geometric definition
right triangle definition
There is a right angled triangle as follows
Then the trigonometric function is defined as follows
| sine | cosine | tangent | cotangent | secant | cosecant |
|---|---|---|---|---|---|
| sin θ = ah \sin\theta=\frac{a}{h}sini=ha | cos θ = bh \cos\theta=\frac{b}{h}cosi=hb | tan θ = ab \tan\theta=\frac{a}{b}tani=ba | cot θ = b a \cot\theta=\frac{b}{a} coti=ab | sec θ = h b \sec\theta=\frac{h}{b} seci=bh | csc θ = h a \csc\theta=\frac{h}{a} csci=ah |
Since it is a right triangle, this definition can only define θ ∈ ( 0 , π 2 ) \theta\in(0,\frac{\pi}{2})i∈(0,2p) range
unit circle definition
There is a unit circle x 2 + y 2 = 1 x^2+y^2=1x2+y2=1 as follows
P ( x , y ) P(x,y) P(x,y ) is a point on the unit circle
Then the trigonometric function is defined as follows
| sine | cosine | tangent | cotangent | secant | cosecant |
|---|---|---|---|---|---|
| sin θ = y 1 = y \sin\theta=\frac{y}{1}=ysini=1y=y | cos θ = x 1 = x \cos\theta=\frac{x}{1}=x cosi=1x=x | tan θ = yx \tan\theta=\frac{y}{x}tani=xy | cot θ = x y \cot\theta=\frac{x}{y} coti=yx | sec θ = 1 x \sec\theta=\frac{1}{x} seci=x1 | csc θ = 1 y \csc\theta=\frac{1}{y}csci=y1 |
At this time θ ∈ ( − ∞ , + ∞ ) \theta\in(-\infty,+\infty)i∈(−∞,+∞)
- θ > 0 \theta>0 i>Rotate counterclockwise at 0 o'clock
- θ < 0 \theta<0 i<Rotate clockwise at 0 o'clock
Function image
sine function
y = sin xy=\sin xy=sinx
Know the sine function from the graph
- The minimum positive period is 2 π 2\pi2 p.m
- The axis of symmetry is x = π 2 + k π , k ∈ Z x=\frac{\pi}{2}+k\pi,\ k\in\mathbb{Z}x=2p+kπ , k∈Z
- The center of symmetry is (k π, 0), k ∈ Z (k\pi,0),\ k\in\mathbb{Z}( kπ ,0), k∈Z
- The values in the first and second quadrants are positive, and the values in the third and fourth quadrants are negative.
cosine function
y = cos xy=\cos xy=cosx
Know the cosine function from the graph
- The minimum positive period is 2 π 2\pi2 p.m
- The axis of symmetry is x = k π , k ∈ Z x=k\pi,\ k\in\mathbb{Z}x=kπ , k∈Z
- The center of symmetry is ( π 2 + k π , 0 ) , k ∈ Z (\frac{\pi}{2}+k\pi,0),\ k\in\mathbb{Z}(2p+kπ ,0), k∈Z
- The values in the first and fourth quadrants are positive, and the values in the second and third quadrants are negative.
tangent function
y = tan xy=\tan xy=tanx
Know the tangent function from the graph
- The minimum positive period is π \piPi
- The center of symmetry is (k π, 0), k ∈ Z (k\pi,0),\ k\in\mathbb{Z}( kπ ,0), k∈Z
- The values in the first and third quadrants are positive, and the values in the second and fourth quadrants are negative.
cotangent function
y = cot x y=\cot x y=cotx
Know the cotangent function from the graph
- The minimum positive period is π \piPi
- The center of symmetry is ( π 2 + k π , 0 ) , k ∈ Z (\frac{\pi}{2}+k\pi,0),\ k\in\mathbb{Z}(2p+kπ ,0), k∈Z
- The values in the first and third quadrants are positive, and the values in the second and fourth quadrants are negative.
secant function
y = sec x y=\sec x y=secx
Know the secant function from the graph
- The minimum positive period is 2 π 2\pi2 p.m
- The axis of symmetry is x = k π , k ∈ Z x=k\pi,\ k\in\mathbb{Z}x=kπ , k∈Z
- The center of symmetry is ( π 2 + k π , 0 ) , k ∈ Z (\frac{\pi}{2}+k\pi,0),\ k\in\mathbb{Z}(2p+kπ ,0), k∈Z
- The values in the first and fourth quadrants are positive, and the values in the second and third quadrants are negative.
