cosx等于多少 必采纳

sinx\u00b7cosx+cosx\u00b7cosx \u7b49\u4e8e\u591a\u5c11\uff1f

\u4e09\u89d2\u51fd\u6570\u516c\u5f0f
\u4e24\u89d2\u548c\u516c\u5f0f
sin(A+B) = sinAcosB+cosAsinB
sin(A-B) = sinAcosB-cosAsinB ?
cos(A+B) = cosAcosB-sinAsinB
cos(A-B) = cosAcosB+sinAsinB
tan(A+B) = (tanA+tanB)/(1-tanAtanB)
tan(A-B) = (tanA-tanB)/(1+tanAtanB)
cot(A+B) = (cotAcotB-1)/(cotB+cotA) ?
cot(A-B) = (cotAcotB+1)/(cotB-cotA)
\u500d\u89d2\u516c\u5f0f
tan2A = 2tanA/(1-tan^2 A)
Sin2A=2SinA•CosA
Cos2A = Cos^2 A--Sin^2 A
=2Cos^2 A\u20141
=1\u20142sin^2 A
\u4e09\u500d\u89d2\u516c\u5f0f
sin3A = 3sinA-4(sinA)^3;
cos3A = 4(cosA)^3 -3cosA
tan3a = tan a • tan(\u03c0/3+a)• tan(\u03c0/3-a)
\u534a\u89d2\u516c\u5f0f
sin(A/2) = \u221a{(1--cosA)/2}
cos(A/2) = \u221a{(1+cosA)/2}
tan(A/2) = \u221a{(1--cosA)/(1+cosA)}
cot(A/2) = \u221a{(1+cosA)/(1-cosA)} ?
tan(A/2) = (1--cosA)/sinA=sinA/(1+cosA)
\u548c\u5dee\u5316\u79ef
sin(a)+sin(b) = 2sin[(a+b)/2]cos[(a-b)/2]
sin(a)-sin(b) = 2cos[(a+b)/2]sin[(a-b)/2]
cos(a)+cos(b) = 2cos[(a+b)/2]cos[(a-b)/2]
cos(a)-cos(b) = -2sin[(a+b)/2]sin[(a-b)/2]
tanA+tanB=sin(A+B)/cosAcosB
\u79ef\u5316\u548c\u5dee
sin(a)sin(b) = -1/2*[cos(a+b)-cos(a-b)]
cos(a)cos(b) = 1/2*[cos(a+b)+cos(a-b)]
sin(a)cos(b) = 1/2*[sin(a+b)+sin(a-b)]
cos(a)sin(b) = 1/2*[sin(a+b)-sin(a-b)]
\u8bf1\u5bfc\u516c\u5f0f
sin(-a) = -sin(a)
cos(-a) = cos(a)
sin(\u03c0/2-a) = cos(a)
cos(\u03c0/2-a) = sin(a)
sin(\u03c0/2+a) = cos(a)
cos(\u03c0/2+a) = -sin(a)
sin(\u03c0-a) = sin(a)
cos(\u03c0-a) = -cos(a)
sin(\u03c0+a) = -sin(a)
cos(\u03c0+a) = -cos(a)
tgA=tanA = sinA/cosA
\u4e07\u80fd\u516c\u5f0f
sin(a) = [2tan(a/2)] / {1+[tan(a/2)]^2}
cos(a) = {1-[tan(a/2)]^2} / {1+[tan(a/2)]^2}
tan(a) = [2tan(a/2)]/{1-[tan(a/2)]^2}
\u5176\u5b83\u516c\u5f0f
a•sin(a)+b•cos(a) = [\u221a(a^2+b^2)]*sin(a+c) [\u5176\u4e2d\uff0ctan(c)=b/a]
a•sin(a)-b•cos(a) = [\u221a(a^2+b^2)]*cos(a-c) [\u5176\u4e2d\uff0ctan(c)=a/b]
1+sin(a) = [sin(a/2)+cos(a/2)]^2;
1-sin(a) = [sin(a/2)-cos(a/2)]^2;;
\u5176\u4ed6\u975e\u91cd\u70b9\u4e09\u89d2\u51fd\u6570
csc(a) = 1/sin(a)
sec(a) = 1/cos(a)
\u53cc\u66f2\u51fd\u6570
sinh(a) = [e^a-e^(-a)]/2
cosh(a) = [e^a+e^(-a)]/2
tg h(a) = sin h(a)/cos h(a)
\u516c\u5f0f\u4e00\uff1a
\u8bbe\u03b1\u4e3a\u4efb\u610f\u89d2\uff0c\u7ec8\u8fb9\u76f8\u540c\u7684\u89d2\u7684\u540c\u4e00\u4e09\u89d2\u51fd\u6570\u7684\u503c\u76f8\u7b49\uff1a
sin\uff082k\u03c0\uff0b\u03b1\uff09= sin\u03b1
cos\uff082k\u03c0\uff0b\u03b1\uff09= cos\u03b1
tan\uff082k\u03c0\uff0b\u03b1\uff09= tan\u03b1
cot\uff082k\u03c0\uff0b\u03b1\uff09= cot\u03b1
\u516c\u5f0f\u4e8c\uff1a
\u8bbe\u03b1\u4e3a\u4efb\u610f\u89d2\uff0c\u03c0+\u03b1\u7684\u4e09\u89d2\u51fd\u6570\u503c\u4e0e\u03b1\u7684\u4e09\u89d2\u51fd\u6570\u503c\u4e4b\u95f4\u7684\u5173\u7cfb\uff1a
