热门日志推荐
人人最热标签
分享这篇日志的人常去
北京千橡网景科技发展有限公司：
文网文[号··京公网安备号·甲测资字
文化部监督电子邮箱：wlwh@··
文明办网文明上网举报电话： 举报邮箱：&&&&&&&&&&&&
请输入手机号，完成注册
请输入验证码
密码必须由6-20个字符组成
下载人人客户端
品评校花校草，体验校园广场Summary of Trigonometric Identities
&Navigation&
&Contact&Graeme&
& Summary of Identities
This page contains a summary of all of the basic trigonometric identities
that you will need to use in a high school trig class, organized by type.
even/odd, co-function and phase shift identities
cos(x) = cos(-x)
sec(x) = 1/cos(x) = 1/cos(-x) = sec(-x)
tan(x) = sin(x)/cos(x) = -sin(-x)/cos(-x) =
sin(x) = -sin(-x)
csc(x) = 1/sin(x) = 1/-sin(-x) = -csc(-x)
cot(x) = cos(x)/sin(x) = cos(-x)/-sin(-x) =
co-functions
sin(x) = cos(π/2-x)
tan(x) = cot(π/2-x)
sec(x) = csc(π/2-x)
cos(x) = sin(π/2-x)
cot(x) = tan(π/2-x)
csc(x) = sec(π/2-x)
phase shift
&&&&&(see also the )
sin(x) = cos(x-π/2) = -sin(x-π) = -cos(x+π/2) =
sin(x+2π)
csc(x) = sec(x-π/2) = -csc(x-π) = -sec(x+π/2) =
csc(x+2π)
tan(x) = -cot(x±π/2) = tan(x+π)
tan(x+π/4) = (1+tan(x))/(1-tan(x))
exact values of trig functions of common angles --
&Brain Dump&
If you're in a high school trig class, and you need to know the exact
values of sin, cos, and tan for these &round number& angles, then I strongly
recommend practicing this brain dump.& On a blank piece of paper, write
down all the degrees that are multiples of 30 or 45, convert them into
radians by multiplying by π/180, and then write down the sin and cos values&
The repetitive nature of these two functions should help a lot.& You
know tan=sin/cos, so just do the math, and write 'em in.
Once you can do this brain dump in two or three minutes, then you will be
ready to do that during every test.& No need to bring a cheat sheet
with you -- make a new one on scrap paper at the very first thing you do
during a test.& (Some teachers may object at first, so it's best to
clear this idea with your teacher first.)
-sqrt(2)/2
-sqrt(3)/2
-sqrt(3)/3
-sqrt(3)/2
-sqrt(2)/2
-sqrt(2)/2
-sqrt(3)/2
-sqrt(3)/2
-sqrt(2)/2
11π/6
-sqrt(3)/3
This idea can be extended using some of the identities presented below.&
Here are a whole lot more .
Pythagorean identity: sin2x
+ cos2x = 1
sin2x + cos2x = 1
Apply the Pythagorean Theorem to this triangle:
+ 1 = sec2x&
Start with the first formula, then divide every
term by cos2x.
Start over, dividing by sin2x.
half angle formulas
sin(x/2) = ±sqrt((1-cos(x))/2)
cos(x) = cos(x/2 + x/2) = cos2(x/2)
- sin2(x/2)
cos(x) = 1 - 2sin2(x/2),
then solve for sin(x/2);
cos(x) = 2cos2(x/2) - 1,
then solve for cos(x/2).
cos(x/2) = ±sqrt((1+cos(x))/2)
tan(x/2) = (1-cos(x))/sin(x)
&&&&&= sin(x)/(1+cos(x))
tan(x/2) = sin(x/2) / cos(x/2)
&&&=&sqrt(1-cos x)/sqrt(1+cos x)
&&&=&(1-cos x)/(sin x),&&by&rationalizing the denominator&
&&&=&(sin x)/(1+cos x),&&by&rationalizing the numerator&
sin(x/2)-cos(x/2) = ±sqrt(1-sin(x)),
cos(x/2)-sin(x/2) = ±sqrt(1-sin(x))
(sin(x/2)-cos(x/2))2
= 1-2sin(x/2)cos(x/2) = 1-sin(x);
take the square root of both sides.
sin(x/2)+cos(x/2) = �sqrt(1+sin(x))
(sin(x/2)+cos(x/2))2
= 1+2sin(x/2)cos(x/2) = 1+sin(x)
tan(x/2+π/4) = (1+sin(x)) / cos(x)
&&&&&= cos(x) / (1-sin(x))
tan(x/2+π/4) = sin(x/2+π/4)/cos(x/2+π/4)
&= (sin(x/2)+cos(x/2))/(cos(x/2)-sin(x/2))
&= sqrt(1+sin x)/sqrt(1-sin x),
and then rationalize, as with tan(x/2), above.
