求数学学霸解题，关于三角函数！ 初三知识，求进 九年级数学，三角函数，数学老师和高手进啊，第6题
\u6c42\u6570\u5b66\u5b66\u9738\u5e2e\u5fd9\u89e3\u7b54\u3002\u521d\u4e09\u51fd\u6570\u9898\u3002\uff082\uff09\u76f4\u7ebfBD\u7684\u89e3\u6790\u5f0f\uff1ay=-x+3
\u8bbeP(a\uff0ca²-4a+3)\u5230BD\u76f4\u7ebf\u7684\u8ddd\u79bb\u4e3ad\u4e140<a<3
d=|a+a²-4a+3-3|/\u221a2=|a²-3a|/\u221a2=|(a-3/2)²-9/4|/\u221a2
\u5f53a=3/2\u65f6\uff0cd\u503c\u6700\u5927
\u5373P(3/2\uff0c-3/4)
\u5728\u4e09\u89d2\u5f62ABC\u4e2d\uff0c\u6709c=acosB+bcosA(\u8bc1\u660e\u4e5f\u633a\u5bb9\u6613\u7684\uff0c\u53ea\u8981\u7531\u70b9C\u5411AB\u4f5c\u9ad8\u7ebfCD\uff0c\u7136\u540eAB=AD+BD,\u518d\u7528\u4e09\u89d2\u51fd\u6570\u5b9a\u4e49\u5c31\u53ef\u4ee5\u4e86)
\u53c8\u2220C=90\u00b0\uff0c\u6240\u4ee5\u2220A+\u2220B=90\u00b0\uff0c\u6240\u4ee5c=acosB+bcosA=acos(90\u00b0-A)+bcos(90\u00b0-B)
=asinA+bsinB.
\u6240\u4ee5\u7b54\u6848\u9009B\u3002
=[cos(90º-40º)]²+sin²(50º)-(1/2)²+1²
=cos²(50º)+sin²(50º)-1/4+1
=2-1/4
=7/4
sin²40°+sin²50°-cos²60°+cot²45°
=cos²50°+sin²50°-1/4+1
=7/4
绛旓細=24-3/8(10-t)² 6.4鈮鈮10 涓哄垎涓轰笁娈电殑鍒嗘鍑芥暟銆
绛旓細鏍规嵁姝ｅ鸡瀹氱悊锛歛/sinA绛変簬b/sinB 3/[(3鈭10)/10]绛変簬b/sinB 鈭10绛変簬b/sinB 鈭碽绛変簬鈭10•sinB 鈭礏鏄柍ABC鐨勫唴瑙 鈭0<sinB鈮1 鈭碽鐨勬渶澶у兼槸鈭10
绛旓細(1)sinA=卤鈭1 - (cosA)^2 =鈭1 - (鈭6/3)^2=卤鈭3/3 鈭礎鏄柍ABC鐨勪竴涓唴瑙掍笖cosA>0 鈭磗inA=鈭3/3 鏍规嵁姝ｅ鸡瀹氱悊锛歛/sinA=b/sinB 3/(鈭3/3)=b/sin(蟺/2 + A)3鈭3=b/cosA锛屽垯b=(3鈭3)脳cosA 鈭碽=3鈭2 (2)S=(1/2)absinC =(1/2)脳3脳(3鈭2)脳sin(A+B)...
绛旓細(1)T=2蟺/(2/3)=3蟺 (2)T=2蟺/4=蟺/2 (3)T=2蟺/2=蟺 鈭垫寮鍑芥暟鐨勫煎煙鏄痆-1锛1]鈭碆
绛旓細sin²(40º)+sin²(50º)-cos²(60º)+cot²(45º)=[cos(90º-40º)]²+sin²(50º)-(1/2)²+1²=cos²(50º)+sin²(50º)-1/4+1 =2-1/4 =7/4 ...
绛旓細瑙ｏ細f(x)=cos^2x-sin^2x+2sinxcosx.=cos2x+sin2x.=鈭2[(鈭2/2)cos2x+(鈭2/2)sin2x]=鈭2[sin2xcos(蟺/4)+cos2xsin(蟺/4)].鈭磃(x)=鈭2sin(2x+蟺/4).(1) f(x)鐨勬渶灏忔鍛ㄦ湡T=2蟺/2=蟺.(2). 鈭祍inx鍦▁鈭圼0,蟺)鍐咃紝鐨勯掑鍖洪棿鏄:[0,蟺/2].f(x)=鈭2sin(2x...
绛旓細缁忓吀渚嬮: 姣忎竴涓緥棰橀兘鏄煡璇嗙殑鍩虹煶锛屽畠浠府鍔╀綘寤虹珛璧涓夎鍑芥暟鐨勫熀纭锛岃浣犲湪鑰冨満涓婂緱蹇冨簲鎵嬨傚彉寮忛: 闈㈠鍗冨彉涓囧寲鐨勯珮鑰冮鍨嬶紝鍙樺紡棰樿浣犲湪鐔熸倝鐨勫熀纭涓婃嫇灞曟濈淮锛屾彁楂樺簲瀵瑰鏉傞鐩殑鑳藉姏銆傚綋鐒讹紝鎴戜滑骞舵湭姝㈡浜庢銆傚鏋滀綘闇瑕佹洿澶氭敮鎸侊紝杩樻湁鍏充簬鍦嗛敟鏇茬嚎鐨勬繁鍏ヨВ鏋愶紝浠ュ強瑙ｉ鎶宸х殑鍒嗕韩锛屽楂樹腑鏁板鐨...
绛旓細鐏垫椿杩愮敤姝ｅ鸡鍑芥暟鐨勫嶈鍏紡锛(sinx/2)^2*(cosx/2)^2=(sin/x*cosx/2)^2 =[(1/2)*2sinx/2cosx/2]^2 =[(1/2)sinx]^2.=(sinx/2)^2
绛旓細=(cos²x + sin²x)² - (1/2)sin²2x + sin2x =1 - (1/2)sin²2x + sin2x =(-1/2)(sin²2x - 2sin2x) + 1 =(-1/2)(sin2x - 1)² + 3/2 鈭碩=2蟺/2=蟺 (2)f(x)min=(-1/2)(-1-1)² + 3/2 =(-1/2)&#...
绛旓細cos(arccosx)=x sin (arccosx)=鏍瑰彿(1-x鐨勫钩鏂)鈭磘an(arccosx)=x/鏍瑰彿(1-x鐨勫钩鏂)tan(蟺-arccosx)=-tan(arccosx)=-x/鏍瑰彿(1-x鐨勫钩鏂)鏄鍑芥暟銆
