求值[2sin50°+sin10°(1+√3tan10°)]*√2(sin80°)*(sin80°)
2个回答
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注:题目似乎多乘了一个sin80°!!!!
解:[2sin50°+sin10°(1+√3tan10°)]*√2(sin80°)
=[2sin50°+sin10°(1+√3tan10°)]*√2(cos10°)
=√2*[2sin50°cos10°+sin10°(cos10°+√3sin10°)]
=√2*[2sin50°cos10°+2sin10°(1/2 *cos10°+√3/2 *sin10°)]
=√2*(2sin50°cos10°+2sin10°cos50°)
=2√2*sin60°
=√6
当然如果题目中是(sin80°)*(sin80°),最后结果为√6cos10° (或√6sin80°)
解:[2sin50°+sin10°(1+√3tan10°)]*√2(sin80°)
=[2sin50°+sin10°(1+√3tan10°)]*√2(cos10°)
=√2*[2sin50°cos10°+sin10°(cos10°+√3sin10°)]
=√2*[2sin50°cos10°+2sin10°(1/2 *cos10°+√3/2 *sin10°)]
=√2*(2sin50°cos10°+2sin10°cos50°)
=2√2*sin60°
=√6
当然如果题目中是(sin80°)*(sin80°),最后结果为√6cos10° (或√6sin80°)
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【2sin50+sin10(1+√3tan10)】*√(2(sin80)^2)
=【2sin50+sin10(1+√3Sin10/Cos10)】*√(2(sin80)^2)
=【2sin50+sin10(Cos10+√3Sin10)/Cos10)】*√(2(sin80)^2)
=【2sin50+2sin10(sin30Cos10+cos30Sin10)/Cos10)】*√(2(sin80)^2)
=【2sin50+2sin10sin40/Cos10)】*√(2(sin80)^2)
=2【(sin50Cos10+sin10sin40)/Cos10】*√(2(sin80)^2)
=2【sin60/Cos10】*√(2(sin80)^2)
=2sin60/Cos10*√2*cos10
=2sin60*√2
=根号6
=【2sin50+sin10(1+√3Sin10/Cos10)】*√(2(sin80)^2)
=【2sin50+sin10(Cos10+√3Sin10)/Cos10)】*√(2(sin80)^2)
=【2sin50+2sin10(sin30Cos10+cos30Sin10)/Cos10)】*√(2(sin80)^2)
=【2sin50+2sin10sin40/Cos10)】*√(2(sin80)^2)
=2【(sin50Cos10+sin10sin40)/Cos10】*√(2(sin80)^2)
=2【sin60/Cos10】*√(2(sin80)^2)
=2sin60/Cos10*√2*cos10
=2sin60*√2
=根号6
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