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确定下列各三角函数数值的符号$\left(( {1} \right )sin{830}^{\circ }$;$\left ( {2} \right )sin\left ( {-\dfrac {35\pi } {17}} \right )$;$\left ( {3} \right )tan{500}^{\circ }$;$\left ( {4} \right )cos\left ( {\dfrac {47\pi } {9}} \right )$.","title_text":"确定下列各三角函数数值的符号$\left ( {1} \right )sin{830}^{\circ }$;$\left ( {2} \right )sin\left ( {-\dfrac {35\pi } {17}} \right )$;$\left ( {3} \right )tan{500}^{\circ }$;$\left ( {4} \right )cos\left ( {\dfrac {47\pi } {9}} \right )$.)

2022-06-19 21:11:56 实事动态来源:
导读 想必现在有很多小伙伴对于确定下列各三角函数数值的符号:$\left ( {1} \right )sin{830}^{\circ }$;$\left ( {2} \right )sin\...

想必现在有很多小伙伴对于确定下列各三角函数数值的符号:$\left ( {1} \right )sin{830}^{\circ }$;$\left ( {2} \right )sin\left ( {-\dfrac {35\pi } {17}} \right )$;$\left ( {3} \right )tan{500}^{\circ }$;$\left ( {4} \right )cos\left ( {\dfrac {47\pi } {9}} \right )$.","title_text":"确定下列各三角函数数值的符号:$\left ( {1} \right )sin{830}^{\circ }$;$\left ( {2} \right )sin\left ( {-\dfrac {35\pi } {17}} \right )$;$\left ( {3} \right )tan{500}^{\circ }$;$\left ( {4} \right )cos\left ( {\dfrac {47\pi } {9}} \right )$.方面的知识都比较想要了解,那么今天小好小编就为大家收集了一些关于确定下列各三角函数数值的符号:$\left ( {1} \right )sin{830}^{\circ }$;$\left ( {2} \right )sin\left ( {-\dfrac {35\pi } {17}} \right )$;$\left ( {3} \right )tan{500}^{\circ }$;$\left ( {4} \right )cos\left ( {\dfrac {47\pi } {9}} \right )$.","title_text":"确定下列各三角函数数值的符号:$\left ( {1} \right )sin{830}^{\circ }$;$\left ( {2} \right )sin\left ( {-\dfrac {35\pi } {17}} \right )$;$\left ( {3} \right )tan{500}^{\circ }$;$\left ( {4} \right )cos\left ( {\dfrac {47\pi } {9}} \right )$.方面的知识分享给大家,希望大家会喜欢哦。

(1)$sin83{0}^{circ }=sinleft ( {72{0}^{circ }+11{0}^{circ }} right )$

$=sin11{0}^{circ }$

$because 11{0}^{circ }$是第二象限角,

$therefore sin11{0}^{circ }gt 0$

即$sin83{0}^{circ }gt 0$

综上所述,结论是:$+$;

(2)$sinleft ( {-dfrac {35pi } {17}} right )=sinleft ( {-2pi -dfrac {pi } {17}} right )$

$=-sindfrac {pi } {17}$

$because dfrac {pi } {17}$是第一象限角,

$therefore -sindfrac {pi } {17}lt 0$

即$sinleft ( {-dfrac {35pi } {17}} right )lt 0$

综上所述,结论是:$-$;

(3)$tan50{0}^{circ }=tanleft ( {3{60}^{circ }+14{0}^{circ }} right )$

$=tan14{0}^{circ }$

$because 14{0}^{circ }$是第二象限角,

$therefore tan14{0}^{circ }lt 0$

$therefore tan50{0}^{circ }lt 0$

综上所述,结论是:$-$;

(4)$cosdfrac {47pi } {9}=cosleft ( {5pi +dfrac {2pi } {9}} right )$

$=-cosdfrac {2pi } {9}$

$because dfrac {2pi } {9}$是第一象限角,

$therefore -cosdfrac {2pi } {9}lt 0$

$therefore cosleft ( {dfrac {47pi } {9}} right )lt 0$

综上所述,结论是:$-$.

本文到此结束,希望对大家有所帮助。


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