Ответ:
a) -2
б) 1/2
Объяснение:
a)
(3cos(2π-B)-sin(π/2+B))/2cos(B-π)
(3cos(B)-cos(B))/-2cos(B)
2cos(B)/-2cos(B)= -2
б)
(2sin(a-2π)-3cos(a-π/2))/2sin(a-3π)
(2sin(a)-3sin(a))/-2sin(a)
-sin(a)/-2sin(a)= -1/-2=1/2
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Verified answer
Ответ:
a) -2
б) 1/2
Объяснение:
a)
(3cos(2π-B)-sin(π/2+B))/2cos(B-π)
(3cos(B)-cos(B))/-2cos(B)
2cos(B)/-2cos(B)= -2
б)
(2sin(a-2π)-3cos(a-π/2))/2sin(a-3π)
(2sin(a)-3sin(a))/-2sin(a)
-sin(a)/-2sin(a)= -1/-2=1/2