题目内容

(1)求
tan39°+tan81°+tan240°
tan39°tan81°
的值;
(2)sin50°(1+
3
sin10°
cos10°
).
考点:两角和与差的正切函数,两角和与差的余弦函数
专题:三角函数的求值
分析:(1)由两角和的正切公式变形可得tan39°+tan81°=tan(39°+81°)(1-tan39°tan81°),把该式代入要求的式子化简可得;
(2)化简可得原式=sin50°
cos10°+
3
sin10°
cos10°
=
2sin50°cos50°
cos10°
=
sin100°
cos10°
,再由诱导公式可得.
解答: 解(1)化简可得
tan39°+tan81°+tan240°
tan39°tan81°

=
tan(39°+81°)(1-tan39°tan81°)+tan60°
tan39°tan81°

=
-
3
(1-tan39°tan81°)+
3
tan39°tan81°

=
3
tan39°tan81°
tan39°tan81°
=
3

(2)sin50°(1+
3
sin10°
cos10°

=sin50°
cos10°+
3
sin10°
cos10°

=sin50°
2(
1
2
cos10°+
3
2
sin10°)
cos10°

=sin50°
2cos(60°-10°)
cos10°

=
2sin50°cos50°
cos10°
=
sin100°
cos10°

=
sin(90°+10°)
cos10°
=
cos10°
cos10°
=1
点评:本题考查三角函数式的化简,涉及两角和与差的正切公式和正余弦公式,属基础题.
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