问题
解答题
(1)化简
(2)计算4
(3)已知tanθ=3,求
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答案
(1)原式=
=-sinα;-sinαcosα cosα
(2)原式=2+log232-log2
=2+log223=2+3=5;9 8
(3)∵tanθ=3,∴原式=
=sin2θ+cos2θ sin2θ-2sinθcosθ
=tan2θ+1 tan2θ-2tanθ
=9+1 9-2
.10 7
(1)化简
(2)计算4
(3)已知tanθ=3,求
|
(1)原式=
=-sinα;-sinαcosα cosα
(2)原式=2+log232-log2
=2+log223=2+3=5;9 8
(3)∵tanθ=3,∴原式=
=sin2θ+cos2θ sin2θ-2sinθcosθ
=tan2θ+1 tan2θ-2tanθ
=9+1 9-2
.10 7