cosecant function
y = csc xy=\csc xy=cscx
Know the cosecant function from the graph
- The minimum positive period is 2 π 2\pi2 p.m
- The axis of symmetry is x = π 2 + k π , k ∈ Z x=\frac{\pi}{2}+k\pi,\ k\in\mathbb{Z}x=2p+kπ , k∈Z
- The center of symmetry is (k π, 0), k ∈ Z (k\pi,0),\ k\in\mathbb{Z}( kπ ,0), k∈Z
- The values in the first and second quadrants are positive, and the values in the third and fourth quadrants are negative.
identity
express each other
It can be seen from the definition of trigonometric functions
- tan θ = sin θ cos θ \tan\theta=\frac{\sin\theta}{\cos\theta}tani=cosisini
- cot θ = 1 tan θ = cos θ sin θ \cot\theta=\frac{1}{\tan\theta}=\frac{\cos\theta}{\sin\theta}coti=tani1=sinicosi
- sec θ = 1 cos θ \sec\theta=\frac{1}{\cos\theta}seci=cosi1
- csc θ = 1 sin θ \csc\theta=\frac{1}{\sin\theta} csci=sini1
Pythagoras Identity
It can be known from the definition of a right triangle and the Pythagorean theorem or the definition of the unit circle
- sin 2 θ + cos 2 θ = 1 \sin^2\theta+\cos^2\theta=1sin2i+cos2i=1
It can be deduced from this
- tan 2 θ + 1 = sec 2 θ \tan^2\theta+1=\sec^2\thetatan2i+1=sec2i
- cot 2 θ + 1 = csc 2 θ \cot^2\theta+1=\csc^2\thetacot2i+1=csc2i
induction formula
形如sin / cos / tan / cot / sec / csc ( θ + k π 2 ) , k ∈ Z \sin/\cos/\tan/\cot/\sec/\csc(\theta+ \frac{k\pi}{2}),\k\in\mathbb{Z}sin/cos/tan/cot/sec/csc ( θ+2kπ), k∈Zwhere "/// "means or.
Transformation formula:odd to even unchanged, look at the quadrants for symbols.
- Odd and even means kkThe parity of k . If it is an odd number, thensin \sinsin andcos \coscos interchange,tan \tantan andcot \cotcot interchange,sec \secsec和csc \csccsc interchange; even numbers remain unchanged
- 将θ \thetaθ is regarded as an acute angle, judgeθ + k π 2 \theta+\frac{k\pi}{2}i+2kπQuadrant. If the trigonometric function takes a negative value in this quadrant, add a negative sign in front of it.
like:
- sin ( θ + π 2 ) = cos θ \sin(\theta+\frac{\pi}{2})=\cos\thetasin ( i+2p)=cosθ
k = 1 k=1 k=1 is an odd number,sin \sinsin changesto cos \coscos
θ + π 2 \theta+\frac{\pi}{2}i+2pIn the second quadrant, sin \sinThe value of sin is positive and the sign is positive - sin ( θ + π ) = − sin θ \sin(\theta+\pi)=-\sin\thetasin ( i+p )=−sinθ
k = 2 k=2 k=2 is an even number, there is no need to change
