题目内容
已知cos(75°+θ)=
,θ为第三象限角,求cos(-225°-θ)+sin(435°+θ)的值.
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考点:运用诱导公式化简求值,两角和与差的正弦函数
专题:三角函数的求值
分析:先求得sin(75°+θ)=-
,故原式化简可得:-[cos(75°+θ)cos30°+sin(75°+θ)sin30°]+sin(75°+θ)代入即可求值.
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解答:
解:∵180°≤θ≤270°
∴255°≤75°+θ≤345°
∴sin(75°+θ)=-
=-
cos(-225°-θ)+sin(435°+θ)
=cos(180°+45°+θ)+sin(75°+θ)
=-cos(75°+θ-30°)+sin(75°+θ)
=-[cos(75°+θ)cos30°+sin(75°+θ)sin30°]+sin(75°+θ)
=-(
×
-
×
)-
=-
.
∴255°≤75°+θ≤345°
∴sin(75°+θ)=-
| 1-cos2(75°+θ) |
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cos(-225°-θ)+sin(435°+θ)
=cos(180°+45°+θ)+sin(75°+θ)
=-cos(75°+θ-30°)+sin(75°+θ)
=-[cos(75°+θ)cos30°+sin(75°+θ)sin30°]+sin(75°+θ)
=-(
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2
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2
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=-
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点评:本题主要考察了运用诱导公式化简求值,两角和与差的正弦函数公式的应用,属于基础题.
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