Task/25060814
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sinx +cosx =√2 ; || : √2
(1/√2)*sinx +(1/√2)*cosx = 1 ;
cos(π/4)*sinx +sin(π/4)*cosx =1 ; * * * sin(π/4)*sinx +cos(π/4)*cosx =1 * * *
sin(x+π/4) =1 ; * * * cos(x -π/4) =1 * * *
x+π/4 =π/2+2π*n ,n∈Z * * * x -π/4 =2π*n ,n∈Z * * *
x =π/4+2π*n , n∈Z . * * * x =π/4 +2π*n ,n∈Z * * *
ответ : x =π/4 +2π*n ,n∈Z .
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формула дополнительного(вспомогательного) угла :
a*sinx +b*cosx =√(a²+b²) sin(x +arctg(b/a)) .