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Verifying mathematics with examples
Formulas of trigonometric functions's derivation and proof

Derived formula: (a+b+c)/(sinA+sinB+sinC)=2R (where r is the radius of the circumscribed circle).

According to sine theorem

a/sinA=b/sinB=c/sinC=2R

therefore

a=2R*sinA

b=2R*sinB

c=2R*sinC

Add up a+b+c=2R*(sinA+sinB+sinC) and bring it in.

(a+b+c)/(sinA+sin b+ sinC)= 2R *(sinA+sin b+ sinC)/(sinA+sin b+ sinC)= 2R

Properties and Derivation of Logarithm

Use 0 to represent the power, and log(a)(b) to represent the logarithm of b with a as the base.

* means multiplication symbol,/means division symbol.

Define formula:

If a n = b(a >;; 0 and a ≠ 1)

Then n=log(a)(b)

Basic nature:

1.a^(log(a)(b))=b

2 . log(a)(MN)= log(a)(M)+log(a)(N);

3 . log(a)(M/N)= log(a)(M)-log(a)(N);

4.log(a)(M^n)=nlog(a)(M)

infer

1. This need not be pushed, but can be obtained directly from the definition (bring [n=log(a)(b)] in the definition into a n = b).

2.MN=M*N

By the basic properties of 1 (replacing m and n)

a^[log(a)(mn)]=a^[log(a)(m)]*a^[log(a)(n)]

According to the nature of the index

a^[log(a)(mn)]=a^{[log(a)(m)]+[log(a)(n)]}

And because exponential function is monotone function, so

log(a)(MN)=log(a)(M)+log(a)(N)

3. Similar to 2.

MN=M/N

By the basic properties of 1 (replacing m and n)

a^[log(a)(m/n)]=a^[log(a)(m)]/a^[log(a)(n)]

According to the nature of the index

a^[log(a)(m/n)]=a^{[log(a)(m)]-[log(a)(n)]}

And because exponential function is monotone function, so

Logarithm (a)(M/N)= Logarithm (a)(M)- Logarithm (a)(N)

4. Similar to 2.

M^n=M^n

From the basic attribute 1 (replace m)

a^[log(a)(m^n)]={a^[log(a)(m)]}^n

According to the nature of the index

a^[log(a)(m^n)]=a^{[log(a)(m)]*n}

And because exponential function is monotone function, so

log(a)(M^n)=nlog(a)(M)

Other attributes:

Attribute 1: bottoming formula

log(a)(N)=log(b)(N)/log(b)(a)

Derived as follows

N=a^[log(a)(N)]

a=b^[log(b)(a)]

By combining the two formulas, it can be concluded that.

n={b^[log(b)(a)]}^[log(a)(n)]=b^{[log(a)(n)]*[log(b)(a)]}

And because n = b [log (b) (n)]

therefore

b^[log(b)(n)]=b^{[log(a)(n)]*[log(b)(a)]}

therefore

log(b)(n)=[log(a)(n)]*[log(b)(a)]

So log(a)(N)=log(b)(N)/log(b)(a)

Nature 2: (I don't know what it's called)

log(a^n)(b^m)=m/n*[log(a)(b)]

Derived as follows

Through the formula [lnx is log (e) (x), and e is called the base of natural logarithm]

log(a^n)(b^m)=ln(a^n)/ln(b^n)

It can be obtained from basic attribute 4.

log(a^n)(b^m)=[n*ln(a)]/[m*ln(b)]=(m/n)*{[ln(a)]/[ln(b)]}

Then according to the bottom changing formula

log(a^n)(b^m)=m/n*[log(a)(b)]

- (

Formula 3:

log(a)(b)= 1/log(b)(a)

Proved as follows:

Log(a)(b)= log(b)(b)/log(b)(a)- Logarithm based on b, log(b)(b)= 1.

