高一数学诱导公式的题 一道高中诱导公式的数学题。
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sin(\u03c0+\u03b1)= - sin\u03b1= - 1/3
sin\u03b1=1/3
cos(5\u03c0+\u03b1)=cos[4\u03c0+(\u03c0+\u03b1)]=cos(\u03c0+\u03b1)= - cos\u03b1
cos\u03b1=\u00b1\u221a(1-sin^2(\u03b1))=\u00b12\u221a2/3
cos(5\u03c0+\u03b1)=\u00b12\u221a2/3
\u2461
cos(\u03c0-\u03b1)= - cos\u03b1= - 1/3
cos\u03b1=1/3
sin(5\u03c0+\u03b1)=sin[4\u03c0+(\u03c0+\u03b1)]=sin(\u03c0+\u03b1)= - sin\u03b1
\u03b1\u662f\u56db\u8c61\u9650\u89d2\uff0c\u6240\u4ee5\uff0csin\u03b1<0
sin\u03b1= - \u221a1-(1/3)^2 = - 2\u221a2/3
\u6240\u4ee5sin(5\u03c0+\u03b1)= - 2\u221a2/3
(a+b)(a^2-ab+b^2)=a^3+b^3
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=3-cos(180-2x)=3+cos2x
或f(cosx)= f(sin(90+x))=3-cos(2(90+x))
=3-cos(180+2x)=3+cos2x
选C
c
倍角公式
Cos2A==1-2Sin^2 a= 2Cos^2 a -1
f(sinx)=3-cos2x
f(sinx)=3-(1-2sin^2 x)
即f(x)=3-(1-2x^2)
f(cosx)= 3-（1-2cos^2 x)
f(cosx)= 3+（2cos^2 x-1)
f(cosx)= 3+cos2x
选C
绛旓細绛旀 锛 sina tana(cosa-sina)+(sina+tana)/(cota+csca)= sina * (cosa - sina) / cosa + (sina + sina/cosa) / (cosa/sina + 1/sina)= sina - (sina)^2 / cosa + (sina)^2 * (1 + 1/cosa) / (1 + cosa)= sina - (sina)^2 / cosa + (sina)^2 * (cosa ...
绛旓細cos120 =cos(180-60)=-cos60=-1/2
绛旓細娉ㄦ剰锛璇卞鍏紡鐨瑙捨变笉鏄0鍒跋/2涔嬮棿 鏄换鎰忚锛屼负浜嗚蹇嗚瀵煎叕寮忥紝鑰佸笀缁欏彛璇锛氬鍙樺伓涓嶅彉锛岀鍙风湅璞￠檺锛屾棤璁何变负浣曡锛屽潎鐪嬫垚閿愯 銆愬墠闈㈢殑绗﹀彿鏄袱杈逛笁瑙掑嚱鏁板肩殑鍏崇郴绗﹀彿銆戣繖閲岀殑伪涓轰换鎰忚锛屼负浜嗙湅娓呮璞￠檺锛屾悶娓呯鍙峰皢伪 鐪嬫垚閿愯 log(8)(1/4)=log₂(1/4)/log₂8=-2/3...
绛旓細3. 鍘熷紡=(-3sina+cosa)/(-4sina+cosa)=(-3tana+1)/(-4tana+1)=2 -3tana+1=-8tana+2 5tana=1 tana=1/5
绛旓細1銆佲垰cos^2(4蟺/3+2蟺)=鈭歝os^2(4蟺/3)=鈭(-1/2锛塣2=1/2 2銆乧os(-80)=cos(80)=k,cos(100)=-cos(80)=-k,sin^2(100)=1-cos^2(100)=1-k^2 鍙堝洜涓簊in100>0锛屾墍浠in(100)=鈭(1-k^2),鎵浠an100=sin100/cos100=-鈭(1-k^2)/k 3銆乼an(a-7蟺)=tan(a)=-3/...
绛旓細鎶夿-C鐪嬫垚涓浣 鍗矨+(B-C)鍜孉-(B-C)鍏紡灞曞紑灏变负 sinAcos(B-C)+cosAsin(B-C)=sinAcos(B-C)-cosAsin(B-C)宸﹀彸涓よ竟娑堝幓sinAcos(B-C锛夋暣鐞嗗緱 cosAsin(B-C)=0 鍗砪osA=0鎴栬卻in(B-C)=0 褰揷osA=0鏃 A涓90搴﹁ 姝ゆ椂鏄洿瑙掍笁瑙掑舰 褰搒in(B-C)=0鏃 B=C 姝ゆ椂鏄瓑鑵颁笁瑙掑舰 绛旀...
绛旓細杩欎釜鏄璇卞鍏紡锛氱涓涓搴旂殑鍏紡涓猴細cos(270-a)=-sina 绗簩涓搴旂殑鍏紡涓猴細sin(180-a)=sina 绗竴缁勶細sin(180+a)=-sina cos(180+a)=-cosa tan(180+a)=tana cot(180+a)=cota 绗簩缁勶細sin(180-a)=sina cos(180-a)=-cosa tan(180-a)=-tana cot(180-a)=-cota 绗笁缁勶細sin(...
绛旓細100搴﹀湪绗簩璞￠檺锛屾鏃秙in鍊间负姝 鎵浠ョ敱cos100=k锛泂in100^2+cos100^2=1 寰梥in100=鈭1-k^2 鎵浠an100=sin100/cos100=锛堚垰1-k^2锛/k 鏍规嵁璇卞鍏紡锛歵an(180-100)=-tan100=tan80 鎵浠an80=-锛堚垰1-k^2锛/k
绛旓細1锛歴in(蟺+A)=1/4,sin(2蟺鈥擜锛夆攃ot(A-蟺锛変箻浠osA 鐨勫=-sinA+cosA*cosA/sinA,鍙坰in(蟺+A)=1/4锛宻inA=-1/4銆傛墍浠ワ紝sin(2蟺鈥擜锛-cot(A-蟺锛変箻浠osA 鐨勫=1/4+[1-锛-1/4锛塣2]/(-1/4)=1/4-4+1/4=-7/2锛2锛氬寲绠锛歵anA(cosA-sinA)+(sinA+tanA)/(cotA+cscA...
绛旓細绛旀閫夋嫨锛坆锛夐鍏堢敱鍓嶄竴涓紡瀛愬緱鈥-sina-sina=-a鈥濇墍浠モ渟ina=锛1/2锛塧鈥濇帴鐫鍖栫畝绗簩涓紡瀛愬緱鈥-sina-2*sina=-3*sina鈥濇墍浠ョ瓟妗堢瓑浜庘滐紙-3/2锛塧鈥濆仛杩欑棰樼洰鍙互鍏堢粨鍚堝浘鍍忥紝鍥惧儚瑕佺啛銆
