已知sinθ=asinΦ;tanθ=btanΦ;θ为锐角,求证cosθ=((a^2-1)/(b^2-1))^(1/2)我的解法是sin^2θ+cos^2θ=1a^2sin^2Φ+a^2/b^2cos^2Φ=1又 sin^2Φ+cos^2Φ=1所以 (a^2-1)sin^2Φ+(a^2-b^2)/b^2cos^2Φ=0得到了a^2=b^2=1,与题目矛盾了,这是肿
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![已知sinθ=asinΦ;tanθ=btanΦ;θ为锐角,求证cosθ=((a^2-1)/(b^2-1))^(1/2)我的解法是sin^2θ+cos^2θ=1a^2sin^2Φ+a^2/b^2cos^2Φ=1又 sin^2Φ+cos^2Φ=1所以 (a^2-1)sin^2Φ+(a^2-b^2)/b^2cos^2Φ=0得到了a^2=b^2=1,与题目矛盾了,这是肿](/uploads/image/z/3710798-62-8.jpg?t=%E5%B7%B2%E7%9F%A5sin%CE%B8%3Dasin%CE%A6%3Btan%CE%B8%3Dbtan%CE%A6%3B%CE%B8%E4%B8%BA%E9%94%90%E8%A7%92%2C%E6%B1%82%E8%AF%81cos%CE%B8%3D%28%28a%5E2-1%29%2F%28b%5E2-1%29%29%5E%281%2F2%29%E6%88%91%E7%9A%84%E8%A7%A3%E6%B3%95%E6%98%AFsin%5E2%CE%B8%2Bcos%5E2%CE%B8%3D1a%5E2sin%5E2%CE%A6%2Ba%5E2%2Fb%5E2cos%5E2%CE%A6%3D1%E5%8F%88+sin%5E2%CE%A6%2Bcos%5E2%CE%A6%3D1%E6%89%80%E4%BB%A5+%28a%5E2-1%29sin%5E2%CE%A6%2B%28a%5E2-b%5E2%29%2Fb%5E2cos%5E2%CE%A6%3D0%E5%BE%97%E5%88%B0%E4%BA%86a%5E2%3Db%5E2%3D1%2C%E4%B8%8E%E9%A2%98%E7%9B%AE%E7%9F%9B%E7%9B%BE%E4%BA%86%2C%E8%BF%99%E6%98%AF%E8%82%BF)
已知sinθ=asinΦ;tanθ=btanΦ;θ为锐角,求证cosθ=((a^2-1)/(b^2-1))^(1/2)我的解法是sin^2θ+cos^2θ=1a^2sin^2Φ+a^2/b^2cos^2Φ=1又 sin^2Φ+cos^2Φ=1所以 (a^2-1)sin^2Φ+(a^2-b^2)/b^2cos^2Φ=0得到了a^2=b^2=1,与题目矛盾了,这是肿
已知sinθ=asinΦ;tanθ=btanΦ;θ为锐角,求证cosθ=((a^2-1)/(b^2-1))^(1/2)
我的解法是
sin^2θ+cos^2θ=1
a^2sin^2Φ+a^2/b^2cos^2Φ=1
又 sin^2Φ+cos^2Φ=1
所以 (a^2-1)sin^2Φ+(a^2-b^2)/b^2cos^2Φ=0
得到了a^2=b^2=1,与题目矛盾了,这是肿么回事?
已明白问题所在。提问关闭。
已知sinθ=asinΦ;tanθ=btanΦ;θ为锐角,求证cosθ=((a^2-1)/(b^2-1))^(1/2)我的解法是sin^2θ+cos^2θ=1a^2sin^2Φ+a^2/b^2cos^2Φ=1又 sin^2Φ+cos^2Φ=1所以 (a^2-1)sin^2Φ+(a^2-b^2)/b^2cos^2Φ=0得到了a^2=b^2=1,与题目矛盾了,这是肿
sinθ=asinΦ 平方
sin^2θ=a^2sin^2Φ
sin^2θ=a^2(1-cos^2Φ)
sin^2θ=a^2-a^2cos^2Φ
1-cos^2θ=a^2-a^2cos^2Φ
cos^2θ=1-a^2+a^2cos^2Φ.1
tanθ=sinθ/cosθ=btanΦ=bsinΦ/cosΦ
sinθ=asinΦ
sinθ/cosθ=asinΦ/cosθ
tanθ=asinΦ/cosθ
asinΦ/cosθ=bsinΦ/cosΦ
cosΦ=bcosθ/a.2
将2式代入1式得
cos^2θ=1-a^2+a^2(b^2cos^2θ/a^2)
cos^2θ=1-a^2+b^2cos^2θ
(1-b^2)cos^2θ=1-a^2
cos^2θ=(1-a^2)/(1-b^2)=(a^2-1)/(b^2-1)
因为θ为锐角
所以cosθ=√[(a^2-1)/(b^2-1)]