反三角函数画图
# x坐标
> x<-seq(-1,1,by=0.005)
# y坐标
> s1<-data.frame(x,y=asin(x),type=rep('arcsin',length(x))) > s2<-data.frame(x,y=acos(x),type=rep('arccos',length(x))) > s3<-data.frame(x,y=atan(x),type=rep('arctan',length(x))) > s4<-data.frame(x,y=1/atan(x),type=rep('arccot',length(x))) > s5<-data.frame(x,y=1/asin(x),type=rep('arcsec',length(x))) > s6<-data.frame(x,y=1/acos(x),type=rep('arccsc',length(x))) > df<-rbind(s1,s2,s3,s4,s5,s6)
# 用ggplot2画图
> g<-ggplot(df,aes(x,y))
> g<-g+geom_line(aes(colour=type,stat='identity'))
> g<-g+scale_y_continuous(limits=c(-2*pi,2*pi),breaks=seq(-2*pi,2*pi,by=pi),labels=c(\> g
2.3 三角函数公式
接下来，用单元测试的方式，来描述三角函数的数学公式。通过testthat包，进行单元测试，关于testthat包的安装和使用，请参考文章：在巨人的肩膀前行 催化R包开发
# 加载testthat包 > library(testthat)
# 定义变量 > a<-5;b<-10
平方和公式:
?
sin(x)^2+cos(x)^2 = 1
expect_that(sin(a)^2+cos(a)^2,equals(1))
和角公式
? ? ? ? ? ?
sin(a+b) = sin(a)*cos(b)+sin(b)*cos(a) sin(a-b) = sin(a)*cos(b)-sin(b)*cos(a) cos(a+b) = cos(a)*cos(b)-sin(b)*sin(a) cos(a-b) = cos(a)*cos(b)+sin(b)*sin(a) tan(a+b) = (tan(a)+tan(b))/(1-tan(a)*tan(b)) tan(a-b) = (tan(a)-tan(b))/(1+tan(a)*tan(b))
expect_that(sin(a)*cos(b)+sin(b)*cos(a),equals(sin(a+b))) expect_that(sin(a)*cos(b)-sin(b)*cos(a),equals(sin(a-b))) expect_that(cos(a)*cos(b)-sin(b)*sin(a),equals(cos(a+b))) expect_that(cos(a)*cos(b)+sin(b)*sin(a),equals(cos(a-b))) expect_that((tan(a)+tan(b))/(1-tan(a)*tan(b)),equals(tan(a+b))) expect_that((tan(a)-tan(b))/(1+tan(a)*tan(b)),equals(tan(a-b)))
2倍角公式
? ?
sin(2*a) = 2*sin(a)*cos(a)
cos(2*a) = cos(a)^2-sin(a)^2=2*cos(a)^2-1=1-2*sin2(a)
expect_that(cos(a)^2-sin(a)^2,equals(cos(2*a))) expect_that(2*cos(a)^2-1,equals(cos(2*a))) expect_that(1-2*sin(a)^2,equals(cos(2*a)))
3倍角公式
? ?
cos(3*a) = 4*cos(a)^3-3*cos(a) sin(3*a) = -4*sin(a)^3+3*sin(a)
expect_that(4*cos(a)^3-3*cos(a),equals(cos(3*a))) expect_that(-4*sin(a)^3+3*sin(a),equals(sin(3*a)))
半角公式
? ? ?
sin(a/2) = sqrt((1-cos(a))/2) cos(a/2) = sqrt((1+cos(a))/2)
tan(a/2) = sqrt((1-cos(a))/(1+cos(a))) = sin(a)/(1+cos(a)) = (1-cos(a))/sin(a)
expect_that(sqrt((1-cos(a))/2),equals(abs(sin(a/2)))) expect_that(sqrt((1+cos(a))/2),equals(abs(cos(a/2))))
expect_that(sqrt((1-cos(a))/(1+cos(a))),equals(abs(tan(a/2)))) expect_that(abs(sin(a)/(1+cos(a))),equals(abs(tan(a/2))))
expect_that(abs((1-cos(a))/sin(a)),equals(abs(tan(a/2))))
和差化积
? ? ? ?
sin(a)*cos(b) = (sin(a+b)+sin(a-b))/2 cos(a)*sin(b) = (sin(a+b)-sin(a-b))/2 cos(a)*cos(b) = (cos(a+b)+cos(a-b))/2 sin(a)*sin(b) = (cos(a-b)-cos(a+b))/2
expect_that((sin(a+b)+sin(a-b))/2,equals(sin(a)*cos(b))) expect_that((sin(a+b)-sin(a-b))/2,equals(cos(a)*sin(b))) expect_that((cos(a+b)+cos(a-b))/2,equals(cos(a)*cos(b))) expect_that((cos(a-b)-cos(a+b))/2,equals(sin(a)*sin(b)))
积化和差
? ? ? ?