double angle formulas
sin(2x)&=&2&sin(x)&cos(x)&
delightfully simple
using an isosceles triangle
cos(2x) = cos2(x)
&&= 1 - 2sin2(x)
&&= 2cos2(x) - 1
From the cos sum formula, using cos(x+x)
tan(2x) = sin(2x)/cos(2x)
& = 2&sin(x)&cos(x)&/&(cos2(x)-sin2(x))
&&= 2 tan(x) / (1-tan2(x))
Use the sin(2x) and cos(2x) formulas, then divide
the numerator and denominator by cos2(x)
triple, quadruple, etc. angle formulas
cos 3x = -3cos+4cos3 &
cos 4x = 1-8cos2+8cos4 &
cos 5x = 5cos-20cos3+16cos5 &
cos 6x = -1+18cos2-48cos4+32cos6 &
cos 7x = -7cos+56cos3-112cos5+64cos7
(&x& on the right hand side is omitted for clarity)
cos((n+1)x)=2cos(x)cos(nx)-cos((n-1)x)
sin 3x = (sin) (-1+4cos2)
sin 4x = (sin) (-4cos+8cos3)
sin 5x = (sin) (1-12cos2+16cos4)
sin 6x = (sin) (6cos-32cos3+32cos5)
sin 7x = (sin) (-1+24cos2-80cos4+64cos6)
. . . . . . use similar recurrence relation as above:&
sin(nx)=2sin((n-1)x)cos(x)-sin((n-2)x)
&In fact, sin(nx)/sinx=gn(cosx) for some polynomials gn
(the Chebyshev polynomial polynomials for
cos((n+1)x) above are called Chebyshev polynomials of the first
kind). They satisfy the same recurrence relation. Easy way how to
see that is just to differentiate the formula cos(nx)=fn(cosx).
continued fraction for tan nx:
&&&n tan x&&&&&&&
(n?-1?)tan?x&&&&&&&
(n?-2?)tan?x&&&&
If n is an integer, the continued fraction
terminates.& .
tan((n+1)x/2) = (sin x + sin 2x + ... + sin
&&&&&&&&&&&&&&&&&&& (cos x + cos 2x + ... + cos nx)
&. . . . . . further reading:
double angle formulas expressed in terms of
These are variously called the &Universal& substitutions and the
, where t=tan(x/2).
Weierstrass
t-substitution
tan(x) = 2 tan(x/2)/(1-tan2(x/2))
tan(x)&=&2t/(1-t2)
from the tan-of-sum formula, using tan(x/2 + x/2)
cos(x) = (1-tan2(x/2))/(1+tan2(x/2))
cos(x)&=&(1-t2)/(1+t2)
cos2(x/2) = (cos(x)+1)/2
(cos(x)+1)/2 = 1/(1+t2)
&. . . . . .& add a row for sin^2(x/2)&
sin(x) = cos(x)tan(x)
&&&& = 2 tan(x/2)/(1+tan2(x/2))
sin(x)&=&2t/(1+t2)
by combining the two formulas above.
dx = 2 cos2(x/2) dt&&&
= (cos(x)+1) dt
dx = ((1-t2)/(1+t2) + 1)
dt&& = 2dt/(1+t2)
t = tan(x/2), so dt/dx = (1/2) sec2(x/2), or
x=2 arctan(t), so dx/dt = 2/(t2+1)
sec(x) - tan(x)
&&&& = (1-tan(x/2))/(1+tan(x/2))
LHS =&(1+tan2(x/2))/(1-tan2(x/2))
- 2 tan(x/2)/(1-tan2(x/2))
&&= (1-tan(x/2))2/(1-tan2(x/2))
&&= (1-tan(x/2)/(1+tan(x/2))
(1-sin(x))/(1+sin(x))
&&&& = (1-tan(x/2))2/(1+tan(x/2))2
&&&&&= (sec(x)-tan(x))2
(1-sin(2x))/(1+sin(2x))
&&= (1-2sin(x)cos(x)) / (1+2sin(x)cos(x))&
&&= (cos(x)-sin(x))2 /
(cos(x)+sin(x))2&
&&= (1-tan(x))2 /
(1+tan(x))2&
cos(x+y) etc. -- cos of sum, sin of sum, tan of sum
cos(x+y) = cos(x)cos(y) - sin(x)sin(y)
sin(x+y) = sin(x)cos(y) + cos(x)sin(y)&
exponential proof of both identities at once:
cos(x+y)+i sin(x+y) = ex+y
&&=(cos(x)+i sin(x))&(cos(y)+i sin(y))
&&=(cos(x)cos(y)-sin(x)sin(y)) + i(sin(x)cos(y)+cos(x)sin(y))
tan(x+y) = (tan(x)+tan(y)) /
(1-tan(x)tan(y))&
cos(x) cos(y) -- sum of cos, etc. -- converting
between a sum and a product of trig functions
are important for solving integrals of
products of trig functions.& The trick is to convert these products
into sums of functions that are easy to integrate.
