Question

求下列各式的值(1)tan^230°+2sin60°+tan45°×sin30°-tan60°+cos^230°

求下列各式的值(1)tan的平方30°+2sin60°+tan45°×sin30°-tan60°+cos的平方30°
（2）cos60°/tan45°-sin60°+tan60°
（3）根号下cos的平方30°-2cos30°+1

Answer 1

tan²30°+2sin60°+tan45°×sin30°-tan60°+cos²30°
=(√3/3)²+2×√3/2+1×1/2-√3+(√3/2)²
=1/3+√3+1/2-√3+3/4
=1/3+1/2+3/4
=4/12+6/12+9/12
=19/12

cos60°/(tan45°-sin60°+tan60°)
=(1/2)/(1-√3/2+√3)
=(1/2)/(2/2-√3/2+2√3/2)
=(1/2)/[(2+√3)/2]
=1/2×2/(2+√3)
=1/(2+√3)
=(2-√3)/[(2-√3)(2+√3)]
=(2-√3)/(2²-√3²)
=(2-√3)/(4-3)
=2-√3

√(cos²30°-2cos30°+1)
=√(1-cos30°)²
=1-cos30°
=1-√3/2
=(2-√3)/2

Answer 2

tan²30°+2sin60°+tan45°×sin30°-tan60°+cos²30°
=(√3/3)²+2×√3/2+1×1/2-√3+(√3/2)²
=1/3+√3+1/2-√3+3/4
=1/3+1/2+3/4
=4/12+6/12+9/12
=19/12

cos60°/(tan45°-sin60°+tan60°)
=(1/2)/(1-√3/2+√3)
=(1/2)/(2/2-√3/2+2√3/2)
=(1/2)/[(2+√3)/2]
=1/2×2/(2+√3)
=1/(2+√3)
=(2-√3)/[(2-√3)(2+√3)]
=(2-√3)/(2²-√3²)
=(2-√3)/(4-3)
=2-√3

√(cos²30°-2cos30°+1)
=√(1-cos30°)²
=1-cos30°
=1-√3/2

Answer 3

你下载一个mathematics输入公式就可以解决