高中三角函数,在锐角三角形ABC中,角A,B,C的对边分别为a,b,c,已知sinA=(2乘根号2)/3.[1]求tan^2[(B+C)/2]+sin^2(A/2)的值,[2]若a=2,S三角形ABC=根号2,求b的值
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![高中三角函数,在锐角三角形ABC中,角A,B,C的对边分别为a,b,c,已知sinA=(2乘根号2)/3.[1]求tan^2[(B+C)/2]+sin^2(A/2)的值,[2]若a=2,S三角形ABC=根号2,求b的值](/uploads/image/z/6857110-46-0.jpg?t=%E9%AB%98%E4%B8%AD%E4%B8%89%E8%A7%92%E5%87%BD%E6%95%B0%2C%E5%9C%A8%E9%94%90%E8%A7%92%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E4%B8%AD%2C%E8%A7%92A%2CB%2CC%E7%9A%84%E5%AF%B9%E8%BE%B9%E5%88%86%E5%88%AB%E4%B8%BAa%2Cb%2Cc%2C%E5%B7%B2%E7%9F%A5sinA%3D%282%E4%B9%98%E6%A0%B9%E5%8F%B72%29%2F3.%5B1%5D%E6%B1%82tan%5E2%5B%28B%2BC%29%2F2%5D%2Bsin%5E2%28A%2F2%29%E7%9A%84%E5%80%BC%2C%5B2%5D%E8%8B%A5a%3D2%2CS%E4%B8%89%E8%A7%92%E5%BD%A2ABC%3D%E6%A0%B9%E5%8F%B72%2C%E6%B1%82b%E7%9A%84%E5%80%BC)
高中三角函数,在锐角三角形ABC中,角A,B,C的对边分别为a,b,c,已知sinA=(2乘根号2)/3.[1]求tan^2[(B+C)/2]+sin^2(A/2)的值,[2]若a=2,S三角形ABC=根号2,求b的值
高中三角函数,在锐角三角形ABC中,角A,B,C的对边分别为a,b,c,已知sinA=(2乘根号2)/3.[1]求tan^2[(B+C)/2]+sin^2(A/2)的值,[2]若a=2,S三角形ABC=根号2,求b的值
高中三角函数,在锐角三角形ABC中,角A,B,C的对边分别为a,b,c,已知sinA=(2乘根号2)/3.[1]求tan^2[(B+C)/2]+sin^2(A/2)的值,[2]若a=2,S三角形ABC=根号2,求b的值
解:sinA=2√2/3,因为是锐角三角形,所以cosA=1/3
tan^2[(B+C)/2]+sin^2 (A/2)=tan^2(π-A)/2+sin^2(A/2)
=cot^2(A/2)+sin^2(A/2)=(cos^2(A/2)/sin^2(A/2)+sin^2(A/2)
=[cos^2(A/2)+sin^2(A/2)*sin^2(A/2)]/sin^2(A/2)
=[cos^2(A/2)+(1-cos^2(A/2)*sin^2(A/2)]/sin^2(A/2)
=[cos^2(A/2)+sin^2(A/2)-cos^2(A/2)*sin^2(A/2)]/sin^2(A/2)
=[1-(1/4)sin^2A]/[(1-cosA)/2]
=[1-(1/4)2√2/3]/[(1-1/3)/2]
=(6-√2)/2
(2)
S=(1/2)bcsinA=(1/2)bc*2√2/3=√2
所以bc=3
根据余弦定理
a^2-b^2-c^2+2bccosA=0,即4-b^2-c^2+2bc*1/3=0
解得:b=c=√3