cos(a+b) cos(a-b) = cos2b - sin2a
cos(x) cos(y) = cos2((x-y)/2)-sin2((x+y)/2),
&& where x=a+b; y=a-b;
&&&&&&& a=(x+y)/2; b=(x-y)/2
cos(a+b) cos(a-b) =
(cos a cos b - sin a sin b) (cos a cos b + sin a sin b) =
cos2a cos2b - sin2a sin2b
(1-sin2a)cos2b - sin2a (1-cos2b)
cos2b - sin2a cos2b - sin2a
+ sin2a cos2b =
cos2b - sin2a&
cos(x) cos(y) = cos(x+y)/2 + cos(x-y)/2
cos(x+y)&=&cos(x)cos(y) - sin(x)sin(y)
cos(x-y)&=&cos(x)cos(y) + sin(x)sin(y)
the proof is to add these equations together.
cos(a) + cos(b) = 2 cos((a-b)/2)
cos((a+b)/2)
from above, where a=x+y; b=x-y; x=(a+b)/2;
y=(a-b)/2;
sin(x) sin(y) = cos(x-y)/2-cos(x+y)/2
cos(x-y)&=&cos(x)cos(y) + sin(x)sin(y)
-cos(x+y)&=&-cos(x)cos(y) + sin(x)sin(y);
the proof is to add these equations together.
cos(b) - cos(a) = 2 sin((a-b)/2)
sin((a+b)/2)
from above, where a=x+y; b=x-y; x=(a+b)/2;
y=(a-b)/2;
cos(x) sin(y) = sin(x+y)/2 - sin(x-y)/2&&
sin(x+y)&=&sin(x)cos(y) + cos(x)sin(y)&
-sin(x-y)&=&-sin(x)cos(y) + cos(x)sin(y);
the proof is to add these equations together.
sin(a) - sin(b) = 2 cos((a+b)/2)
sin((a-b)/2)
sin(a) + sin(b) = 2 cos((a-b)/2) sin((a+b)/2)
from above, where a=x+y; b=x-y; x=(a+b)/2;
y=(a-b)/2;
2nd equation replaces b with -b.
(cos(b)-cos(a)) / (sin(a)+sin(b)) =
tan((a-b)/2)
(cos(b)-cos(a)) / (sin(a)-sin(b)) = tan((a+b)/2)
divide the cos(b)-cos(a) and sin(a)+sin(b)
identities, above.
(sin(a)+sin(b)) / (cos(a)+cos(b)) =
tan((a+b)/2)
(sin(a)-sin(b)) / (cos(a)+cos(b)) = tan((a-b)/2)
divide the sin(a)+sin(b) and cos(a)+cos(b)
identities, above.
cos(a−b) cos(t+u) − cos(a+b) cos(t−u)
= sin(u+a) sin(b−t) − sin(u−a) sin(b+t)
.& This identity is used in the proof of .
generalized phase shift
a cos(x-A) + b cos(x-B) = r cos(x-θ),&where
&&&&&r = sqrt(a2 + b2
+ 2ab cos(A-B)), and
&&&&&θ = atan2(a cos A + b cos B, a sin A + b sin B)
y = a cos(x-A) + b cos(x-B)
&&= cos x (a cos A + b cos B) + sin x (a sin A + b sin B)
&&= cos x (r cos θ) + sin x (r sin θ)
&&= r cos(x-θ)
explains this more completely.
&geometric progression& identities
I call these &geometric progression& identities, because if a, b, c are
in geometric progression then a/b = b/c...
(1+sin(x)) / cos(x) = cos(x) / (1-sin(x))
1-sin2(x)=cos2(x)
(1+sin(x))(1-sin(x))=cos2(x)
(1+sin(x))/cos(x)=cos(x)/(1-sin(x))
Note: this equals tan(x/2+π/4) -- see the half angle formula section
of this page.