θ + π \theta+\pii+π is in the third quadrant,sin \sinThe value of sin is negative and the sign is negative - cos ( θ + 3 π 2 ) = sin θ \cos(\theta+\frac{3\pi}{2})=\sin\thetacos ( i+23 p.m)=sinθ
k = 3 k=3 k=3 is an odd number,cos \coscos becomessin \sinsin
θ + 3 π 2 \theta+\frac{3\pi}{2}i+23 p.mIn the fourth quadrant, cos \cosThe cos value is positive and the sign is positive - tan ( θ − π 2 ) = − cot θ \tan(\theta-\frac{\pi}{2})=-\cot\thetatan ( θ−2p)=−cotθ
k = − 1 k=-1 k=− 1 is an odd number,tan \tantan becomescot \cotcot
θ − π 2 \theta-\frac{\pi}{2}i−2pIn the fourth quadrant, tan \tanThe tan value is negative and the sign is negative - sin ( − θ ) = − sin θ \sin(-\theta)=-\sin\thetasin ( − θ )=−sinθ
k = 0 k=0 k=0 is an even number, and there is no need to change
− θ -\theta− θ is in the fourth quadrant,sin \sinThe value of sin is negative and the sign is negative
The above conclusion can be easily drawn from the function graph
sum and difference formula
sine
- sin ( θ 1 + θ 2 ) = sin θ 1 cos θ 2 + cos θ 1 sin θ 2 \sin(\theta_1+\theta_2)=\sin \theta_1\cos \theta_2+\cos \theta_1\sin \theta_2sin ( i1+i2)=sini1cosi2+cosi1sini2
- sin ( θ 1 − θ 2 ) = sin θ 1 cos θ 2 − cos θ 1 sin θ 2 \sin(\theta_1-\theta_2)=\sin \theta_1\cos \theta_2-\cos \theta_1 \sin\theta_2sin ( i1−i2)=sini1cosi2−cosi1sini2
Assume a unit circle θ 1 2 + θ 2 2 = 1 \theta_1^2+\theta_2^2=1i12+i22=1
则∣ AG ∣ = sin θ 2 , ∣ OG ∣ = cos θ 2 |AG|=\sin\theta_2,\ |OG|=\cos\theta_2∣AG∣=sini2, ∣OG∣=cosi2
∴ ∣ MP ∣ = ∣ GR ∣ = ∣ OG ∣ sin θ 1 = sin θ 1 cos θ 2 \therefore |MP|=|GR|=|OG|\sin\theta_1=\sin\theta_1\cos\ . theta_2∴∣MP∣=∣GR∣=∣OG∣sini1=sini1cosi2
∵ △ O N P ∼ △ G N M \because\triangle ONP\sim\triangle GNM ∵△ONP∼△ GNM (one angle of a right triangle is equal)
∴ ∠ NGM = ∠ NOP = θ 1 \therefore\angle NGM=\angle NOP=\theta_1∴∠NGM=∠NOP=i1
∵ ∠ N G P + ∠ A G M = ∠ M A G + ∠ A G M = 90 ° \because\angle NGP+\angle AGM=\angle MAG+\angle AGM=90\degree ∵∠NGP+∠AGM=∠MAG+∠AGM=90°
∴ ∠ MAG = ∠ NGM = θ 1 \therefore\angle MAG=\angle NGM=\theta_1∴∠MAG=∠NGM=i1
∴ ∣ AM ∣ = ∣ AG ∣ cos θ 1 = cos θ 1 sin θ 2 \therefore |AM|=|AG|\cos\theta_1=\cos\theta_1\sin\theta_2∴∣AM∣=∣AG∣cosi1=cosi1sini2
∴ \therefore ∴
sin ( θ 1 + θ 2 ) = ∣ AP ∣ = ∣ AM ∣ + ∣ MP ∣ = sin θ 1 cos θ 2 + cos θ 1 sin θ 2 \begin{aligned}\sin(\theta_1+ \theta_2)&=|AP|=|AM|+|MP|\\&=\sin\theta_1\cos\theta_2+\cos\theta_1\sin\theta_2\end{aligned}sin ( i1+i2)=∣AP∣=∣AM∣+∣MP∣=sini1cosi2+cosi1sini2
令θ 2 = − θ 2 \theta_2=-\theta_2i2=− i2则
