Question

化简(2sin50+sin10(1+根号3tan10))根号(2sin^280)

Answer 1

解：
原式=[ 2sin50°+sin10°(1＋√3·tan10°)]√2cos10°
=2√2sin50°cos10°+√2 sin10°cos10°+√6 sin^210°
=2√2sin50°cos10°+√2/2 sin20°+√6/2(1-cos20°)
=2√2sin50°cos10°+√2/2 sin20°-√6/2cos20°+√6/2
=2√2sin50°cos10°-√2(√3/2cos20°-1/2sin20°)+√6/2
=2√2sin50°cos10°-√2sin(60°-20°)+√6/2
=2√2sin50°cos10°-√2sin40°+√6/2
=2√2sin50°cos10°-√2sin(50°-10°)+√6/2
=2√2sin50°cos10°-√2sin50°cos10°+√2cos50°sin10°+√6/2
=√2sin50°cos10°+√2cos50°sin10°+√6/2
=√2sin60°+√6/2
=√6

追问

sin^2(80)怎么变成cos10了

追答

sin80=sin(90-10)=cos10
√sin^2(80)=cos10

Answer 2

解：
原式=[ 2sin50°+sin10°(1＋√3·tan10°)]√2sin(90-10)
=[ 2sin50°+sin10°(1＋√3·tan10°)]√2cos10°
=2√2sin50°cos10°+√2 sin10°cos10°+√6 sin^210°
=2√2sin50°cos10°+√2/2 sin20°+√6/2(1-cos20°)
=2√2sin50°cos10°+√2/2 sin20°-√6/2cos20°+√6/2
=2√2sin50°cos10°-√2(√3/2cos20°-1/2sin20°)+√6/2
=2√2sin50°cos10°-√2sin(60°-20°)+√6/2
=2√2sin50°cos10°-√2sin40°+√6/2
=2√2sin50°cos10°-√2sin(50°-10°)+√6/2
=2√2sin50°cos10°-√2sin50°cos10°+√2cos50°sin10°+√6/2
=√2sin50°cos10°+√2cos50°sin10°+√6/2
=√2sin60°+√6/2
=√6

Answer 3

最后那个是sin的280次方？？？

追问

(sin80)^2