(1+sin(x)) / cos(x) = cos(x) / (1-sin(x))
&= (1+cos(x)+sin(x)) / (1+cos(x)-sin(x))
Linear Combination of Fractions:
If A/B = C/D, then (rA+sC)/(rB+sD) = C/D ()
(1-cos(x)) / sin(x) = sin(x) / (1+cos(x))
&= (1+sin(x)-cos(x)) / (1+sin(x)+cos(x))
Note: this equals tan(x/2) -- half angle.
and Linear Combination of Fractions.
(sec(x)-1) / tan(x) = tan(x) / (sec(x)+1))
&= (sec(x)+tan(x)-1) / (sec(x)+tan(x)+1)
and Linear Combination of Fractions.
(sec(x)-tan(x)) = 1 / (sec(x)+tan(x))
&= (1+sec(x)-tan(x))/(1+sec(x)+tan(x))
ln Isec(x)-tan(x)I = -ln Isec(x)+tan(x)I
and Linear Combination of F
Watch out for equivalent solutions of integrals!
(csc(x)-1) / cot(x) = cot(x) / (csc(x)+1))
Any difference of squares can be turned
into a geometric progression identity!
—& (A-B)/C = C/(A+B)
(csc(x)-cot(x)) = 1/ (csc(x)+cot(x))
Triangle identities -- sin(arccos(x)), etc.
I've noticed some students have trouble understanding the meaning of
sin(arccos(x)), so I'll dissect it for you in this paragraph.& Let an
angle, A, represented by the red line in the triangles, below, have a cosine
of x.& That is cos(A)=x, or in other words, A=arccos(x).& So then
sin(arccos(x))=sin(A).& To rephrase the whole thing, sin(arccos(x)) is
saying &what is the sine of the angle whose cosine is x?&
sin(arccos(x)) = ±sqrt(1-x2)&
sqrt(1-x2)
tan(arccos(x)) = ±sqrt(1-x2)/x&
cos(arcsin(x)) = ±sqrt(1-x2)&
sqrt(1-x2)
tan(arcsin(x)) = ±x/sqrt(1-x2)&
cos(arctan(x)) = ±1/sqrt(1+x2)
sqrt(1+x2)
sin(arctan(x)) = ±x/sqrt(1+x2)
sin(arcsec(x)) = ±sqrt(x2-1)/x&
sqrt(x2-1)
tan(arcsec(x)) = ±sqrt(x2-1)&
cos(arccsc(x)) = ±sqrt(x2-1)/x&
sqrt(x2-1)
tan(arccsc(x)) = ±1/sqrt(x2-1)&
cos(arccot(x)) = ±x/sqrt(1+x2)
sqrt(1+x2)
sin(arccot(x)) = ±1/sqrt(1+x2)
Euler identities and hyperbolic functions
ex&=&1&+&x/1!&+&x2/2!&+&x3/3!&+&...&
cos(x) = 1&&-&x2/2!&+&x4/4!&-&x6/6!&+&...&
sin(x) = x/1!&-&x3/3!&+&x5/5!&-&...&
cosh(x) = 1&&+&x2/2!&+&x4/4!&+&x6/6!&+&...&
sin(x) = x/1!&+&x3/3!&+&x5/5!&+&...&
Euler's formula for ex
cosh(x) = (ex+e-x)/2
sinh(x) = (ex-e-x)/2
tanh(x) = (ex-e-x)/(ex+e-x)
eix&=&cos(x) + i sin(x) =
cosh(ix) + sinh(ix)
ex&=&cos(ix) - i sin(ix) = cosh(x) + sinh(x)
Euler's formula for ex,
then replace x with -ix
cosh(ix) = cos(x)
sinh(ix) = i sin(x)
cos(ix) = cosh(x)
sin(ix) = i sinh(x)
arcsinh(x) = ln(x + sqrt(x? + 1))
arccosh(x) = ln(x + sqrt(x? - 1))
arctanh(x) = (1/2) (ln(1+x) - ln(1-x))
cosh?(x) - sinh?(x) = 1
Additional factoids
If additional trig identities come to light (send me an email if yours is
missing from this page!) then I'll add them here, until I can find a &home&
for them in the other sections of this page.
tan(x-y) + tan(y-z) + tan(z-x) = tan(x-y) *tan(y-z)*tan(z-x)
Tan-1(1/2) + Tan-1(1/5) + Tan-1(1/8)
= π/4
tan-1(1/2) + tan-1(1/5) + tan-1(1/8)=π/4.
See the diagram to the right.
Gregory's Formula for Arctan, Machin's formula for π/4
arctan x = x - x3/3 + x5/5 - x7/7 + x9/9
where -1 & x & 1
This was discovered in 1672 by James Gregory ().& It's a useful formula
because it converges quickly when x is small.& π/4 is arctan 1, so it provides a
formula for π/4, which is
π/4 = 1 - 1/3 + 1/5 - 1/7 + 1/9 - ...