sin ( θ 1 − θ 2 ) = sin [ θ 1 + ( − θ 2 ) ] = sin θ 1 cos ( − θ 2 ) + cos θ 1 sin ( − θ 2 ) = sin θ 1 cos θ 2 − cos θ 1 sin θ 2 \begin{aligned}\sin(\theta_1-\theta_2)&=\sin[\theta_1+(-\theta_2)]\\&=\sin\theta_1 \cos(-\theta_2)+\cos\theta_1\sin(-\theta_2)\\&=\sin\theta_1\cos\theta_2-\cos\theta_1\sin\theta_2\end{aligned}sin ( i1−i2)=sin [ i1+( − i2)]=sini1cos ( − θ2)+cosi1sin ( − θ2)=sini1cosi2−cosi1sini2
cosine
- cos ( θ 1 + θ 2 ) = cos θ 1 cos θ 2 − sin θ 1 sin θ 2 \cos(\theta_1+\theta_2)=\cos \theta_1\cos \theta_2-\sin \theta_1\ without \theta_2cos ( i1+i2)=cosi1cosi2−sini1sini2
- cos ( θ 1 − θ 2 ) = cos θ 1 cos θ 2 + sin θ 1 sin θ 2 \cos(\theta_1-\theta_2)=\cos \theta_1\cos \theta_2+\sin \theta_1\ without \theta_2cos ( i1−i2)=cosi1cosi2+sini1sini2
Let θ 2 = θ 2 + π 2 \theta_2=\theta_2+\frac{\pi}{2}i2=i2+2p则
cos ( θ 1 + θ 2 ) = sin ( θ 1 + θ 2 + π 2 ) = sin θ 1 cos ( θ 2 + π 2 ) + cos θ 1 sin ( θ 2 + π ) = cos θ 1 cos θ 2 − sin θ 1 sin θ 2 \begin{aligned}\cos(\theta_1+\theta_2)&=\sin(\theta_1+\theta_2+\frac{\pi}{2}; )\\&=\sin\theta_1\cos(\theta_2+\frac{\pi}{2})+\cos\theta_1\sin(\theta_2+\frac{\pi}{2})\\&=\cos \theta_1\cos\theta_2-\sin\theta_1\sin\theta_2\end{aligned}cos ( i1+i2)=sin ( i1+i2+2p)=sini1cos ( i2+2p)+cosi1sin ( i2+2p)=cosi1cosi2−sini1sini2
令θ 2 = − θ 2 \theta_2=-\theta_2i2=− i2则
cos ( θ 1 − θ 2 ) = cos [ θ 1 + ( − θ 2 ) ] = cos θ 1 cos ( − θ 2 ) − sin θ 1 sin ( − θ 2 ) = cos θ 1 cos θ 2 + sin θ 1 sin θ 2 \begin{aligned}\cos(\theta_1-\theta_2)&=\cos[\theta_1+(-\theta_2)]\\&=\cos\theta_1 \cos(-\theta_2)-\sin\theta_1\sin(-\theta_2)\\&=\cos\theta_1\cos\theta_2+\sin\theta_1\sin\theta_2\end{aligned}cos ( i1−i2)=cos [ i1+( − i2)]=cosi1cos ( − θ2)−sini1sin ( − θ2)=cosi1cosi2+sini1sini2
The ratio
tan ( θ 1 + θ 2 ) = tan θ 1 + tan θ 2 1 − tan θ 1 tan θ 2 \tan(\theta_1+\theta_2)=\frac{\tan \theta_1+\tan \theta_2 }{1-\tan \theta_1\tan \theta_2}tan ( θ1+i2)=1−tani1tani2tani1+tani2
tan ( θ 1 − θ 2 ) = tan θ 1 − tan θ 2 1 + tan θ 1 tan θ 2 \tan(\theta_1-\theta_2)=\frac{\tan \theta_1-\tan \ theta_2}{1+\tan \theta_1\tan \theta_2}tan ( θ1−i2)=1+tani1tani2tani1−tani2