sin\uff08\u03c0\uff0b\u03b1\uff09= -sin\u03b1
cos\uff08\u03c0\uff0b\u03b1\uff09= -cos\u03b1
tan\uff08\u03c0\uff0b\u03b1\uff09= tan\u03b1
cot\uff08\u03c0\uff0b\u03b1\uff09= cot\u03b1
\u516c\u5f0f\u4e09\uff1a
\u4efb\u610f\u89d2\u03b1\u4e0e -\u03b1\u7684\u4e09\u89d2\u51fd\u6570\u503c\u4e4b\u95f4\u7684\u5173\u7cfb\uff1a
sin\uff08-\u03b1\uff09= -sin\u03b1
cos\uff08-\u03b1\uff09= cos\u03b1
tan\uff08-\u03b1\uff09= -tan\u03b1
cot\uff08-\u03b1\uff09= -cot\u03b1
\u516c\u5f0f\u56db\uff1a
\u5229\u7528\u516c\u5f0f\u4e8c\u548c\u516c\u5f0f\u4e09\u53ef\u4ee5\u5f97\u5230\u03c0-\u03b1\u4e0e\u03b1\u7684\u4e09\u89d2\u51fd\u6570\u503c\u4e4b\u95f4\u7684\u5173\u7cfb\uff1a
sin\uff08\u03c0-\u03b1\uff09= sin\u03b1
cos\uff08\u03c0-\u03b1\uff09= -cos\u03b1
tan\uff08\u03c0-\u03b1\uff09= -tan\u03b1
cot\uff08\u03c0-\u03b1\uff09= -cot\u03b1
\u516c\u5f0f\u4e94\uff1a
\u5229\u7528\u516c\u5f0f-\u548c\u516c\u5f0f\u4e09\u53ef\u4ee5\u5f97\u52302\u03c0-\u03b1\u4e0e\u03b1\u7684\u4e09\u89d2\u51fd\u6570\u503c\u4e4b\u95f4\u7684\u5173\u7cfb\uff1a
sin\uff082\u03c0-\u03b1\uff09= -sin\u03b1
cos\uff082\u03c0-\u03b1\uff09= cos\u03b1
tan\uff082\u03c0-\u03b1\uff09= -tan\u03b1
cot\uff082\u03c0-\u03b1\uff09= -cot\u03b1
\u516c\u5f0f\u516d\uff1a
\u03c0/2\u00b1\u03b1\u53ca3\u03c0/2\u00b1\u03b1\u4e0e\u03b1\u7684\u4e09\u89d2\u51fd\u6570\u503c\u4e4b\u95f4\u7684\u5173\u7cfb\uff1a
sin\uff08\u03c0/2+\u03b1\uff09= cos\u03b1
cos\uff08\u03c0/2+\u03b1\uff09= -sin\u03b1
tan\uff08\u03c0/2+\u03b1\uff09= -cot\u03b1
cot\uff08\u03c0/2+\u03b1\uff09= -tan\u03b1
sin\uff08\u03c0/2-\u03b1\uff09= cos\u03b1
cos\uff08\u03c0/2-\u03b1\uff09= sin\u03b1
tan\uff08\u03c0/2-\u03b1\uff09= cot\u03b1
cot\uff08\u03c0/2-\u03b1\uff09= tan\u03b1
sin\uff083\u03c0/2+\u03b1\uff09= -cos\u03b1
cos\uff083\u03c0/2+\u03b1\uff09= sin\u03b1
tan\uff083\u03c0/2+\u03b1\uff09= -cot\u03b1
cot\uff083\u03c0/2+\u03b1\uff09= -tan\u03b1
sin\uff083\u03c0/2-\u03b1\uff09= -cos\u03b1
cos\uff083\u03c0/2-\u03b1\uff09= -sin\u03b1
tan\uff083\u03c0/2-\u03b1\uff09= cot\u03b1
cot\uff083\u03c0/2-\u03b1\uff09= tan\u03b1
(\u4ee5\u4e0ak\u2208Z)
\u8fd9\u4e2a\u7269\u7406\u5e38\u7528\u516c\u5f0f\u6211\u8d39\u4e86\u534a\u5929\u7684\u52b2\u624d\u8f93\u8fdb\u6765,\u5e0c\u671b\u5bf9\u5927\u5bb6\u6709\u7528
A•sin(\u03c9t+\u03b8)+ B•sin(\u03c9t+\u03c6) =
\u221a{(A^2 +B^2 +2ABcos(\u03b8-\u03c6)} • sin{ \u03c9t + arcsin[ (A•sin\u03b8+B•sin\u03c6) / \u221a{A^2 +B^2; +2ABcos(\u03b8-\u03c6)} }
\u221a\u8868\u793a\u6839\u53f7,\u5305\u62ec{\u2026\u2026}\u4e2d\u7684\u5185\u5bb9

 

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