三角学 示例

x के लिये हल कीजिये 1-(sin(x)+cos(x))^2=-sin(2x)
1-(sin(x)+cos(x))2=-sin(2x)1(sin(x)+cos(x))2=sin(2x)
解题步骤 1
化简左边。
点击获取更多步骤...
解题步骤 1.1
化简 1-(sin(x)+cos(x))21(sin(x)+cos(x))2
点击获取更多步骤...
解题步骤 1.1.1
化简每一项。
点击获取更多步骤...
解题步骤 1.1.1.1
(sin(x)+cos(x))2(sin(x)+cos(x))2 重写为 (sin(x)+cos(x))(sin(x)+cos(x))(sin(x)+cos(x))(sin(x)+cos(x))
1-((sin(x)+cos(x))(sin(x)+cos(x)))=-sin(2x)1((sin(x)+cos(x))(sin(x)+cos(x)))=sin(2x)
解题步骤 1.1.1.2
使用 FOIL 方法展开 (sin(x)+cos(x))(sin(x)+cos(x))(sin(x)+cos(x))(sin(x)+cos(x))
点击获取更多步骤...
解题步骤 1.1.1.2.1
运用分配律。
1-(sin(x)(sin(x)+cos(x))+cos(x)(sin(x)+cos(x)))=-sin(2x)1(sin(x)(sin(x)+cos(x))+cos(x)(sin(x)+cos(x)))=sin(2x)
解题步骤 1.1.1.2.2
运用分配律。
1-(sin(x)sin(x)+sin(x)cos(x)+cos(x)(sin(x)+cos(x)))=-sin(2x)1(sin(x)sin(x)+sin(x)cos(x)+cos(x)(sin(x)+cos(x)))=sin(2x)
解题步骤 1.1.1.2.3
运用分配律。
1-(sin(x)sin(x)+sin(x)cos(x)+cos(x)sin(x)+cos(x)cos(x))=-sin(2x)1(sin(x)sin(x)+sin(x)cos(x)+cos(x)sin(x)+cos(x)cos(x))=sin(2x)
1-(sin(x)sin(x)+sin(x)cos(x)+cos(x)sin(x)+cos(x)cos(x))=-sin(2x)1(sin(x)sin(x)+sin(x)cos(x)+cos(x)sin(x)+cos(x)cos(x))=sin(2x)
解题步骤 1.1.1.3
化简并合并同类项。
点击获取更多步骤...
解题步骤 1.1.1.3.1
化简每一项。
点击获取更多步骤...
解题步骤 1.1.1.3.1.1
乘以 sin(x)sin(x)sin(x)sin(x)
点击获取更多步骤...
解题步骤 1.1.1.3.1.1.1
sin(x)sin(x) 进行 11 次方运算。
1-(sin1(x)sin(x)+sin(x)cos(x)+cos(x)sin(x)+cos(x)cos(x))=-sin(2x)1(sin1(x)sin(x)+sin(x)cos(x)+cos(x)sin(x)+cos(x)cos(x))=sin(2x)
解题步骤 1.1.1.3.1.1.2
sin(x)sin(x) 进行 11 次方运算。
1-(sin1(x)sin1(x)+sin(x)cos(x)+cos(x)sin(x)+cos(x)cos(x))=-sin(2x)1(sin1(x)sin1(x)+sin(x)cos(x)+cos(x)sin(x)+cos(x)cos(x))=sin(2x)
解题步骤 1.1.1.3.1.1.3
使用幂法则 aman=am+naman=am+n 合并指数。
1-(sin(x)1+1+sin(x)cos(x)+cos(x)sin(x)+cos(x)cos(x))=-sin(2x)1(sin(x)1+1+sin(x)cos(x)+cos(x)sin(x)+cos(x)cos(x))=sin(2x)
解题步骤 1.1.1.3.1.1.4
1111 相加。
1-(sin2(x)+sin(x)cos(x)+cos(x)sin(x)+cos(x)cos(x))=-sin(2x)1(sin2(x)+sin(x)cos(x)+cos(x)sin(x)+cos(x)cos(x))=sin(2x)
1-(sin2(x)+sin(x)cos(x)+cos(x)sin(x)+cos(x)cos(x))=-sin(2x)1(sin2(x)+sin(x)cos(x)+cos(x)sin(x)+cos(x)cos(x))=sin(2x)
解题步骤 1.1.1.3.1.2
乘以 cos(x)cos(x)cos(x)cos(x)
点击获取更多步骤...