tan ( θ 1 + θ 2 ) = sin ( θ 1 + θ 2 ) cos ( θ 1 + θ 2 ) = sin θ 1 cos θ 2 + cos θ 1 sin θ 2 cos θ 1 cos θ 2 − sin θ 1 sin θ 2 = sin θ 1 cos θ 2 cos θ 1 cos θ 2 + cos θ 1 sin θ 2 cos θ 1 cos θ 2 cos θ 2 1 cos θ 2 cos θ 1 cos θ 2 − sin θ 1 sin θ 2 cos θ 1 cos θ 2 = sin θ 1 cos θ 1 + sin θ 2 cos θ 2 1 − sin θ 1 cos θ 1 sin θ 2 cos θ 2 = tan θ 1 + tan θ 2 1 − tan θ 1 tan θ 2\begin{aligned}\tan(\theta_1+\theta_2)&=\frac{\sin(\theta_1+\theta_2)}{\cos(\theta_1+\theta_2)}\\&=\frac{\sin\theta_1\cos\theta_2+\cos\theta_1\sin\theta_2}{\cos\theta_1\cos\theta_2-\sin\theta_1\sin\theta_2}\\&=\frac{\frac{\sin\theta_1\cos\theta_2}{\cos\theta_1\cos\theta_2}+\frac{\cos\theta_1\sin\theta_2}{\cos\theta_1\cos\theta_2}}{\frac{\cos\theta_1\cos\theta_2}{\cos\theta_1\cos\theta_2}-\frac{\sin\theta_1\sin\theta_2}{\cos\theta_1\cos\theta_2}}\\&=\frac{\frac{\sin\theta_1}{\cos\theta_1}+\frac{\sin\theta_2}{\cos\theta_2}}{1-\frac{\sin\theta_1}{\cos\theta_1}\frac{\sin\theta_2}{\cos\theta_2}}\\&=\frac{\tan\theta_1+\tan\theta_2}{1-\tan\theta_1\tan\theta_2}\end{aligned}tan ( θ1+i2)=cos ( i1+i2)sin ( i1+i2)=cosi1cosi2−sini1sini2sini1cosi2+cosi1sini2=cosi1cosi2cosi1cosi2−cosi1cosi2sini1sini2cosi1cosi2sini1cosi2+cosi1cosi2cosi1sini2=1−cosi1sini1cosi2sini2cosi1sini1+cosi2sini2=1−tani1tani2tani1+tani2
令θ 2 = − θ 2 \theta_2=-\theta_2i2=− i2则
tan ( θ 1 − θ 2 ) = tan [ θ 1 + ( − θ 2 ) ] = tan θ 1 + tan ( − θ 2 ) 1 − tan θ 1 tan ( − θ 2 ) = tan θ 1 − tan θ 2 1 + tan θ 1 tan θ 2 \begin{aligned}\begin(\theta_1-\theta_2)&=\begin[\theta_1+(-\theta_2)]\\&= \frac{\tan\theta_1+\tan(-\theta_2)}{1-\tan\theta_1\tan(-\theta_2)}\\&=\frac{\tan\theta_1-\tan\theta_2}{1+ \tan\theta_1\tan\theta_2}\end{aligned}tan ( θ1−i2)=tan [ θ1+( − i2)]=1−tani1tan ( − θ2)tani1+tan ( − θ2)=1+tani1tani2tani1−tani2
double angle formula
It can be deduced from the sum-difference formula
- sin 2 θ = 2 sin θ cos θ = 2 sin θ cos θ sin 2 θ + cos 2 θ = 2 tan θ 1 + tan 2 θ \begin{aligned}\sin 2\theta&; =2\sin\theta\cos\theta\\&=\frac{2\sin\theta\cos\theta}{\sin^2\theta+\cos^2\theta}\\&=\frac{2\ tan\theta}{1+\tan^2\theta}\end{aligned}sin2 i=2sinicosi=sin2i+cos2i2sinicosi=1+tan2i2tani
- cos 2 θ = cos 2 θ − sin 2 θ = 1 − tan 2 θ 1 + tan 2 θ = 2 cos 2 θ − 1 = 1 − 2 sin 2 θ \begin{aligned}\cos 2\theta&=\cos^2\theta-\sin^2\theta\\&=\frac{1-\tan^2\theta}{1+\tan^2\theta}\\&=2\cos ^2 \theta-1\\&=1-2\sin^2\theta\end{aligned}cos2 i=cos2i−sin2i=1+tan2i1−tan2i=2cos2i−1=1−2sin2i
- tan 2 θ = 2 tan θ 1 − tan 2 θ \tan 2\theta=\frac{2\tan \theta}{1-\tan^2 \theta};tan2 i=1−tan2i2tani
fall official
It is easy to derive from the double angle formula
- sin 2 θ = 1 − cos 2 θ 2 \sin^2\theta=\frac{1-\cos2\theta}{2}sin2i=21−cos2 i
