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виталик0128
@виталик0128
August 2021
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sin (pi/6-x)-cos(pi/3+x)
Как решить ?
по какой формуле ... ?
спасибо
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sedinalana
Verified answer
Sinπ/6*cosx-cosπ/6*sinx-cosπ/3*cosx+sinπ/3*sinx=
=1/2*cosx-√3/2*sinx-1/2*cosx+√3/2*sinx=0
0 votes
Thanks 2
oganesbagoyan
Verified answer
Или
sin(π/6 -x) - cos(π/3 +x) =sin( π/2 - ( π/3 +x) )
- cos(π/3 +x) =
cos(π/3 +x) -
cos(π/3 +x) =0.
------
sin( π/2 - α) =cosα
.
1 votes
Thanks 0
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Answers & Comments
Verified answer
Sinπ/6*cosx-cosπ/6*sinx-cosπ/3*cosx+sinπ/3*sinx==1/2*cosx-√3/2*sinx-1/2*cosx+√3/2*sinx=0
Verified answer
Илиsin(π/6 -x) - cos(π/3 +x) =sin( π/2 - ( π/3 +x) ) - cos(π/3 +x) =
cos(π/3 +x) - cos(π/3 +x) =0.
------
sin( π/2 - α) =cosα .