1) sin5a × cos2a 2) sin8a × cos12a 3) cos5a × cos7a - вопрос №15228123574615561411948168 от Samvel7777 24.11.2020 02:06

Question

Алгебра: 1) sin5a × cos2a 2) sin8a × cos12a 3) cos5a × cos7a 4) cos6a × cos(-15a) 5)sin6a × sin14a 6)sin3a ×sin(-21a)​

Samvel7777 · Answer

1)= 1/2(sin(5a+2a)+sin(5a-2a))=sin(7a)/2 + sin(3a)/2

2)=1/2(sin(8a+12a)+sin(8a-12a))=sin(20a)/2 + sin(-4a)/2

3)=1/2(cos(5a-7a)+cos(5a+7a))=cos(-2a)/2 + cos(12a)/2

4)=1/2 (cos(6a-(-15a))+cos(6a+(-15a))=cos(21a)/2 + cos(-9)/2

5)=1/2(cos(6a-14a)-cos(6a+14a))=cos(8a)/2 - cos(20a)/2

6)=1/2(cos(3a-(-21a))-cos(3a+(-21a))=cos(24)/2 - cos(-18)/2

Объяснение:

1) sin5a × cos2a 2) sin8a × cos12a 3) cos5a × cos7a 4) cos6a × cos(-15a) 5)sin6a × sin14a 6)sin3a ×sin(-21a)​