解题步骤 1.1.1.3.1.2.1
cos(x)cos(x) 进行 11 次方运算。
1-(sin2(x)+sin(x)cos(x)+cos(x)sin(x)+cos1(x)cos(x))=-sin(2x)1(sin2(x)+sin(x)cos(x)+cos(x)sin(x)+cos1(x)cos(x))=sin(2x)
解题步骤 1.1.1.3.1.2.2
cos(x)cos(x) 进行 11 次方运算。
1-(sin2(x)+sin(x)cos(x)+cos(x)sin(x)+cos1(x)cos1(x))=-sin(2x)1(sin2(x)+sin(x)cos(x)+cos(x)sin(x)+cos1(x)cos1(x))=sin(2x)
解题步骤 1.1.1.3.1.2.3
使用幂法则 aman=am+naman=am+n 合并指数。
1-(sin2(x)+sin(x)cos(x)+cos(x)sin(x)+cos(x)1+1)=-sin(2x)1(sin2(x)+sin(x)cos(x)+cos(x)sin(x)+cos(x)1+1)=sin(2x)
解题步骤 1.1.1.3.1.2.4
1111 相加。
1-(sin2(x)+sin(x)cos(x)+cos(x)sin(x)+cos2(x))=-sin(2x)1(sin2(x)+sin(x)cos(x)+cos(x)sin(x)+cos2(x))=sin(2x)
1-(sin2(x)+sin(x)cos(x)+cos(x)sin(x)+cos2(x))=-sin(2x)1(sin2(x)+sin(x)cos(x)+cos(x)sin(x)+cos2(x))=sin(2x)
1-(sin2(x)+sin(x)cos(x)+cos(x)sin(x)+cos2(x))=-sin(2x)1(sin2(x)+sin(x)cos(x)+cos(x)sin(x)+cos2(x))=sin(2x)
解题步骤 1.1.1.3.2
重新排序 sin(x)cos(x)sin(x)cos(x) 的因式。
1-(sin2(x)+cos(x)sin(x)+cos(x)sin(x)+cos2(x))=-sin(2x)1(sin2(x)+cos(x)sin(x)+cos(x)sin(x)+cos2(x))=sin(2x)
解题步骤 1.1.1.3.3
cos(x)sin(x)cos(x)sin(x)cos(x)sin(x)cos(x)sin(x) 相加。
1-(sin2(x)+2cos(x)sin(x)+cos2(x))=-sin(2x)1(sin2(x)+2cos(x)sin(x)+cos2(x))=sin(2x)
1-(sin2(x)+2cos(x)sin(x)+cos2(x))=-sin(2x)1(sin2(x)+2cos(x)sin(x)+cos2(x))=sin(2x)
解题步骤 1.1.1.4
移动 cos2(x)cos2(x)
1-(sin2(x)+cos2(x)+2cos(x)sin(x))=-sin(2x)1(sin2(x)+cos2(x)+2cos(x)sin(x))=sin(2x)
解题步骤 1.1.1.5
使用勾股恒等式。
1-(1+2cos(x)sin(x))=-sin(2x)1(1+2cos(x)sin(x))=sin(2x)
解题步骤 1.1.1.6
化简每一项。
点击获取更多步骤...
解题步骤 1.1.1.6.1
2cos(x)2cos(x)sin(x)sin(x) 重新排序。
1-(1+sin(x)(2cos(x)))=-sin(2x)1(1+sin(x)(2cos(x)))=sin(2x)
解题步骤 1.1.1.6.2
sin(x)sin(x)22 重新排序。
1-(1+2sin(x)cos(x))=-sin(2x)1(1+2sin(x)cos(x))=sin(2x)
解题步骤 1.1.1.6.3
使用正弦倍角公式。
1-(1+sin(2x))=-sin(2x)1(1+sin(2x))=sin(2x)
1-(1+sin(2x))=-sin(2x)1(1+sin(2x))=sin(2x)
解题步骤 1.1.1.7
运用分配律。
1-11-sin(2x)=-sin(2x)111sin(2x)=sin(2x)
解题步骤 1.1.1.8
-11 乘以 11
1-1-sin(2x)=-sin(2x)11sin(2x)=sin(2x)
1-1-sin(2x)=-sin(2x)11sin(2x)=sin(2x)
解题步骤 1.1.2
通过减去各数进行化简。
点击获取更多步骤...
解题步骤 1.1.2.1
11 中减去 11
0-sin(2x)=-sin(2x)0sin(2x)=sin(2x)
解题步骤 1.1.2.2
00 中减去 sin(2x)sin(2x)
-sin(2x)=-sin(2x)sin(2x)=sin(2x)
-sin(2x)=-sin(2x)sin(2x)=sin(2x)
-sin(2x)=-sin(2x)sin(2x)=sin(2x)
-sin(2x)=-sin(2x)sin(2x)=sin(2x)
解题步骤 2
将所有包含 sin(2x)sin(2x) 的项移到等式左边。
点击获取更多步骤...
解题步骤 2.1
在等式两边都加上 sin(2x)sin(2x)
-sin(2x)+sin(2x)=0sin(2x)+sin(2x)=0
解题步骤 2.2
-sin(2x)sin(2x)sin(2x)sin(2x) 相加。
0=00=0
0=00=0
解题步骤 3
因为 0=00=0,所以方程对于 xx 的所有值将恒成立。
所有实数
解题步骤 4
结果可以多种形式表示。
所有实数
区间计数法:
(-,)(,)
 [x2  12  π  xdx ]  x2  12  π  xdx