三角函数 积化和差、和差化积公式





积化和差公式



公式1

s i n A ⋅ s i n B = − 1 2 [ c o s ( A + B ) − c o s ( A − B ) ] \mathrm{sin}A\cdot\mathrm{sin}B=-\dfrac{1}{2}[\mathrm{cos}(A+B)-\mathrm{cos}(A-B)] sinAsinB=21[cos(A+B)cos(AB)].


− 1 2 [ c o s ( A + B ) − c o s ( A − B ) ] -\dfrac{1}{2}[\mathrm{cos}(A+B)-\mathrm{cos}(A-B)] 21[cos(A+B)cos(AB)]
= − 1 2 ( c o s A ⋅ c o s B − s i n A ⋅ s i n B − c o s A ⋅ c o s B − s i n A ⋅ s i n B ) =-\dfrac{1}{2}(\mathrm{cos}A\cdot\mathrm{cos}B-\mathrm{sin}A\cdot\mathrm{sin}B-\mathrm{cos}A\cdot\mathrm{cos}B-\mathrm{sin}A\cdot\mathrm{sin}B) =21(cosAcosBsinAsinBcosAcosBsinAsinB)
= s i n A ⋅ s i n B =\mathrm{sin}A\cdot\mathrm{sin}B =sinAsinB



公式2

c o s A ⋅ c o s B = 1 2 [ c o s ( A + B ) + c o s ( A − B ) ] \mathrm{cos}A\cdot\mathrm{cosB}=\dfrac{1}{2}[\mathrm{cos}(A+B)+\mathrm{cos}(A-B)] cosAcosB=21[cos(A+B)+cos(AB)].


1 2 [ c o s ( A + B ) + c o s ( A − B ) ] \dfrac{1}{2}[\mathrm{cos}(A+B)+\mathrm{cos}(A-B)] 21[cos(A+B)+cos(AB)]
= 1 2 ⋅ ( c o s A ⋅ c o s B − s i n A ⋅ s i n B + c o s A ⋅ c o s B + s i n A ⋅ s i n B ) =\dfrac{1}{2}\cdot(\mathrm{cos}A\cdot\mathrm{cos}B-\mathrm{sin}A\cdot\mathrm{sin}B+\mathrm{cos}A\cdot\mathrm{cos}B+\mathrm{sin}A\cdot\mathrm{sin}B) =21(cosAcosBsinAsinB+cosAcosB+sinAsinB)
= c o s A ⋅ c o s B =\mathrm{cos}A\cdot\mathrm{cosB} =cosAcosB



公式3

s i n A ⋅ c o s B = 1 2 ⋅ [ s i n ( A + B ) + s i n ( A − B ) ] \mathrm{sinA}\cdot\mathrm{cos}B=\dfrac{1}{2}\cdot[\mathrm{sin}(A+B)+\mathrm{sin}(A-B)] sinAcosB=21[sin(A+B)+sin(AB)].


1 2 ⋅ [ s i n ( A + B ) + s i n ( A − B ) ] \dfrac{1}{2}\cdot[\mathrm{sin}(A+B)+\mathrm{sin}(A-B)] 21[sin(A+B)+sin(AB)]
= 1 2 ⋅ ( s i n A ⋅ c o s B + c o s A ⋅ s i n B + s i n A ⋅ c o s B − c o s A ⋅ s i n B ) =\dfrac{1}{2}\cdot(\mathrm{sin}A\cdot\mathrm{cos}B+\mathrm{cos}A\cdot\mathrm{sin}B+\mathrm{sin}A\cdot\mathrm{cos}B-\mathrm{cos}A\cdot\mathrm{sin}B) =21(sinAcosB+cosAsinB+sinAcosBcosAsinB)
= s i n A ⋅ c o s B =\mathrm{sinA}\cdot\mathrm{cos}B =sinAcosB



公式4

c o s A ⋅ s i n B = 1 2 ⋅ [ s i n ( A + B ) − s i n ( A − B ) ] \mathrm{cos}A\cdot\mathrm{sin}B=\dfrac{1}{2}\cdot[\mathrm{sin}(A+B)-\mathrm{sin}(A-B)] cosAsinB=21[sin(A+B)sin(AB)].