- cos 2 θ = cos 2 θ + 1 2 \cos^2\theta=\frac{\cos2\theta+1}{2}cos2i=2cos2 i+1
It is easy to derive from the above two formulas
- tan 2 θ = 1 − cos 2 θ 1 + cos 2 θ \tan^2\theta=\frac{1-\cos2\theta}{1+\cos2\theta};tan2i=1+cos2 i1−cos2 i
half-width formula
The infinitive set
tan θ 2 = 1 − cos θ sin θ = sin θ 1 + cos θ \tan\frac{\theta}{2}=\frac{1-\cos \theta}{ \sin \theta}=\frac{\sin\theta}{1+\cos\theta}tan2i=sini1−cosi=1+cosisini
∵ cos θ = 1 − 2 sin 2 θ 2 \because \cos \theta=1-2\sin^2\frac{\theta}{2}∵cosi=1−2sin22i
∴ 1 − cos θ = 2 sin 2 θ 2 \therefore 1-\cos \theta=2\sin^2\frac{\theta}{2}∴1−cosi=2sin22i
∵ sin θ = 2 sin θ 2 cos θ 2 \because \sin \theta=2\sin\frac{\theta}{2}\cos\frac{\theta}{2}∵sini=2sin2icos2i
∴ \therefore ∴
1 − cos θ sin θ = 2 sin 2 θ 2 2 sin θ 2 cos θ 2 = sin θ 2 cos θ 2 = tan θ 2 \begin{aligned}\frac{1-\ cos \theta}{\sin \theta}&=\frac{2\sin^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta }{2}}\\&=\frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}}\\&=\tan\frac{\theta}{ 2}\end{aligned}sini1−cosi=2sin2icos2i2sin22i=cos2isin2i=tan2i
∵ cos θ = 2 cos 2 θ 2 − 1 \because \cos \theta=2\cos^2\frac{\theta}{2}-1∵cosi=2cos22i−1
∴ 1 + cos θ = 2 cos 2 θ 2 \therefore 1+\cos \theta=2\cos^2\frac{\theta}{2}∴1+cosi=2cos22i
∴ \therefore ∴
sin θ 1 + cos θ = 2 sin θ 2 cos θ 2 2 cos 2 θ 2 = sin θ 2 cos θ 2 = tan θ 2 \begin{aligned}\frac{\sin\ theta}{1+\cos\theta}&=\frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\cos^2\frac{\theta }{2}}\\&=\frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}}\\&=\tan\frac{\theta}{ 2}\end{aligned}1+cosisini=2cos22i2sin2icos2i=cos2isin2i=tan2i
Product and difference formula
- sin θ 1 sin θ 2 = − 1 2 [ cos ( θ 1 + θ 2 ) − cos ( θ 1 − θ 2 ) ] \sin\theta_1\sin\theta_2=-\frac{1}{2 }[\cos(\theta_1+\theta_2)-\cos(\theta_1-\theta_2)]sini1sini2=−21[ cos ( i1+i2)−cos ( i1−i2)]
- sin θ 1 cos θ 2 = 1 2 [ sin ( θ 1 + θ 2 ) + sin ( θ 1 − θ 2 ) ] \sin\theta_1\cos\theta_2=\frac{1}{2}[ \sin(\theta_1+\theta_2)+\sin(\theta_1-\theta_2)]sini1cosi2=21[ sin ( i1+i2)+sin ( i1−i2)]
- cos θ 1 sin θ 2 = 1 2 [ sin ( θ 1 + θ 2 ) − sin ( θ 1 − θ 2 ) ] \cos\theta_1\sin\theta_2=\frac{1}{2}[ \sin(\theta_1+\theta_2)-\sin(\theta_1-\theta_2)]cosi1sini2=21[ sin ( i1+i2)−sin ( i1−i2)]
- cos θ 1 cos θ 2 = 1 2 [ cos ( θ 1 + θ 2 ) + cos ( θ 1 − θ 2 ) ] \cos\theta_1\cos\theta_2=\frac{1}{2}[ \cos(\theta_1+\theta_2)+\cos(\theta_1-\theta_2)]cosi1cosi2=21[ cos ( i1+i2)+cos ( i1−i2)]