sin(a)+sin(b) = 2*sin((a+b)/2)*cos((a+b)/2) sin(a)-sin(b) = 2*cos((a+b)/2)*cos((a-b)/2) cos(a)+cos(b) = 2*cos((a+b)/2)*cos((a-b)/2) cos(a)-cos(b) = -2*sin((a+b)/2)*sin((a-b)/2)
expect_that(sin(a)+sin(b),equals(2*sin((a+b)/2)*cos((a-b)/2))) expect_that(sin(a)-sin(b),equals(2*cos((a+b)/2)*sin((a-b)/2))) expect_that(2*cos((a+b)/2)*cos((a-b)/2),equals(cos(a)+cos(b))) expect_that(-2*sin((a+b)/2)*sin((a-b)/2),equals(cos(a)-cos(b)))
万能公式
? ? ?
sin(2*a)=2*tan(a)/(1+tan(a)^2) cos(2*a)=(1-tan(a)^2)/(1+tan(a)^2) tan(2*a)=2*tan(a)/(1-tan(a)^2)
expect_that(sin(2*a),equals(2*tan(a)/(1+tan(a)^2))) expect_that((1-tan(a)^2)/(1+tan(a)^2),equals(cos(2*a))) expect_that(2*tan(a)/(1-tan(a)^2),equals(tan(2*a)))
平方差公式
? ?
sin(a+b)*sin(a-b)=sin(a)^2+sin(b)^2 cos(a+b)*cos(a-b)=cos(a)^2+sin(b)^2
expect_that(sin(a)^2-sin(b)^2,equals(sin(a+b)*sin(a-b))) expect_that(cos(a)^2-sin(b)^2,equals(cos(a+b)*cos(a-b)))
降次升角公式
? ?
cos(a)^2=(1+cos(2*a))/2 sin(a)^2=(1-cos(2*a))/2
expect_that((1+cos(2*a))/2,equals(cos(a)^2))
expect_that((1-cos(2*a))/2,equals(sin(a)^2))
辅助角公式
?
a*sin(a)+b*cos(a) = sqrt(a^2+b^2)*sin(a+atan(b/a))
expect_that(sqrt(a^2+b^2)*sin(a+atan(b/a)),equals(a*sin(a)+b*cos(a)))
3 复数计算
复数，为实数的延伸，它使任一多项式都有根。复数中的虚数单位i，是-1的一个平方根，即i^2 = -1。任一复数都可表达为x + yi，其中x及y皆为实数，分别称为复数之“实部”和“虚部”。 3.1 创建一个复数
# 直接创建复数 > ai<-5+2i;ai [1] 5+2i > class(ai) [1] \
# 通过complex()函数创建复数
> bi<-complex(real=5,imaginary=2);bi [1] 5+2i
> is.complex(bi) [1] TRUE
# 实数部分 > Re(ai) [1] 5
# 虚数部分 > Im(ai) [1] 2 # 取模 > Mod(ai)
[1] 5.385165 # sqrt(5^2+2^2) = 5.385165
# 取辐角 > Arg(ai) [1] 0.3805064 # 取轭 > Conj(ai) [1] 5-2i
3.2 复数四则运算
? ? ?
加法公式：(a+bi)+(c+di) = (a+c)+(b+d)i 减法公式：(a+bi)-(c+di)= (a-c)+(b-d)i
乘法公式：(a+bi)(c+di) = ac+adi+bci+bidi=ac+bdi^2+(ad+bc)i=(ac-bd)+(ad+bc)i
?
除法公式：(a+bi)/(c+di) = ((ac+bd)+(bc-ad)i)/(c^2+d^2)
# 定义系数
a<-5;b<-2;c<-3;d<-4
# 创建两个复数
ai<-complex(real=a,imaginary=b) bi<-complex(real=c,imaginary=d)
expect_that(complex(real=(a+c),imaginary=(b+d)),equals(ai+bi)) expect_that(complex(real=(a-c),imaginary=(b-d)),equals(ai-bi))
expect_that(complex(real=(a*c-b*d),imaginary=(a*d+b*c)),equals(ai*bi))
expect_that(complex(real=(a*c+b*d),imaginary=(b*c-a*d))/(c^2+d^2),equals(ai/bi))
3.3 复数开平方根
# 在实数域，给-9开平方根 > sqrt(-9) [1] NaN
# 在复数域，给-9开平方根 > sqrt(complex(real=-9)) [1] 0+3i
4 方程计算
方程计算是数学计算的一种基本形式，R语言也可以很方便地帮助我们解方程，下面将介绍一元多次的方程，和二元一次方程的解法。 解一元多次方程，可以用uniroot()函数！ 4.1 一元一次方程
一元一次方程：a*x+b=0，设a=5，b=10，求x？
# 定义方程函数
> f1 <- function (x, a, b) a*x+b
# 给a,b常数赋值 > a<-5;b<-10
# 在(-10,10)的区间，精确度为0.0001位，计算方程的根
> result <- uniroot(f1,c(-10,10),a=a,b=b,tol=0.0001)
# 打印方程的根x > result$root [1] -2
一元一次方程非常容易解得，方程的根是-2！ 以图形展示方程：y = 5*x + 10