题目内容
(1)求
的值;
(2)sin50°(1+
).
| tan39°+tan81°+tan240° |
| tan39°tan81° |
(2)sin50°(1+
| ||
| 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°
=
=
,再由诱导公式可得.
(2)化简可得原式=sin50°
cos10°+
| ||
| cos10° |
| 2sin50°cos50° |
| cos10° |
| sin100° |
| cos10° |
解答:
解(1)化简可得
=
=
=
=
;
(2)sin50°(1+
)
=sin50°
=sin50°
=sin50°
=
=
=
=
=1
| tan39°+tan81°+tan240° |
| tan39°tan81° |
=
| tan(39°+81°)(1-tan39°tan81°)+tan60° |
| tan39°tan81° |
=
-
| ||||
| tan39°tan81° |
=
| ||
| tan39°tan81° |
| 3 |
(2)sin50°(1+
| ||
| cos10° |
=sin50°
cos10°+
| ||
| cos10° |
=sin50°
2(
| ||||||
| cos10° |
=sin50°
| 2cos(60°-10°) |
| cos10° |
=
| 2sin50°cos50° |
| cos10° |
| sin100° |
| cos10° |
=
| sin(90°+10°) |
| cos10° |
| cos10° |
| cos10° |
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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