数学三角函数问题求详解 2.4. 高中数学三角函数问题,给值求值。求详解

\u6c42\u6570\u5b66\u9ad8\u624b\u8be6\u89e3\u4e09\u89d2\u51fd\u6570\u95ee\u9898\uff01\uff01

\u7528\u4f59\u5f26\u5b9a\u7406
\u89e3\uff1a\u5728\u4e09\u89d2\u5f62ABC\u4e2d\uff0ccosC =(a^2+b^2-c^2)/(2ab)
cosB =(a^2+c^2-b^2)/(2ac)
\u6240\u4ee5\uff0cbcosC \uff1d\uff082a \u2014c \uff09cosB\u53ef\u4ee5\u8f6c\u5316\u4e3a\uff1a
b*[(a^2+b^2-c^2)/(2ab)]=(2a-c)[(a^2+c^2-b^2)/(2ac)]
\u6574\u7406\u6709\uff1a
a^2+b^2-c^2=\uff082a-c)(a^2+c^2-b^2)/c
\u6240\u4ee5
a^2+b^2-c^2=2a(a^2+c^2-b^2)/c -(a^2+c^2-b^2)
\u6574\u7406\u6709
2a^2=2a(a^2+c^2-b^2)/c
\u6240\u4ee5\uff0c\u4e24\u8fb9\u521d\u4ee54a^2\u6709
1/2=(a^2+c^2-b^2)/\uff082ac\uff09=cosB
\u6240\u4ee5\u89d2B=60\u00b0

\u8fd9\u4e00\u7c7b\u578b\u7684\u9898\u76ee\uff0c\u90fd\u662f\u7528\u201c\u6b63\u4f59\u5f26\u5b9a\u7406\u201d\u6765\u505a\u5230\u201c\u8fb9\u5316\u89d2\u201d\u6216\u201c\u89d2\u5316\u8fb9\u201d\uff01\u800c\u6c42\u89e3\u7684\uff01

\u4f46\u662f\u600e\u6837\u80fd\u60f3\u5230\u7528\u8fd9\u4e9b\u65b9\u6cd5\u5462\uff1f\u6211\u7684\u610f\u601d\u662f\u9047\u5230\u54ea\u79cd\u95ee\u9898\uff0c\u8be5\u7528\u54ea\u79cd\u65b9\u6cd5\uff0c\u600e\u6837\u53ef\u4ee5\u5feb\u901f\u5224\u65ad\uff1f
\u7b54\u597d\u4e86\u52a0\u5206\uff01

\u4e0d\u662f\u8bf4\u4e86\u561b\uff01\uff01\u5475\u5475\uff01\uff01\uff01
\u4e00\u822c\u4e09\u89d2\u5f62\u7684\u95ee\u9898\uff0c\u90fd\u9996\u5148\u8003\u8651\u201c\u6b63\u4f59\u5f26\u5b9a\u7406\u201d\uff0c\u7279\u522b\u662f\u4e09\u89d2\u51fd\u6570\u548c\u8fb9\u957f\u6df7\u4e00\u8d77\u7684\u8868\u8fbe\u5f0f\uff0c\u66f4\u8981\u7528\u8fd9\u4e24\u4e2a\u5b9a\u7406\u628a\u8fb9\u957f\u5316\u4e3a\u4e09\u89d2\u51fd\u6570\uff01

\u6839\u636e\u9898\u610f\u6c42\u51facos\u03b1=3\u221a10/20 \uff0c sin\uff08\u03b1\uff0d\u03b2\uff09=sin\uff5b\uff0d\uff08\u03b2\uff0d\u03b1\uff09\uff5d=\uff0dsin\uff08\u03b2\uff0d\u03b1\uff09=\u221a10/10
\u2235\u03b1\u3001\u03b2\u662f\u9510\u89d2 \u4e14sin\uff08\u03b2\uff0d\u03b1\uff09\uff1c0
\u6240\u4ee5\u6c42\u51fasin\u03b1=3/5 \uff0ccos\uff08\u03b1\uff0d\u03b2\uff09=cos\uff08\u03b2\uff0d\u03b1\uff09=3\u221a10/10
\u2235\u03b2=\uff08\u03b2\uff0d\u03b1\uff09+\u03b1
\u2234cos\u03b2=cos\uff5b\uff08\u03b2\uff0d\u03b1\uff09+\u03b1\uff5d=coa\uff08\u03b2\uff0d\u03b1\uff09cos\u03b1\uff0dsin\uff08\u03b2\uff0d\u03b1\uff09sin\u03b1
=9\u221a10/50
\u5269\u4e0b\u81ea\u5df1\u7b97~~\u52a0\u6df1\u5370\u8c61~~~

\u4e0d\u61c2\u518d\u95ee\u5427\uff01

两题都用到诱导公式：
2、cos(-2019π)=cos(2019π)
=cos(2018π + π)
=cos(2π•1009 + π)
=cosπ=-1
4、sin160º=sin(180º-20º)
=sin20º=a
cos340º=cos(360º-20º)=cos20º
∵sin²20º + cos²20º=1
∴cos20º=√1 - sin²20º=√1-a²

(2). cos(-2010π)=cos(2019π)=cos(1009×2π+π)=cosπ=-1; 选B;
(4). sin160°=sin(180°-20°)=sin20°=a，则cos340°=cos(360°-20°)=cos20°=√(1-a²);选B

题头看不清。

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