题目内容
已知cos(75°+θ)=
,θ为第三象限角,求cos(255°+θ)+(435°+θ)的值.
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考点:运用诱导公式化简求值,两角和与差的余弦函数
专题:三角函数的求值
分析:先由条件求得sin(75°+θ)=-
,把原式化简为-[cos(75°+θ)cos30°+sin(75°+θ)sin30°]+sin(75°+θ),代入即可求值.
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解答:
解:∵θ为第三象限角,不妨假设180°≤θ≤270°,则 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)
=-(
×
-
×
)-
=
.
sin(75°+θ)=-
| 1-cos(75°+θ)2 |
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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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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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