一道关于三角函数的题目
\u4e00\u9053\u4e09\u89d2\u51fd\u6570\u7684\u9898\u76ee\u3002
\u795d\uff1a\u697c\u4e3b\u5b66\u4e1a\u8fdb\u6b65\uff01
\u56e0\u4e3a\u9898\u76ee\u4e2d\u8bf4\uff1a\u8bb0\u5411\u4e0a\u9707\u52a8\u7684\u4f4d\u79fb\u662f\u6b63\u65b9\u5411
\u5f39\u7c27\u9635\u5b50\u4ece\u9707\u52a8\u5230\u6700\u4f4e\u70b9\u5f00\u59cb\u8ba1\u65f6\uff0c\u5f53t=0\u65f6\uff0c\u4f4d\u79fb\u4e3a\u8d1f
=sin70°根号( 2(cos20°)^2)
=根号2(sin70°)^2
=根号2/2(1-cos140°)
=根号2/2(1+cos40°)
cos40°+sin50°（1+√3*tan10°)
=cos40°+2sin50°（sin30°cos10°+cos30°sin10°)/cos10°
=cos40°+2sin50°sin40°/cos10°
=cos40°+sin100°/cos10°
=cos40°+1
原式=根号2/2
你的表达式不清晰
不过可以把40＝30＋10
50＝60－10代入
[COS40+SIN50(1+根号3*TAN10)]/SIN70*[根号（1+COS40）] =[sin50 +sin50*(1+tg60tg10)]/sin70*根2*cos20 =[sin50+sin50*(cos60cos10+sin60sin10)/(cos60cos10)]/sin70*根2*cos20 =[sin50+2sin50*cos50/cos10]/sin70*根2*cos20 =(sin50+sin100/cos10)/sin70*根2*cos20 = (1+sin50)/[根2/2 *（sin90+sin50) =根号2
同意答案--根号2
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绛旓細=1-cos2伪+sin2伪 =1+鈭2sin(2伪-蟺/4)锛-蟺/4<2伪-蟺/4鈮3蟺/4锛-鈭2/2<sin(2伪-蟺/4)鈮1锛0<S鈮1+鈭2.浜诧紝姝ゅ皬棰樼殑鍑犱綍鎰忎箟鏄細宸茬煡涓夎褰㈢殑澶栨帴鍦嗙殑蟺/4鍦嗗懆瑙扐鎵澶瑰鸡闀緼C涓2锛屾眰涓夎褰BC闈㈢Н鐨勬渶澶у笺傚彲鐭ュ綋涓夎褰BC涓虹瓑鑵颁笁瑙掑舰鏃堕潰绉渶澶с
绛旓細3 f锛-蟺/6锛=0 鎵浠锛坸锛鍏充簬锛-蟺/6,0锛夊绉 3姝ｇ‘ 4f锛-x锛=4sin锛-2x+蟺/3锛=-4sin锛2x-蟺/3锛夊崟璋冨鍖洪棿鏄/2+2k蟺<=2x-蟺/3<=3蟺/2+2k蟺 绛旀缁欑殑鍗曡皟鍑忓尯闂翠笉绛夊紡锛屽洜涓-2x+蟺/3 x鐨勭郴鏁版槸璐熺殑鎵浠ヨ鎶婄郴鏁板彉鎴愭鍚庡湪鎸塻inx鐨勮寰嬫潵鍋氥傛墍浠4閿欒 缁间笂鎵...
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