1 2 ⋅ [ s i n ( A + B ) − s i n ( A − B ) ] \dfrac{1}{2}\cdot[\mathrm{sin}(A+B)-\mathrm{sin}(A-B)] 21[sin(A+B)sin(AB)]
= 1 2 ⋅ ( s i n A ⋅ c o s B + c o s A ⋅ s i n B − s i n A ⋅ c o s B + c o s A ⋅ s i n B ) =\dfrac{1}{2}\cdot(\mathrm{sin}A\cdot\mathrm{cos}B+\mathrm{cos}A\cdot\mathrm{sin}B-\mathrm{sin}A\cdot\mathrm{cos}B+\mathrm{cos}A\cdot\mathrm{sin}B) =21(sinAcosB+cosAsinBsinAcosB+cosAsinB)
= c o s A ⋅ s i n B =\mathrm{cos}A\cdot\mathrm{sin}B =cosAsinB





和差化积公式



公式1

s i n A + s i n B = 2 ⋅ s i n A + B 2 ⋅ c o s A − B 2 . \mathrm{sin}A+\mathrm{sin}B=2\cdot\mathrm{sin}\dfrac{A+B}{2}\cdot\mathrm{cos}\dfrac{A-B}{2}. sinA+sinB=2sin2A+Bcos2AB.


2 ⋅ s i n A + B 2 ⋅ c o s A − B 2 2\cdot\mathrm{sin}\dfrac{A+B}{2}\cdot\mathrm{cos}\dfrac{A-B}{2} 2sin2A+Bcos2AB

= 2 ⋅ ( s i n A 2 ⋅ c o s B 2 + c o s A 2 ⋅ s i n B 2 ) ⋅ ( c o s A 2 ⋅ c o s B 2 + s i n A 2 ⋅ s i n B 2 ) =2\cdot(\mathrm{sin}\dfrac{A}{2}\cdot\mathrm{cos}\dfrac{B}{2}+\mathrm{cos}\dfrac{A}{2}\cdot\mathrm{sin}\dfrac{B}{2})\cdot(\mathrm{cos}\dfrac{A}{2}\cdot\mathrm{cos}\dfrac{B}{2}+\mathrm{sin}\dfrac{A}{2}\cdot\mathrm{sin}\dfrac{B}{2}) =2(sin2Acos2B+cos2Asin2B)(cos2Acos2B+sin2Asin2B)

= 2 ⋅ ( s i n A 2 c o s A 2 c o s 2 B 2 + c o s 2 A 2 s i n B 2 c o s B 2 + s i n 2 A 2 s i n B 2 c o s B 2 + s i n 2 B 2 s i n A 2 c o s A 2 ) =2\cdot(\mathrm{sin}\dfrac{A}{2}\mathrm{cos}\dfrac{A}{2}\mathrm{cos}^2\dfrac{B}{2}+\mathrm{cos}^2\dfrac{A}{2}\mathrm{sin}\dfrac{B}{2}\mathrm{cos}\dfrac{B}{2}+\mathrm{sin}^2\dfrac{A}{2}\mathrm{sin}\dfrac{B}{2}\mathrm{cos}\dfrac{B}{2}+\mathrm{sin}^2\dfrac{B}{2}\mathrm{sin}\dfrac{A}{2}\mathrm{cos}\dfrac{A}{2}) =2(sin2Acos2Acos22B+cos22Asin2Bcos2B+sin22Asin2Bcos2B+sin22Bsin2Acos2A)

= 2 s i n A 2 c o s A 2 ⋅ ( s i n 2 B 2 + c o s 2 B 2 ) + 2 s i n B 2 c o s B 2 ⋅ ( s i n 2 A 2 + c o s 2 A 2 ) =2\mathrm{sin}\dfrac{A}{2}\mathrm{cos}\dfrac{A}{2}\cdot(\mathrm{sin}^2\dfrac{B}{2}+\mathrm{cos}^2\dfrac{B}{2})+2\mathrm{sin}\dfrac{B}{2}\mathrm{cos}\dfrac{B}{2}\cdot(\mathrm{sin}^2\dfrac{A}{2}+\mathrm{cos}^2\dfrac{A}{2}) =2sin2Acos2A(sin22B+cos22B)+2sin2Bcos2B(sin22A+cos22A)

= s i n A + s i n B =\mathrm{sin}A+\mathrm{sin}B =sinA+sinB



公式2

s i n A − s i n B = 2 ⋅ c o s A + B 2 ⋅ s i n A − B 2 . \mathrm{sin}A-\mathrm{sin}B=2\cdot\mathrm{cos}\dfrac{A+B}{2}\cdot\mathrm{sin}\dfrac{A-B}{2}. sinAsinB=2cos2A+Bsin2AB.


