1)cos425°; 2)cos2alpha,sin alpha=4/5; 3)cos2x×cos - вопрос №142522453223129209 от MN0G0ZNAAL 14.09.2020 21:16

Question

Алгебра: 1)cos425°; 2)cos2alpha,sin alpha=4/5; 3)cos2x×cos x=sin2x×sin x

MN0G0ZNAAL · Answer

1)cos425⁰=cos(360⁰+65⁰)=cos(2π+65⁰)=cos65⁰;
2)cos2α=cos²α-sin²α=1-2sin²α=1-2(4/5)²=1-2·16/25=1-32/25=-7/25;
3)cos2x·cosx=sin2x·sinx ⇒cos2x·cosx-sin2x·sinx=0 ⇒
cos(2x+x)=0;⇒cos3x=0 ⇒
3x=π/2+kπ;k∈Z
x=π/6+kπ/3;k∈Z