(Gregory never explicitly wrote down this formula but another famous
mathematician of the time, Gottfried Leibniz (), mentioned it in print
first in 1682, and so this special case of Gregory's series is usually called
Leibnitz Formula for π.)
This formula converges very slowly.& It is better to note that&
π/4 = arctan(1) = 4 arctan(1/5) - arctan(1/239)
(This formula is called Machin's formula, discovered In 1706 by John Machin
Derivation of the formula:
Remember that tan x+y = (tan x + tan y) / (1 - tan x tan y)
so tan 2x = (2 tan x) / (1 - tan2 x)
let z be arctan 1/5.
tan 2z = (2 tan z) / (1 - tan2 z) = (2/5)/(1-1/25) =
(2/5)/(24/25) = 50/120 = 5/12
Now let w be arctan 5/12
tan 2w = (2 tan 2) / (1 - tan2 w) = (5/6)/(1-25/144) =
(5/6)/(119/144) = 720/714 = 120/119
So arctan 120/119 is an angle 4 times as large as arctan 1/5.
Now let x be arctan 120/119 and y be arctan -1/239
tan x+y = (120/119 - 1/239) / (1 + (120/119)(1/239)) = (120*239-119) /
In other words, the sum of 4 arctan 1/5 and arctan -1/239 is equal to
the arctan of 1, i.e. π/4.
How can you discover your own group of small angles with rational
tangents whose sum is π/4?
In other words, if you have an angle arctan a, what is the value of b
for which arctan a + arctan b = π/4?
Let x = arctan a, and
let y = arctan b, and let x + y = π/4
tan x+y = (tan x + tan y) / (1 - tan x tan y)
1 = (a + b) / (1 - ab)
a + b = 1 - ab
b + ab = 1 - a
b(1+a) = (1-a)
b = (1-a) / (1+a)
After John Machin calculated that 4 arctan 1/5 is equal to arctan
120/119, then he presumably set a=120/119 and used this formula for b
b = (1-a) / (1 + a)
= (1 - 120/119) / (1 + 120/119)
= (-1/119) / (239/119)
Here's another one that adds up to π/4: 3 arctan 1/4 + arctan 5/99
Internet references
SOS Math: &
(b. 1646, d. 1716) was a German philosopher, mathematician, and logician who is
probably most well known for having invented the differential and integral
calculus (independently of Sir Isaac Newton).
Wikipedia: &
Mathworld article: .
Mathworld: , including formulas for ,
Mathworld:
formulas, which convert sums of sines and cosines into products of sines and
cosines, and vice-versa.
Related Pages in this website
functions of any multiple of an angle.
calculus section of this website
is a clever trig substitution that lets you solve the kind of integrals that
naturally come up in the polar form of certain functions. The substitution is
t=tan(x/2).
sinh(x) and cosh(x), which, together with exp(x) and the circular functions
sin(x) and cos(x) form a family of functions.
by an arbitrary
derivations of various special integrals requires extensive use of the trig
identities on this page.
The webmaster and author of this查看: 3348|回复: 6
如何在公式里表示#N/A
阅读权限30
在线时间 小时
如何在公式里表示#N/A
阅读权限30
在线时间 小时
isna(单元格区域），若单元格为#n/a,返回true,否则返回false
阅读权限95
在线时间 小时
一般 计算值 不存在的时候！会出现 ＃N/A
阅读权限95
在线时间 小时
以下是引用latary在 10:54:43的发言：如何在公式里表示#N/A=NA()
阅读权限30
在线时间 小时
match(L20,$K$20:$K$22,)&&#n/a那这个公式如何表达呢？
阅读权限20
在线时间 小时
把公式中的＃N/A变为"＃N/A"
阅读权限200
在线时间 小时
回复：（latary）如何在公式里表示#N/A
match(L20,$K$20:$K$22,)&&#n/a表达为：not(isna(match(L20,$K$20:$K$22,)))用在IF()函数中：if(not(isna(match(L20,$K$20:$K$22,))),表达式一，表达式二)&&#n/a时返回表达式一，=#n/a时返回表达式二通常公式写为：if(isna(match(L20,$K$20:$K$22,)),表达式二，表达式一)
玩命加载中，请稍候
玩命加载中，请稍候
Powered by
本论坛言论纯属发表者个人意见，任何违反国家相关法律的言论，本站将协助国家相关部门追究发言者责任！ & & 本站特聘法律顾问：徐怀玉律师 李志群律师