三道三角函数证明题，求高手解答
\u5e2e\u5fd9\u89e3\u4e09\u9053\u4e09\u89d2\u51fd\u6570\u8bc1\u660e\u9898\uff0c\u8981\u89e3\u7b54\u8fc7\u7a0b\uff0c\u9644\u56fe\uff0c\u53ef\u4ee5\u62ff\u7b14\u5199\uff0120\u5206LZ\u8bf7\u6ce8\u610f\u5206\u8fa8\uff0c\u697c\u4e0adrug2009\u7684\u56de\u7b54\u6709\u95ee\u9898\u3002
\u7b2c1\u9898\u6700\u540e\u201c=cos(x/2)cos(x/4)cos(x/8)..cos(x/2^n)*2^n/x\u201d\u8fd9\u4e2a\u5f0f\u5b50\u662f\u9519\u7684\uff0c\u4e0d\u53ef\u80fd\u5f97\u5230\u7ed3\u8bba\u3002
\u7b2c3\u9898\u662fWallis\u516c\u5f0f\uff0c\u600e\u4e48\u53ef\u80fd\u4e24\u6b65\u5c31\u63a8\u51fa\u6765\uff1f\u7b2c\u4e00\u4e2a\u7b49\u5f0f\u5e94\u8be5\u5229\u7528\u4e86Euler\u7684\u7ecf\u5178\u7ed3\u679c\uff0c\u4f46\u90a3\u4e2a\u7ed3\u679c\u7684\u8bc1\u660e\u53ef\u4e0d\u5bb9\u6613\uff01\u4e0d\u80fd\u62ff\u6765\u76f4\u63a5\u7528\u3002
1. \u8bb0T(n)=cos(x/2)*cos(x/4)*\u2026*cos(x/2^n). \u5219\u53cd\u590d\u5229\u7528\u4e8c\u500d\u89d2\u516c\u5f0f\uff1a2^n*sin(x/(2^n))T(n)=(2^n)cos(x/2)*cos(x/4)*\u2026*cos(x/2^n))*sin(x/(2^n))=sinx. t->0\u65f6sint~t, \u6240\u4ee5n->\u221e\u65f6(2^n)sin(x/(2^n))->x. \u6545n->\u221e\u65f6T(n)->sinx/x.
3. \u8fd9\u5c31\u662fWallis\u516c\u5f0f[(2n)!!]^2/[(2n-1)!!]^2->\u03c0/2. Wallis\u6709\u6bd4\u8f83\u7b80\u5355\u7684\u8bc1\u6cd5\uff0c\u53ef\u4ee5\u53c2\u770b\u767e\u5ea6\u767e\u79d1\u201cWallis\u516c\u5f0f\u201d\u8bcd\u6761\uff0c\u90a3\u91cc\u5199\u5f97\u5f88\u6e05\u695a\u3002
2. \u6b64\u9898\u8f83\u96be\u3002\u5bb9\u6613\u770b\u51fax=\u03c0/2+k\u03c0\u65f6\u5de6\u53f3\u5747\u4e3a0\uff0c\u7531\u6b64\u53ef\u4ee5\u770b\u51fa\u6b64\u5f0f\u662f\u5982\u4f55\u60f3\u5230\u7684\u3002\u6b64\u9898\u4e0eEuler\u7684\u7ecf\u5178\u7ed3\u679csinx=x(1-x²/\u03c0²)(1-x²/2\u03c0²)(1-x²/3\u03c0²)\u2026\u2026\u5f62\u5f0f\u4e0a\u7c7b\u4f3c\uff0c\u8bc1\u6cd5\u4e5f\u7c7b\u4f3c\u3002\u800c\u90a3\u662f\u4e00\u4e2a\u6bd4\u8f83\u56f0\u96be\u7684\u95ee\u9898(\u5f53\u5e74Euler\u53ea\u662f\u731c\u51fa\u6765\u4e86\uff0c\u6ca1\u6709\u4e25\u683c\u8bc1\u660e)\uff0c\u6709\u521d\u7b49\u89e3\u6cd5\uff0c\u4f46\u8f83\u957f\u3002LZ\u53ef\u4ee5\u53c2\u770b\u300a\u6570\u5b66\u5206\u6790\u4e60\u9898\u8bfe\u8bb2\u4e49\u300b(\u8c22\u60e0\u6c11\u7b49\u56db\u4eba\u8457\uff0c\u9ad8\u7b49\u6559\u80b2\u51fa\u7248\u793e)\u4e0b\u518c\u4f8b13.4.3. \u6b64\u9898\u7684\u5177\u4f53\u8fc7\u7a0b\u592a\u957f\u6211\u5c31\u4e0d\u5199\u4e86\uff0cLZ\u53ef\u4ee5\u770b\u6211\u8bf4\u7684\u90a3\u672c\u4e66\u3002\u5982\u679c\u6ca1\u6709\u7684\u8bdd\u53ef\u4ee5\u8ffd\u95ee\u6211\uff0c\u6211\u5199\u4e2a\u7b80\u8981\u601d\u8def:-)
∴tanA＋tanB＝1－tanA·tanB ∴tan(A＋B)＝1
∵A、B为锐角 ∴0＜A＋B＜π ∴A＋B＝π/4
37、tan(θ＋ψ)＝(tanθ＋tanψ)/(1－tanθtanψ)＝1
∵θ、ψ为锐角 ∴0º＜θ＋ψ＜180º ∴θ＋ψ＝45º
38、
绛旓細鈭礎銆丅涓洪攼瑙 鈭0锛淎锛婤锛溝 鈭碅锛婤锛澫/4 37銆乼an(胃锛嬒)锛(tan胃锛媡an蠄)/(1锛峵an胃tan蠄)锛1 鈭滴搞佅堜负閿愯 鈭0º锛溛革紜蠄锛180º 鈭次革紜蠄锛45º38銆
绛旓細=(sina^4cosa^4-2sina^2*cosa^2+2)/(sina^2*cosa^2)浠ina^2*cosa^2=x,鍗硏=sin2a^2/4,(0鈮鈮1/4)鍘熷紡=(x^2-2x+2)/x=x+2/x-2,(0鈮鈮1/4)寰堝鏄璇佹槑鍑芥暟y=x+2/x-2,鍦0鈮鈮も垰2涔嬮棿鏄掑噺鍑芥暟銆傛墍浠ワ紝鍘熷紡鈮1/4+8-2=25/4 鎵浠ワ紝鍘熷紡鎴愮珛 ...