The formula
{ sin ( θ 1 + θ 2 ) = sin θ 1 cos θ 2 + cos θ 1 sin θ 2 ( 1 ) sin ( θ 1 − θ 2 ) = sin θ cos θ 2 − cos θ 1 sin θ 2 ( 2 ) cos ( θ 1 + θ 2 ) = cos θ 1 cos θ 2 − sin θ 1 sin θ 2 ( 3 ) cos ( θ 1 − θ 2 ) = cos θ 1 cos θ 2 + sin θ 1 sin θ 2 ( 4 ) \begin{cases} \sin(\theta_1+\theta_2)=\sin\theta_1\cos\theta_2+\cos \theta_1\sin\theta_2&&&&&(1)\\ \sin(\theta_1-\theta_2)=\sin\theta_1\cos\theta_2-\cos\theta_1\sin\theta_2&&&&&(2)\\ \cos(\theta_1+\theta_2 )=\cos\theta_1\cos\theta_2-\sin\theta_1\sin\theta_2&&&&&(3)\\ \cos(\theta_1-\theta_2)=\cos\theta_1\cos\theta_2+\sin\theta_1\sin\theta_2&&&&& (4)\\\end{cases}⎩ ⎨ ⎧sin ( i1+i2)=sini1cosi2+cosi1sini2sin ( i1−i2)=sini1cosi2−cosi1sini2cos ( i1+i2)=cosi1cosi2−sini1sini2cos ( i1−i2)=cosi1cosi2+sini1sini2(1)(2)(3)(4)
Formula (4) − (4)-( 4 ) −Formula (3) (3)( 3 )得
sin θ 1 sin θ 2 = 1 2 [ cos ( θ 1 − θ 2 ) − cos ( θ 1 + θ 2 ) ] \sin\theta_1\sin\theta_2=\frac{1} {2}[\cos(\theta_1-\theta_2)-\cos(\theta_1+\theta_2)]sini1sini2=21[ cos ( i1−i2)−cos ( i1+i2)]
Formula(4) + (4)+( 4 ) + formula(3) (3)( 3 )得
cos θ 1 cos θ 2 = 1 2 [ cos ( θ 1 + θ 2 ) + cos ( θ 1 − θ 2 ) ] \cos\theta_1\cos\theta_2=\frac{1} {2}[\cos(\theta_1+\theta_2)+\cos(\theta_1-\theta_2)]cosi1cosi2=21[ cos ( i1+i2)+cos ( i1−i2)]
Formula(1) + (1)+( 1 ) + formula(2) (2)( 2 )得
sin θ 1 cos θ 2 = 1 2 [ sin ( θ 1 + θ 2 ) + sin ( θ 1 − θ 2 ) ] \sin\theta_1\cos\theta_2=\frac{1} {2}[\sin(\theta_1+\theta_2)+\sin(\theta_1-\theta_2)]sini1cosi2=21[ sin ( i1+i2)+sin ( i1−i2)]
Formula(1) − (1)-( 1 ) −Formula (2) (2)( 2 )得
cos θ 1 sin θ 2 = 1 2 [ sin ( θ 1 + θ 2 ) − sin ( θ 1 − θ 2 ) ] \cos\theta_1\sin\theta_2=\frac{1} {2}[\sin(\theta_1+\theta_2)-\sin(\theta_1-\theta_2)]cosi1sini2=21[ sin ( i1+i2)−sin ( i1−i2)]
sum-difference product formula
- sin θ 1 + sin θ 2 = 2 sin ( θ 1 + θ 2 2 ) cos ( θ 1 − θ 2 2 ) \sin\theta_1+\sin\theta_2=2\sin(\frac{\theta_1+\ theta_2}{2})\cos(\frac{\theta_1-\theta_2}{2})sini1+sini2=2sin(2i1+i2)cos(2i1−i2)
- sin θ 1 − sin θ 2 = 2 cos ( θ 1 + θ 2 2 ) sin ( θ 1 − θ 2 2 ) \sin\theta_1-\sin\theta_2=2\cos(\frac{\theta_1+ \theta_2}{2})\sin(\frac{\theta_1-\theta_2}{2})sini1−sini2=2cos(2i1+i2)sin(2i1−i2)
- cos θ 1 + cos θ 2 = 2 cos ( θ 1 + θ 2 2 ) cos ( θ 1 − θ 2 2 ) \cos\theta_1+\cos\theta_2=2\cos(\frac{\theta_1+\ theta_2}{2})\cos(\frac{\theta_1-\theta_2}{2})cosi1+cosi2=2cos(2i1+i2)cos(2i1−i2)
