Trigonometric identities(Sin(alpha)&Cos(alpha))

Trigonometry Tutorial

Trigonometric identities(Sin(alpha)&Cos(alpha))

Intro

Sin(alpha)

  • sin(alpha)=y/r
  • Basic rules of sin(alpha)
    • -sin(alpha)=sin(-alpha) -> -alpha is alpha in inverse of rotation direction.So x value will stay same but y value will be negative of current y.
    • in a triangle,there is a sine function rule with every angle of triangle.Lets say there is a triangle with alpha,beta and thita and the a,b and c values that represent one of triangle edge.So if alpha angle is between b edge and c edge,so beta angle is between a edge and c edge too,thita agle is between a edge and b edge then a/sin(alpha)=b/sin(beta)=c/sin(thita)
    • you can calculate sin(alpha) as sin(alpha)=(ei*alpha-e-i*alpha)/(2*i)

Cos(alpha)

  • cos(alpha)=x/r
  • Basic rules of cos(alpha)
    • cos(alpha)=cos(-alpha) -> -alpha is alpha in inverse of rotation direction.So x value will stay same.
    • cos(alpha)+cos(beta)=2*cos((alpha+beta)/2)*cos((alpha-beta)/2)->if you draw a coorditane system you can calculate it
    • you can calculate cos(alpha) as cos(alpha)=(ei*alpha+e-i*alpha)/2

Some formula contains sin(alpha) also cos(alpha)

  • alpha+beta=90o then sin(alpha)=cos(beta)
  • sin2(alpha)+cos2(alpha)=1
  • sin(alpha+beta)=sin(alpha)*cos(beta)+sin(beta)*cos(alpha)
    and cos(alpha+beta)=cos(alpha)*cos(beta)-sin(alpha)*sin(beta)
  • sin(alpha-beta)=sin(alpha)*cos(beta)-sin(beta)*cos(alpha) and cos(alpha-beta)=cos(alpha)*cos(beta)+sin(alpha)*sin(beta)
  • sin(alpha)+sin(beta)=2*sin((alpha+beta)/2)*cos((alpha-beta)/2)
  • sin(alpha)-sin(beta)=2*sin((alpha-beta)/2)*cos((alpha+beta)/2)
  • cos(alpha)-cos(beta)=-2*sin((alpha+beta)/2)*sin((alpha-beta)/2)
  • sin(2*alpha)=2*sin(alpha)*cos(alpha) and cos(2*alpha)=cos2(alpha)-sin2(alpha)

Sample Problem

  1. Calculate sin(75)+sin(15) .
  2. Calculate cos(75)+cos(15) .
  3. Find sin(2*alpha)=2*sin(alpha)cos(alpha) with sin(alpha+beta) formula .

Solution

  1. sin(75)+sin(15)=2*sin((75+15)/2)*cos((75-15)/2))
    • sin((75+15)/2)=sin(45)=1/21/2
    • cos((75-15)/2)=cos(30)/2=31/2/2
    • 2*1/21/2*31/2/2=(3/2)1/2
    • answer is (3/2)1/2
  2. cos(75)+cos(15)=2*cos((75+15)/2)*cos((75-15)/2))
    • cos((75+15)/2)=cos(45)=1/21/2
    • cos((75-15)/2)=cos(30)/2=31/2/2
    • 2*1/21/2*31/2/2=(3/2)1/2
    • answer is (3/2)1/2
  3. sin(alpha+beta)=sin(alpha)*cos(beta)+cos(alpha)*sin(beta)
    • sin(alpha+alpha)=sin(alpha)*cos(alpha)+cos(alpha)*sin(alpha)=2*sin(alpha)cos(alpha)



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