= 1/log(b)(a)

Also deformable:

log(a)(b)*log(b)(a)= 1

Two-angle sum formula

sin(A+B) = sinAcosB+cosAsinB

sin(A-B) = sinAcosB-cosAsinB

cos(A+B) = cosAcosB-sinAsinB

cos(A-B) = cosAcosB+sinAsinB

tan(A+B)= 1

Tan ()

cot(A+B)= 1

Kurt (A-B)

Double angle formula

tan2A =

Sin2A=2SinA? Kosa

Cos2A = Cos2A-Sin2A = 2 Cos2A- 1 = 1-2 Sin2A

Triple angle formula

sin3A = 3sinA-4(sinA)3

cos3A = 4(cosA)3-3cosA

tan3a = tana tan(+a) tan(-a)

half-angle formula

sin()=

cos()=

Tan () =

cot()=

tan()==

Sum difference product

sina+sinb=2sincos

sina-sinb=2cossin

cosa+cosb = 2coscos

cosa-cosb =-2 sin

tana+tanb=

Sum and difference of products

sinasinb = -[cos(a+b)-cos(a-b)]

cosacosb = [cos(a+b)+cos(a-b)]

sinacosb = [sin(a+b)+sin(a-b)]

cosasinb = [sin(a+b)-sin(a-b)]

Inductive formula

Sin(-a)=- Sina

cos(-a) = cosa

sin(-a) = cosa

Cos(-a) = Sina

sin(+a) = cosa

Cos(+a)=- Sina

sin(π-a) = sina

cos(π-a) = -cosa

sin(π+a) = -sina

cos(π+a) = -cosa

tgA=tanA =

General formula of trigonometric function

Sina =

cosa=

tana=

Other formulas

Answer? Sina +b? Cosa =×sin(a+c)[ where tanc=]

Answer? Crime (A)-B? Cos(a)=×cos(a-c)[ where tan(c)=]

1+sin(a) =(sin+cos)2

1-sin(a) = (sin-cos)2

Other non-critical trigonometric functions

CSC(a)= 1

Seconds (a) = seconds

Hyperbolic function

sinh(a)= 1

cosh(a)= 1

TG h(a)= 1

Formula 1:

Let α be an arbitrary angle, and the values of the same trigonometric function with the same angle of the terminal edge are equal:

sin(2kπ+α)= sinα

cos(2kπ+α)= cosα

tan(2kπ+α)= tanα

cot(2kπ+α)= cotα

Equation 2:

Let α be an arbitrary angle, and the relationship between the trigonometric function value of π+α and the trigonometric function value of α;

Sine (π+α) =-Sine α

cos(π+α)= -cosα

tan(π+α)= tanα

cot(π+α)= cotα

Formula 3:

The relationship between arbitrary angle α and the value of-α trigonometric function;

Sine (-α) =-Sine α

cos(-α)= cosα

tan(-α)= -tanα

Kurt (-α) =-Kurt α

Equation 4:

The relationship between π-α and the trigonometric function value of α can be obtained by Formula 2 and Formula 3:

Sine (π-α) = Sine α

cos(π-α)= -cosα

tan(π-α)= -tanα

cot(π-α)=-coα

Formula 5:

The relationship between the trigonometric function values of 2π-α and α can be obtained by Formula-and Formula 3:

Sine (2π-α)=- Sine α

cos(2π-α)= cosα

tan(2π-α)= -tanα

Kurt (2π-α)=- Kurt α

Equation 6:

α and the relationship between α and the trigonometric function value of α;

sin(+α)= cosα

cos(+α)= -sinα

tan(+α)= -cotα

cot(+α)= -tanα

sin(-α)= cosα

cos(-α)= sinα

tan(-α)= cotα

cot(-α)= tanα

sin(+α)= -cosα

cos(+α)= sinα

tan(+α)= -cotα

cot(+α)= -tanα

sin(-α)= -cosα

cos(-α)= -sinα

tan(-α)= cotα

cot(-α)= tanα

(higher than k∈Z)

It took me a long time to input this common formula in physics, hoping it will be useful to everyone.

