观察下列恒等式:∵tan2a-1tanα=-2(1-tan2α)2tanα,∴tanα-1tanα=-2tan2α①∴tan2α-1tan2α=-2tan4α②tan4α-1tan4α=-2tan8α

题目简介

观察下列恒等式:∵tan2a-1tanα=-2(1-tan2α)2tanα,∴tanα-1tanα=-2tan2α①∴tan2α-1tan2α=-2tan4α②tan4α-1tan4α=-2tan8α

题目详情

观察下列恒等式:
tan2a-1
tanα
=-
2(1-tan2α)
2tanα

∴tanα-
1
tanα
=-
2
tan2α

∴tan2α-
1
tan2α
=-
2
tan4α

tan4α-
1
tan4α
=-
2
tan8α

由此可知:tan
π
32
+2tan
π
16
+4tan
π
8
-
1
tan
π
32
=(  )
A.-2B.-4C.-6D.-8
题型:单选题难度:偏易来源:不详

答案

∵tanclass="stub"π
32
-class="stub"1
tanclass="stub"π
32
=-class="stub"2
tanclass="stub"π
16

∴原式=2tanclass="stub"π
16
-class="stub"2
tanclass="stub"π
16
+4tanclass="stub"π
8

=class="stub"-4
tanclass="stub"π
8
+4tanclass="stub"π
8
=4×(-class="stub"2
tanclass="stub"π
4
)=-8.
选D.

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