(Ⅰ)已知tanθ=2,求1-sin2θ1+cos2θ的值;(Ⅱ)化简:sin2αsin2β+cos2αcos2β-12cos2αcos2β.-数学

题目简介

(Ⅰ)已知tanθ=2,求1-sin2θ1+cos2θ的值;(Ⅱ)化简:sin2αsin2β+cos2αcos2β-12cos2αcos2β.-数学

题目详情

(Ⅰ)已知tanθ=2,求
1-sin2θ
1+cos2θ
的值;
(Ⅱ)化简:sin2αsin2β+cos2αcos2β-
1
2
cos2αcos2β.
题型:解答题难度:中档来源:不详

答案

(Ⅰ)∵tanθ=2,
class="stub"1-sin2θ
1+cos2θ
=
sin2θ+cos2θ-2sinθcosθ
1+2cos2θ-1
(3分)
=
sin2θ+cos2θ-2sinθcosθ
2cos2θ

=
tan2θ+1-2tanθ
2
(7分)
=class="stub"4+1-2×2
2
=class="stub"1
2
;(8分)

(Ⅱ) sin2αsin2β+cos2αcos2β-class="stub"1
2
cos2αcos2β
=class="stub"1-cos2α
2
•class="stub"1-cos2β
2
+class="stub"1+cos2α
2
•class="stub"1+cos2β
2
-class="stub"1
2
cos2αcos2β(13分)
=class="stub"1
4
[(1-cos2α)(1-cos2β)+(1+cos2α)(1+cos2β)]-class="stub"1
2
cos2αcos2β
=class="stub"1
4
[2+2cos2αcos2β]-class="stub"1
2
cos2αcos2β
=class="stub"1
2
+class="stub"1
2
cos2αcos2β-class="stub"1
2
cos2αcos2β
=class="stub"1
2
.(16分)

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