- cos θ 1 − cos θ 2 = − 2 sin ( θ 1 + θ 2 2 ) sin ( θ 1 − θ 2 2 ) \cos\theta_1-\cos\theta_2=-2\sin(\frac{ \theta_1+\theta_2}{2})\sin(\frac{\theta_1-\theta_2}{2})cosi1−cosi2=−2sin(2i1+i2)sin(2i1−i2)
∵ \because ∵
θ 1 = θ 1 + θ 2 2 + θ 1 − θ 2 2 , θ 2 = θ 1 + θ 2 2 − θ 1 − θ 2 2 \theta_1=\frac{\theta_1+\theta_2}{2}+\ frac{\theta_1-\theta_2}{2},\\theta_2=\frac{\theta_1+\theta_2}{2}-\frac{\theta_1-\theta_2}{2}i1=2i1+i2+2i1−i2, i2=2i1+i2−2i1−i2
∴ \therefore ∴
sin θ 1 = sin ( θ 1 + θ 2 2 ) cos ( θ 1 − θ 2 2 ) + cos ( θ 1 + θ 2 2 ) sin ( θ 1 − θ 2 2 ) sin θ 2 = sin ( θ 1 + θ 2 2 ) cos ( θ 1 − θ 2 2 ) − cos ( θ 1 + θ 2 2 ) sin ( θ 1 − θ 2 2 ) cos θ 1 = cos ( θ 1 + θ 2 2 ) cos ( θ 1 − θ 2 2 ) − sin ( θ 1 + θ 2 2 ) sin ( θ 1 − θ 2 2 ) cos θ 2 = cos ( θ 1 + θ 2 2 ) cos ( θ 1 − θ 2 2 ) + sin ( θ 1 + θ 2 2 ) sin ( θ 1 − θ 2 2 )\begin{aligned}&\sin\theta_1=\sin(\frac{\theta_1+\theta_2}{2})\cos(\frac{\theta_1-\theta_2}{2})+\cos(\frac{\theta_1+\theta_2}{2})\sin(\frac{\theta_1-\theta_2}{2})\\&\sin\theta_2=\sin(\frac{\theta_1+\theta_2}{2})\cos(\frac{\theta_1-\theta_2}{2})-\cos(\frac{\theta_1+\theta_2}{2})\sin(\frac{\theta_1-\theta_2}{2})\\&\cos\theta_1=\cos(\frac{\theta_1+\theta_2}{2})\cos(\frac{\theta_1-\theta_2}{2})-\sin(\frac{\theta_1+\theta_2}{2})\sin(\frac{\theta_1-\theta_2}{2})\\&\cos\theta_2=\cos(\frac{\theta_1+\theta_2}{2})\cos(\frac{\theta_1-\theta_2}{2})+\sin(\frac{\theta_1+\theta_2}{2})\sin(\frac{\theta_1-\theta_2}{2})\end{aligned}sini1=sin(2i1+i2)cos(2i1−i2)+cos(2i1+i2)sin(2i1−i2)sini2=sin(2i1+i2)cos(2i1−i2)−cos(2i1+i2)sin(2i1−i2)cosi1=cos(2i1+i2)cos(2i1−i2)−sin(2i1+i2)sin(2i1−i2)cosi2=cos(2i1+i2)cos(2i1−i2)+sin(2i1+i2)sin(2i1−i2)
∴ \therefore ∴
sin θ 1 + sin θ 2 = 2 sin ( θ 1 + θ 2 2 ) cos ( θ 1 − θ 2 2 ) \sin\theta_1+\sin\theta_2=2\sin(\frac{\theta_1+ \theta_2}{2})\cos(\frac{\theta_1-\theta_2}{2})sini1+sini2=2sin(2i1+i2)cos(2i1−i2)
Others can be proved in the same way
Auxiliary angle formula
a sin θ + b cos θ = a 2 + b 2 sin ( θ + φ ) a\sin\theta+b\cos\theta=\sqrt{a^2+b^2}\sin(\theta+ \varphi)asini+bcosi=a2+b2sin ( i+φ )
and
sin ( φ ) = ba 2 + b 2 , cos ( φ ) = aa 2 + b 2 \sin(\varphi)=\frac{b}{\sqrt{a^2+b^2} },~\cos(\varphi)=\frac{a}{\sqrt{a^2+b^2}}sin ( φ )=a2+b2b, cos ( φ )=a2+b2a
reference
[1] Wikipedia Trigonometric Functions https://zh.wikipedia.org/wiki/Trigonometric Functions
[2] Baidu Encyclopedia Trigonometric Function Formulas https://baike.baidu.com/item/Trigonometric Function Formulas/4374733