2 ⋅ c o s A + B 2 ⋅ s i n A − B 2 2\cdot\mathrm{cos}\dfrac{A+B}{2}\cdot\mathrm{sin}\dfrac{A-B}{2} 2cos2A+Bsin2AB

= 2 ⋅ ( c o s A 2 c o s B 2 − s i n A 2 s i n B 2 ) ⋅ ( s i n A 2 c o s B 2 − c o s A 2 s i n B 2 ) =2\cdot(\mathrm{cos}\dfrac{A}{2}\mathrm{cos}\dfrac{B}{2}-\mathrm{sin}\dfrac{A}{2}\mathrm{sin}\dfrac{B}{2})\cdot(\mathrm{sin}\dfrac{A}{2}\mathrm{cos}\dfrac{B}{2}-\mathrm{cos}\dfrac{A}{2}\mathrm{sin}\dfrac{B}{2}) =2(cos2Acos2Bsin2Asin2B)(sin2Acos2Bcos2Asin2B)

= 2 ⋅ ( s i n A 2 c o s A 2 c o s 2 B 2 − s i n 2 A 2 s i n B 2 c o s B 2 − c o s 2 A 2 s i n B 2 c o s B 2 + s i n A 2 c o s A 2 s i n 2 B 2 ) =2\cdot(\mathrm{sin}\dfrac{A}{2}\mathrm{cos}\dfrac{A}{2}\mathrm{cos}^2\dfrac{B}{2}-\mathrm{sin}^2\dfrac{A}{2}\mathrm{sin}\dfrac{B}{2}\mathrm{cos}\dfrac{B}{2}-\mathrm{cos}^2\dfrac{A}{2}\mathrm{sin}\dfrac{B}{2}\mathrm{cos}\dfrac{B}{2}+\mathrm{sin}\dfrac{A}{2}\mathrm{cos}\dfrac{A}{2}\mathrm{sin}^2\dfrac{B}{2}) =2(sin2Acos2Acos22Bsin22Asin2Bcos2Bcos22Asin2Bcos2B+sin2Acos2Asin22B)

= 2 s i n A 2 c o s A 2 ⋅ ( s i n 2 B 2 + c o s 2 B 2 ) − 2 s i n B 2 c o s B 2 ⋅ ( s i n 2 A 2 + c o s 2 A 2 ) =2\mathrm{sin}\dfrac{A}{2}\mathrm{cos}\dfrac{A}{2}\cdot(\mathrm{sin}^2\dfrac{B}{2}+\mathrm{cos}^2\dfrac{B}{2})-2\mathrm{sin}\dfrac{B}{2}\mathrm{cos}\dfrac{B}{2}\cdot(\mathrm{sin}^2\dfrac{A}{2}+\mathrm{cos}^2\dfrac{A}{2}) =2sin2Acos2A(sin22B+cos22B)2sin2Bcos2B(sin22A+cos22A)

= s i n A − s i n B =\mathrm{sin}A-\mathrm{sin}B =sinAsinB



公式3

c o s A + c o s B = 2 ⋅ c o s A + B 2 c o s A − B 2 \mathrm{cos}A+\mathrm{cos}B=2\cdot\mathrm{cos}\dfrac{A+B}{2}\mathrm{cos}\dfrac{A-B}{2} cosA+cosB=2cos2A+Bcos2AB.


2 ⋅ c o s A + B 2 c o s A − B 2 2\cdot\mathrm{cos}\dfrac{A+B}{2}\mathrm{cos}\dfrac{A-B}{2} 2cos2A+Bcos2AB

= 2 ⋅ ( c o s A 2 c o s B 2 − s i n A 2 s i n B 2 ) ⋅ ( c o s A 2 c o s B 2 + s i n A 2 s i n B 2 ) =2\cdot(\mathrm{cos}\dfrac{A}{2}\mathrm{cos}\dfrac{B}{2}-\mathrm{sin}\dfrac{A}{2}\mathrm{sin}\dfrac{B}{2})\cdot(\mathrm{cos}\dfrac{A}{2}\mathrm{cos}\dfrac{B}{2}+\mathrm{sin}\dfrac{A}{2}\mathrm{sin}\dfrac{B}{2}) =2(cos2Acos2Bsin2Asin2B)(cos2Acos2B+sin2Asin2B)

= 2 c o s 2 A 2 c o s 2 B 2 − 2 s i n 2 A 2 s i n 2 B 2 =2\mathrm{cos}^2\dfrac{A}{2}\mathrm{cos}^2\dfrac{B}{2}-2\mathrm{sin}^2\dfrac{A}{2}\mathrm{sin}^2\dfrac{B}{2} =2cos22Acos22B2sin22Asin22B

