Trigonometric Formulas

These are the common formulas used in Chapter 9 to 14 of Textbook of Algebra and Trigonometry Class XI, Punjab Textbook Board Lahore. This handout is very helpful to remember the formulas. All these formulas are given for real valued and defined trigonometric functions. A PDF file can be downloaded for high quality printing and a word file is also given if you wish to modify the contents or credit as you need.

  • sin2θ+cos2θ=1sin2θ+cos2θ=1
  • 1+tan2θ=sec2θ1+tan2θ=sec2θ
  • 1+cot2θ=csc2θ1+cot2θ=csc2θ
  • sin(θ)=sinθsin(θ)=sinθ
  • cos(θ)=cosθcos(θ)=cosθ
  • tan(θ)=tanθtan(θ)=tanθ
  • sin(α+β)=sinαcosβ+cosαsinβsin(α+β)=sinαcosβ+cosαsinβ
  • sin(αβ)=sinαcosβcosαsinβsin(αβ)=sinαcosβcosαsinβ
  • cos(α+β)=cosαcosβsinαsinβcos(α+β)=cosαcosβsinαsinβ
  • cos(αβ)=cosαcosβ+sinαsinβcos(αβ)=cosαcosβ+sinαsinβ
  • tan(α+β)=tanα+tanβ1tanαtanβtan(α+β)=tanα+tanβ1tanαtanβ
  • tan(αβ)=tanαtanβ1+tanαtanβtan(αβ)=tanαtanβ1+tanαtanβ
  • sin2θ=2sinθcosθsin2θ=2sinθcosθ
  • cos2θ=cos2θsin2θcos2θ=cos2θsin2θ
  • tan2θ=2tanθ1tan2θtan2θ=2tanθ1tan2θ
  • sin2θ2=1cosθ2sin2θ2=1cosθ2
  • cos2θ2=1+cosθ2cos2θ2=1+cosθ2
  • tan2θ2=1cosθ1+cosθtan2θ2=1cosθ1+cosθ
  • sin3θ=3sinθ4sin3θsin3θ=3sinθ4sin3θ
  • cos3θ=4cos3θ3cosθcos3θ=4cos3θ3cosθ
  • tan3θ=3tanθtan3θ13tan2θtan3θ=3tanθtan3θ13tan2θ
  • sin2θ=2tanθ1+tan2θsin2θ=2tanθ1+tan2θ
  • cos2θ=1tan2θ1+tan2θcos2θ=1tan2θ1+tan2θ
  • tan2θ=2tanθ1tan2θtan2θ=2tanθ1tan2θ
  • sin(α+β)+sin(αβ)=2sinαcosβsin(α+β)+sin(αβ)=2sinαcosβ
  • sin(α+β)sin(αβ)=2cosαsinβ
  • cos(α+β)+cos(αβ)=2cosαcosβ
  • cos(α+β)cos(αβ)=2sinαsinβ
  • sinθ+sinϕ=2sinθ+ϕ2cosθϕ2
  • sinθsinϕ=2cosθ+ϕ2sinθϕ2
  • cosθ+cosϕ=2cosθ+ϕ2cosθϕ2
  • cosθcosϕ=2sinθ+ϕ2sinθϕ2
  • sin1A+sin1B=sin1(A1B2+B1A2)
  • sin1Asin1B=sin1(A1B2B1A2)
  • cos1A+cos1B=cos1(AB(1A2)(1B2))
  • cos1Acos1B=cos1(AB+(1A2)(1B2))
  • tan1A+tan1B=tan1A+B1AB
  • tan1Atan1B=tan1AB1+AB

Three Steps to solve sin(nπ2±θ)

Step I: First check that n is even or odd.

Step II: If n is even then the answer will be in sin and if the n is odd then sin will be converted to cos and vice versa (i.e. cos will be converted to sin).

Step III: Now check in which quadrant nπ2±θ is lying if it is in Ist or IInd quadrant the answer will be positive as sin is positive in these quadrants and if it is in the IIIrd or IVth quadrant the answer will be negative.
e.g. sin667 =sin(7(90)+37)
Since n=7 is odd so answer will be in cos and 667 is in IVth quadrant and sin is –ive in IVth quadrant therefore answer will be in negative. i.e sin667=cos37.
Similar technique is used for other trigonometric ratios. i.e. tancot and seccsc.