△ABC中,
0 = cos(2B+C) + 2sinAsinB
= cos(B + B + C) + 2sinAsinB
= cos(B + PI - A) + 2sinAsinB
= -cos(B - A) + 2sinAsinB
= -[cosBcosA + sinAsinB] + 2sinAsinB
= -[cosBcosA - sinAsinB]
= -cos(A + B)
= -cos(PI -C)
= cosC
因,0 < C < PI,
所以, C = PI/2.
△ABC一定是直角三角形。
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