= 2 c o s 2 A 2 ( 1 − s i n 2 B 2 ) − 2 s i n 2 B 2 ( 1 − c o s 2 A 2 ) =2\mathrm{cos}^2\dfrac{A}{2}(1-\mathrm{sin}^2\dfrac{B}{2})-2\mathrm{sin}^2\dfrac{B}{2}(1-\mathrm{cos}^2\dfrac{A}{2}) =2cos22A(1sin22B)2sin22B(1cos22A)

= 2 c o s 2 A 2 − 2 c o s 2 A 2 s i n 2 B 2 − 2 s i n 2 B 2 + 2 c o s 2 A 2 s i n 2 B 2 =2\mathrm{cos}^2\dfrac{A}{2}-2\mathrm{cos}^2\dfrac{A}{2}\mathrm{sin}^2\dfrac{B}{2}-2\mathrm{sin}^2\dfrac{B}{2}+2\mathrm{cos}^2\dfrac{A}{2}\mathrm{sin}^2\dfrac{B}{2} =2cos22A2cos22Asin22B2sin22B+2cos22Asin22B

= 2 c o s 2 A 2 − 1 + 1 − 2 s i n 2 B 2 =2\mathrm{cos}^2\dfrac{A}{2}-1+1-2\mathrm{sin}^2\dfrac{B}{2} =2cos22A1+12sin22B

= c o s A + c o s B =\mathrm{cos}A+\mathrm{cos}B =cosA+cosB



公式4

c o s A − c o s B = − 2 s i n A + B 2 s i n A − B 2 \mathrm{cos}A-\mathrm{cos}B=-2\mathrm{sin}\dfrac{A+B}{2}\mathrm{sin}\dfrac{A-B}{2} cosAcosB=2sin2A+Bsin2AB.


− 2 s i n A + B 2 s i n A − B 2 -2\mathrm{sin}\dfrac{A+B}{2}\mathrm{sin}\dfrac{A-B}{2} 2sin2A+Bsin2AB

− 2 ⋅ ( s i n A 2 c o s B 2 + c o s A 2 s i n B 2 ) ⋅ ( s i n A 2 c o s B 2 − c o s A 2 s i n B 2 ) -2\cdot(\mathrm{sin}\dfrac{A}{2}\mathrm{cos}\dfrac{B}{2}+\mathrm{cos}\dfrac{A}{2}\mathrm{sin}\dfrac{B}{2})\cdot(\mathrm{sin}\dfrac{A}{2}\mathrm{cos}\dfrac{B}{2}-\mathrm{cos}\dfrac{A}{2}\mathrm{sin}\dfrac{B}{2}) 2(sin2Acos2B+cos2Asin2B)(sin2Acos2Bcos2Asin2B)

= 2 c o s 2 A 2 s i n 2 B 2 − 2 s i n 2 A 2 c o s 2 B 2 =2\mathrm{cos}^2\dfrac{A}{2}\mathrm{sin}^2\dfrac{B}{2}-2\mathrm{sin}^2\dfrac{A}{2}\mathrm{cos}^2\dfrac{B}{2} =2cos22Asin22B2sin22Acos22B

= 2 c o s 2 A 2 ( 1 − c o s 2 B 2 ) − 2 c o s 2 B 2 ( 1 − c o s 2 A 2 ) =2\mathrm{cos}^2\dfrac{A}{2}(1-\mathrm{cos}^2\dfrac{B}{2})-2\mathrm{cos}^2\dfrac{B}{2}(1-\mathrm{cos}^2\dfrac{A}{2}) =2cos22A(1cos22B)2cos22B(1cos22A)

= 2 c o s 2 A 2 − 2 c o s 2 A 2 c o s 2 B 2 − 2 c o s 2 B 2 + 2 c o s 2 A 2 c o s 2 B 2 =2\mathrm{cos}^2\dfrac{A}{2}-2\mathrm{cos}^2\dfrac{A}{2}\mathrm{cos}^2\dfrac{B}{2}-2\mathrm{cos}^2\dfrac{B}{2}+2\mathrm{cos}^2\dfrac{A}{2}\mathrm{cos}^2\dfrac{B}{2} =2cos22A2cos22Acos22B2cos22B+2cos22Acos22B

= 2 c o s 2 A 2 − 1 + 1 − 2 c o s 2 B 2 =2\mathrm{cos}^2\dfrac{A}{2}-1+1-2\mathrm{cos}^2\dfrac{B}{2} =2cos22A1+12cos22B

= c o s A − c o s B =\mathrm{cos}A-\mathrm{cos}B =cosAcosB





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