^\circ =2\cdot \dfrac{70^\circ-10^\circ }{2}\cdot cos\dfrac{70^\circ +10^\circ }{2}=2\cdot sin30^\circ \cdot cos40^\circ =\\\\=2\cdot \dfrac{1}{2}\cdot cos40^\circ =cos40^\circ \\\\2\ -\ a$

$cos80^\circ -cos20^\circ =-2\cdot sin\dfrac{80^\circ +20^\circ }{2}\cdot sin\dfrac{80^\circ -20^\circ }{2}=\\\\=-2\cdot sin50^\circ \cdot sin30^\circ =-2\cdot sin50^\circ \cdot \dfrac{1}{2}=-sin50^\circ \\\\3\ -\ b$

Answer 2

Объяснение:

9) tga=V5/ 5,  (V-корень),  по формуле   tg2a=2tga/(1-tg^2 a)=

2*(V5/5) /(1-1/5)=2V5/ 5  :4/5=2V5 /4=V5/ 2,  tg2a=V5/ 2,

2tg^2 2a=2*(V5/ 2)^2=2*5/4=5/2=2,5

10.  1) sin10+sin50=2sin(60/2)*cos(-20)=2*sin30*cos20=2*1/2*

cos20=cos20

    2) sin70-sin10=2sin(60/2)*cos(80/2)=2*1/2*cos40=cos40

   3)  cos80-cos20=-2sin(100/2)*sin(60/2)=-2sin50*sin30=-2sin50*1/2=

-sin50