Answer? sin(ωt+θ)+ B? sin(ωt+φ) =×sin

Formulas of trigonometric functions Certificate (All)

Formula expression

Multiplication and factorization A2-B2 = (a+b) (a-b) A3+B3 = (a+b) (A2-AB+B2) A3-B3 = (a-b) (A2+AB+B2)

Trigonometric inequality | A+B |≤| A |+B||||| A-B|≤| A |+B || A |≤ B < = > -b≤a≤b

|a-b|≥|a|-|b| -|a|≤a≤|a|

The solution of the unary quadratic equation-b+√ (B2-4ac)/2a-b-b+√ (B2-4ac)/2a

The relationship between root and coefficient x1+x2 =-b/ax1* x2 = c/a Note: Vieta theorem.

Discriminant b2-4a=0 Note: The equation has two equal real roots.

B2-4ac >0 Note: The equation has real roots.

B2-4ac & lt; 0 Note: The equation has multiple yokes.

formulas of trigonometric functions

Two-angle summation formula sin (a+b) = sinacosb+cosasinbsin (a-b) = sinacosb-sinbcosa.

cos(A+B)= cosa cosb-Sina sinb cos(A-B)= cosa cosb+Sina sinb

tan(A+B)=(tanA+tanB)/( 1-tanA tanB)tan(A-B)=(tanA-tanB)/( 1+tanA tanB)

ctg(A+B)=(ctgActgB- 1)/(ctg B+ctgA)ctg(A-B)=(ctgActgB+ 1)/(ctg B-ctgA)

The angle doubling formula tan2a = 2tana/(1-tan2a) ctg2a = (ctg2a-1)/2ctga.

cos2a = cos2a-sin2a = 2 cos2a- 1 = 1-2 sin2a

Half-angle formula sin (a/2) = √ ((kloc-0/-COSA)/2) sin (a/2) =-√ ((kloc-0/-COSA)/2).

cos(A/2)=√(( 1+cosA)/2)cos(A/2)=-√(( 1+cosA)/2)

tan(A/2)=√(( 1-cosA)/(( 1+cosA))tan(A/2)=-√(( 1-cosA)/(( 1+cosA))

ctg(A/2)=√(( 1+cosA)/(( 1-cosA))ctg(A/2)=-√(( 1+cosA)/(( 1-cosA))

Sum-difference product 2sina cosb = sin (a+b)+sin (a-b) 2cosasinb = sin (a+b)-sin (a-b)

2 cosa cosb = cos(A+B)-sin(A-B)-2 sinasinb = cos(A+B)-cos(A-B)

sinA+sinB = 2 sin((A+B)/2)cos((A-B)/2 cosA+cosB = 2 cos((A+B)/2)sin((A-B)/2)

tanA+tanB = sin(A+B)/cosa cosb tanA-tanB = sin(A-B)/cosa cosb

ctgA+ctgBsin(A+B)/Sina sinb-ctgA+ctgBsin(A+B)/Sina sinb

The sum of the first n terms in some sequences is1+2+3+4+5+6+7+8+9+…+n = n (n+1)/21+3+5+7+9+/kloc-0.

2+4+6+8+ 10+ 12+ 14+…+(2n)= n(n+ 1) 12+22+32+42+52+62+72+82+…+N2 = n(n+ 1)(2n+ 1)/6

13+23+33+43+53+63+…n3 = N2(n+ 1)2/4 1 * 2+2 * 3+3 * 4+4 * 5+5 * 6+6 * 7+…+n(n+ 1)= n(n+ 1)(n+2)/3

Sine theorem a/sinA=b/sinB=c/sinC=2R Note: where r represents the radius of the circumscribed circle of a triangle.

Cosine Theorem b2=a2+c2-2accosB Note: Angle B is the included angle between side A and side C..

