函数f(x)=sin2x-cos2x，怎么化简到f(x)=√2sin(2x-π/4)的..求解 设函数F(X)=√2/2cos(2x+π/4)+sin...
\u7531f\uff08x\uff09=sin2x+cos2x=\u221a2sin\uff082x+\u03c0/4\uff09\uff0c\u6c42\u8be6\u7ec6\u7684\u5316\u7b80\u8fc7\u7a0bf(x)=sin(2x)+cos(2x)
=\u221a2[(\u221a2/2)sin(2x)+(\u221a2/2)cos(2x)]
=\u221a2[sin(2x)cos(\u03c0/4)+cos(2x)sin(\u03c0/4)]
=\u221a2sin(2x+\u03c0/4)
\u5148\u5316\u7b80cos(2x+\u03c0/4)\uff0csin^2x
\u2192cos(2x+\u03c0/4)=\u221a2/2*cos2x-\u221a2/2*sin2x
sin^2x=-cos2x/2+1/2
\u2192\u6240\u4ee5F(x)=cos2x/2-sin2x/2-cos2x/2+1/2=-sin2x/2+1/2
\u2192\u6240\u4ee5:-\u03c0/2+2k\u03c0\u22642x\u2264\u03c0/2+2k\u03c0
\u6240\u4ee5\uff0c\u5355\u51cf\u533a\u95f4:[-\u03c0/4+k\u03c0\uff0c\u03c0/4+k\u03c0]\uff0ck\u2208Z
\u2192\u03c0/2+2k\u03c0\u22642x\u22643\u03c0/2+2k\u03c0
\u6240\u4ee5\u5355\u589e\u533a\u95f4\uff0c[\u03c0/4+k\u03c0\uff0c3\u03c0/4+k\u03c0]\uff0ck\u2208Z
=√2*[sin2x*(√2)/2 - cos2x*(√2)/2]
=√2*[sin2x*cos(π/4) -cos2x*sin(π/4)]
=√2*sin(2x-π/4)
f(x)=sin2x—cos2x=sin2xcos(π/4)—cos2xsin(π/4)=√2sin(2x-π/4)
绛旓細瑙ｄ箣寰 - 蟺 3 +k蟺鈮鈮 蟺 6 +k蟺 锛坘鈭圸锛夛紝鈭鍑芥暟f锛坸锛鐨勯掑鍖洪棿鏄 [- 蟺 3 +k蟺锛 蟺 6 +k蟺] 锛宬鈭圸锛庯紙鈪★級鐢眆锛坸锛夆墺2锛屽緱 2sin(2x+ 蟺 6 )+1鈮2 锛坘鈭圸锛夛紝鍗 sin(2x+ 蟺 6 )鈮 1 2 ...
绛旓細濡傚浘
绛旓細f(x)=sin(2x+蟺/3)+sin(2x-蟺/3)+2cos^2x-1 =sin2xcos蟺/3+cos2xsin蟺/3+sin2xcos蟺/3-cos2xsin蟺/3+cos2x =2sin2xcos蟺/3+cos2x =sin2x+cos2x =鈭2*(鈭2/2*sin2x+鈭2/2*cos2x)=鈭2*(sin2xcos蟺/4+cos2xsin蟺/4)=鈭2*sin(2x+蟺/4)T=2蟺/2=蟺 x鈭...
绛旓細锛1锛 f(x)=2sin(2x+ 蟺 6 )+1-m 锛屸埓 m=2sin(2x+ 蟺 6 )+1 鍦 [0锛 蟺 2 ] 鍐呮湁瑙ｂ埖 0鈮鈮 蟺 2 锛屸埓 蟺 6 鈮2x+ 蟺 6 鈮 7蟺 6 鈭 0鈮2sin(2x+ 蟺 6 )+1鈮3 锛屸埓0鈮鈮3锛2...
绛旓細姹傚寰楋細f鈥锛坸锛=-4sinxcosx+2 3 cos2x=-2sin2x+2 3 cos2x=4sin锛 蟺 3 -2x锛夛紝浠鈥诧紙x锛=0锛屽緱鍒皒= 蟺 6 锛屸埖f锛0锛=2+a锛宖锛 蟺 2 锛=a锛宖锛 蟺 6 锛=3+a锛屸埓鍑芥暟鐨勬渶灏忓间负a锛屽張鍑芥暟鍖洪棿 [0锛 蟺 2 ...
绛旓細f(x)=2cos(2x+ 蟺/3)+4鈭3sinxcosx =2[(1/2)cos2x-(鈭3/2)sin2x]+4鈭3sinxcosx =cos2x-鈭3sin2x+2鈭3sin2x =cos2x+鈭3sin2x =2[(1/2)cos2x+(鈭3/2)sin2x]=2sin(2x+ 蟺/6)鈶犲綋0鈮鈮は/2鏃讹紝蟺/6鈮2x+ 蟺/6鈮7蟺/6 姝ゆ椂锛宻in2x鈭圼-1/2锛1]鈭鍑芥暟f(x)...
绛旓細锛堟湰灏忛婊″垎14鍒嗭級锛1锛f锛坸锛=1- 3 sin2x+2cos 2 x=cos2x- 3 sin2x+2 锛2鍒嗭級=2cos锛2x+ 蟺 3 锛+2锛岋紙4鍒嗭級鈭-1鈮os锛2x+ 蟺 3 锛夆墹1锛屸埓0鈮2cos锛2x+ 蟺 3 锛+2鈮4锛屸埓f锛坸锛夌殑鏈澶у间负4锛岋紙5鍒嗭級褰2x+ 蟺 ...
绛旓細瑙ｂ埖 f(x)= 5 2 sin2x-5 3 1+cos2x 2 = 5 2 sin2x- 5 3 2 cos2x- 5 3 2 = 5( 1 2 sin2x- 3 2 cos2x)- 5 3 2 = 5sin(2x- 蟺 3 )- 5 ...
绛旓細锛圛锛f锛坸锛=sin2x+cos2x= 2 ( 2 2 sin2x+ 2 2 cos2x) = 2 sin(2x+ 蟺 4 ) 锛屸埓 T= 2蟺 2 =蟺 锛庯紙II锛夌敱 - 蟺 2 +2k蟺鈮2x+ 蟺 4 鈮 蟺 2 +2k蟺 锛岃В寰 - 3蟺 8 +k蟺鈮...
绛旓細鐢卞凡鐭ワ細 f(x)= 3 sin2x-cos2x=2( 3 2 sin2x- 1 2 cos2x)=2sin(2x- 蟺 6 ) 锛岋紙1锛夊綋 2x- 蟺 6 =2k蟺+ 蟺 2 锛宬鈭圸 锛屽嵆锛 sin(2x- 蟺 6 )=1 鏃讹紝f锛坸锛夊彇鏈澶у2锛庢鏃秞鐨勯泦鍚堜负锛 {x|x=k蟺+ ...