绛旓細鈭1+sinx=[sin(x/2) + cos(x/2)]^2 锛 1-sinx=[cos(x/2) - sin(x/2)]^2 鈭礳osx=[cos(x/2)]^2 - [sin(x/2)]^2 =[cos(x/2) + sin(x/2)]脳[cos(x/2) - sin(x/2)]鈭寸瓑寮忓乏杈瑰垎瀛=[sin(x/2) + cos(x/2)]脳{[sin(x/2) + cos(x/2)]+[cos(x/...
绛旓細锛1-cos^2 伪)/(sin伪-cos伪)-(sin伪+cos伪)/(tan^2 伪-1)=sin^2伪(sin伪+cos伪)/(sin^2伪锛峜os^2伪)-(sin伪+cos伪)cos^2伪/(sin^2伪锛峜os^2伪)=(sin伪+cos伪)*(sin^2伪锛峜os^2伪)/(sin^2伪锛峜os^2伪)=sin伪+cos伪 (2)(2-cos^2 伪)(2+tan^2 伪)=(1+sin...
绛旓細=1/2*cos2b 鍙坈os2a=1/2*cos2b 鍗虫湁:宸﹁竟=鍙宠竟 [1-(tana/2)^2]/[1+(tana/2)^2]=[1-(tanb)^2]/2[1+(tanb)^2]涓囪兘鍏紡璇佹槑:cosa=[(cosa/2)^2-(sina/2)^2]/[(cosa/2)^2+(sina/2)^2] ,(鍒嗗瓙,鍒嗘瘝鍚岄櫎(cosa/2)^2寰 =[1-(tana/2)^2]/[1+(tana/2)^2]...
绛旓細鍥炵瓟锛1=sinasina+cosacosa鍙樻崲 鍒嗗瓙瀹屽叏骞虫柟寮,鍜屽垎姣嶇害鍒,鍐嶅垎瀛愬垎姣嶅悓闄や互cosa
绛旓細sin^2胃cos^2胃+sin^2胃+cos^4胃=1 宸﹁竟=cos^2胃锛坰in^2胃+cos^2胃锛+sin^2胃 =cos^2胃+sin^2胃 =1=鍙宠竟锛 鎵浠ワ紝绛夊紡鎴愮珛 (1-cosa)/sina=sina/(1+cosa)宸﹁竟=(1-cosa)(1+cosa)/sina(1+cosa)=(1-cos^2a)/sina(1+cosa)=sin^2a/sina(1+cosa)=sina/(1+cosa)= 鍙宠竟锛...
绛旓細1鎸夊畬鍏ㄥ钩鏂瑰睍寮锛屼袱涓钩鏂归」鐩稿姞寰1锛岃屼腑闂撮」姝ｅソ鏄簩鍊嶈鍏紡锛屼笉缁嗚浜2杩閬撻鎶婂彸杈瑰涓囪兘鍏紡搴旇闈炲父绠鍗曪紝鎻愮ず涓鐐瑰氨鏄簲鐢ㄤ竾鑳藉叕寮忔椂锛屽彲浠ュ湪杩欓噷鎶妕an a璁炬垚x锛屾瘮杈冨鏄撲功鍐欙紝鐢ㄤ竾鑳藉叕寮忚В棰橈紝閫氬父鎴戦兘杩欐牱鍋氾紒3杩欓亾棰樺師棰樹綘娌″姞鎷彿锛岃繖绉嶄功鍐欎笉姝ｇ‘銆傝В鏃讹紝鍏堟妸鍒嗘瘝涔樺埌鍙宠竟鍘伙紝鎴愪负...
绛旓細cos(4伪)+4cos(2伪)+3 =2cos²(2伪)-1+4cos(2伪)+3 =2cos²(2伪)+4cos(2伪)+2 =2[cos(2伪)+1]²=2(2cos²伪-1+1)²=2(2cos²伪)²=2路4cos⁴伪 =8cos⁴伪 2.[1+sin(2伪)]/[2cos²伪+sin(2伪)]=(...
绛旓細杩閬撻鐩灏辨槸鎺屾彙涓浜涓夎鍑芥暟鍜屽樊鍊嶈鍏紡绛夌瓑 鍊掓帹 2sin2b-sin2a=sin2(a+b)4sinbcosb-2sinacosa=2sin锛坅+b锛塩os锛坅+b锛2sinbcosb-sinacosa=锛坰inacosb+cosasinb锛*锛坈osacosb-sinasinb锛2sinbcosb-sinacosa=sinacosa(cosb)^2-sinbcosb(sina)^2 +sinbcosb(cosa)^2-sinacosa(sinb)^2 ...