Tangent theorem:

[(a+b)/(a-b)]= {[Tan(a+b)/2]/[Tan(a-b)/2]}

The standard equation of a circle (x-a)2+(y-b)2=r2 Note: (A, B) is the center coordinate.

General equation of circle x2+y2+Dx+Ey+F=0 Note: D2+E2-4f > 0

Parabolic standard equation y2=2px y2=-2px x2=2py x2=-2py

Lateral area of a straight prism S=c*h lateral area of an oblique prism s = c' * h.

Lateral area of a regular pyramid S= 1/2c*h' lateral area of a regular prism S= 1/2(c+c')h'

The lateral area of the frustum of a cone S = 1/2(c+c')l = pi(R+R)l The surface area of the ball S=4pi*r2.

Lateral area of cylinder S=c*h=2pi*h lateral area of cone s =1/2 * c * l = pi * r * l.

The arc length formula l=a*r a is the radian number r > of the central angle; 0 sector area formula s= 1/2*l*r

Conical volume formula V= 1/3*S*H Conical volume formula V= 1/3*pi*r2h

Oblique prism volume V=S'L Note: where s' is the straight cross-sectional area and l is the side length.

Cylinder volume formula V=s*h cylinder V=pi*r2h

-trigonometric function product sum and difference product formula.

If you can't remember it, push it yourself, and use the sine and cosine of the sum of two angles:

cos(A+B)=cosAcosB-sinAsinB

cos(A-B)=cosAcosB+sinAsinB

Add or subtract these two formulas to get the sum and difference of two sets of products:

Addition: cosAcosB=[cos(A+B)+cos(A-B)]/2

Subtraction: sinasinb =-[cos (a+b)-cos (a-b)]/2

sin(A+B)=sinAcosB+sinBcosA

sin(A-B)=sinAcosB-sinBcosA

By adding or subtracting these two formulas, the sum and difference of two sets of products can be obtained:

Addition: sinAcosB=[sin(A+B)+sin(A-B)]/2.

Subtraction: sinBcosA=[sin(A+B)-sin(A-B)]/2

Such ***4 groups are sum-difference integrals, and then the sum-difference integrals are reversed.

I don't know if it's to make you remember the exam, but you can also infer it temporarily when you really can't remember the exam.

Jia Zhengqian

Positive subtraction and positive remainder are in front.

What's more, what's more.

There is nothing left to reduce, and it is still negative.

The positive surplus is positive and JUNG WOO is negative.

Plus, plus, plus, minus, plus, minus.

.

3. Some conclusions in the triangle: (no need to remember)

( 1)anA+tan b+ tanC = tanA tanB tanC

(2)sinA+tsin B+sinC = 4 cos(A/2)cos(B/2)cos(C/2)

(3)cosA+cosB+cosC = 4 sin(A/2)sin(B/2)sin(C/2)+ 1

(4)sin2A+sin2B+sin2C = 4 Sina sinB sinC

(5)cos2A+cos2B+cos2C =-4 cosacosbcosc- 1

...........................

It is known that sin α = m sin (α+2 β), | m | < 1, and verified that tan (α+β) = (1+m)/(kloc-0/-m) tan β.

Solution: sinα=m sin(α+2β)

sin(a+β-β)=msin(a+β+β)

sin(a+β)cosβ-cos(a+β)sinβ= msin(a+β)cosβ+mcos(a+β)sinβ

sin(a+β)cosβ( 1-m)= cos(a+β)sinβ(m+ 1)

tan(α+β)=( 1+m)/( 1-m)tanβ

Reduced power formula

(sin^2)x= 1-cos2x/2

(cos^2)x=i=cos2x/2

General formula of trigonometric function

Let tan(a/2)=t

sina=2t/( 1+t^2)

cosa=( 1-t^2)/( 1+t^2)

tana=2t/( 1-t^2)