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中考三角函数的公式推导

中考三角函数的公式推导

更新时间:2023-12-08 15:43:11

中考三角函数的公式推导

以下是三角函数的所有公式:

正弦函数公式:

sin(A±B) = sinA cosB ± cosA sinB

sin2A = 2sinA cosA

sinA±B/2 = ±√[(1±cosB)/2] sin(A±B)/2

余弦函数公式:

cos(A±B) = cosA cosB ∓ sinA sinB

cos2A = cos^2 A - sin^2 A = 2cos^2 A - 1 = 1-2sin^2 A

cosA±B/2 = ±√[(1±cosB)/2] cos(A±B)/2

正切函数公式:

tan(A±B) = (tanA ± tanB) / (1 ∓ tanA tanB)

tan2A = 2tanA / (1 - tan^2 A)

tan(A±B/2) = sinB / (cosA ± 1)

余切函数公式:

cotA + cotB = (sinA sinB) / (cosA sinB + sinA cosB)

cot2A = (cot^2 A - 1) / 2 = (1 - tan^2 A) / 2cotA

cot(A±B/2) = cosB / (sinA ± 1)

割函数公式:

secA = 1/cosA,cscA = 1/sinA

以下是三角函数公式的推导公式:

1.双角和差公式:

sin(A±B) = sinA cosB ± cosA sinB (以sin(A+B)的证明为例)

将两个角分别表示成其一半角的正弦和余弦:

sin(A+B) = sin[(A/2)+(A/2)+B]

= sin(A/2)cos(A/2)+cos(A/2)sin(B+A/2)

= 2sin(A/2)(cos(A/2) + 1/2sin(B+A/2))

= 2sin(A/2)(cos(A/2) + cos(B/2)sin(A/2))

= 2sin(A/2)cos(B/2)cos(A/2) + 2(sin(A/2))^2sin(B/2)

= 2sin(A/2)cos(B/2)cos(A/2) + (1-cos^2(A/2))sin(B/2)

= 2sin(A/2)cos(B/2)cos(A/2) + sin(B/2) - cos^2(A/2)sin(B/2)

= sinAcosB+cosAsinB

2.双角公式:

sin2A = 2sinA cosA

(sin2A的证明)

sin2A = sin(A+A) =sinAcosA + cosAsinA= 2sinAcosA

3.半角公式:

sin(A/2) = ±√[(1±cosB)/2],cos(A/2) = ±√[(1±cosB)/2]

以sin(A/2+B/2) = √[(1+cosA)/2]√[(1+cosB)/2] - sin(A/2-B/2)的证明为例

将sin(A/2+B/2)换成sin[(A+B)/2],然后将其表示成cosA和sinB的函数,再代入√[(1+cosA)/2]√[(1+cosB)/2] - sin(A/2-B/2),得到:

√[(1+cosA)/2]√[(1+cosB)/2] - sin(A/2-B/2) = 2sin[(A+B)/4]cos[(A-B)/4]

= √[(1 + cosA)/2]sinB/2 + cosB/2 √[(1 - cosA)/2] - √[(1 + cosA)/2]cosB/2 + sinB/2 √[(1 - cosA)/2]

将两个平方根化简:

√[(1 + cosA)/2]√[(1 - cosA)/2] =√[(1 - cos ^2A)/4] = sinA/2

√[(1 + cosB)/2]√[(1 - cosB)/2] =√[(1 - cos ^2B)/4] = sinB/2

再将其代入上式,整理得:

sin(A/2 + B/2) = √[(1 + cosA)/2]√[(1 + cosB)/2] - sin(A/2 - B/2)

= sin(A/2)cos(B/2) + cos(A/2)sin(B/2)

4.和差化积公式:

cos(A±B) = cosA cosB ∓ sinA sinB

(cos(A-B)的证明)

cos(A-B) = cosAcos(-B) - sinAsin(-B)

= cosAcosB + sinAsinB

和sina=(e^ix - e^-ix)/2, cosa=(e^ix + e^-ix)/2的欧拉公式可以得到exp(ix) = cosa + i sina

cosAcosB + sinAsinB = (e^ia + e^-ia)/2(e^ib + e^-ib)/2 + i(e^ia - e^-ia)/2(e^ib - e^-ib)/2

= 1/2 [(e^ia cosb + e^ib cosa + e^i(a+b) + e^-i(a+b))]/2 + i [(e^ia cosb - e^ib cosa + e^i(a-b) - e^-i(a-b))]/2

= cos(a+b) +0i+ i sin(a-b)+0i

= cos (A-B)+ i sin (A-B)

将等号两边的实部和虚部分别相等,即得:

cos(A-B) = cosA cosB + sinA sinB (实部)

sin(A-B) = sinA cosB - cosA sinB (虚部)

5.商品化公式:

tan(A±B) = (tanA ± tanB) / (1 ∓ tanA tanB)

(tan(A+B)的证明)

tan(A+B) = sin(A+B)/cos(A+B)

= [sinAcosB + cosAsinB]/[cosAcosB - sinAsinB]

= (tanA + tanB)/(1 - tanA tanB)

6.万能欧拉公式:

e^ix = cosx + i sinx

1.当x=π时,得到欧拉公式:e^iπ+1=0

2.取x=-ix,代入欧拉公式,得e^-x = cosx - i sinx

3.对1式两边同时除以e^ix,得到cotx + 1 = tanx,对2式两边同时除以e^ix,得到cosx = (e^ix + e^-ix)/2及sinx = (e^ix - e^-ix)/2i,由此可以得到上述公式。

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