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    • 2.11.1: Review of Trigonometry
      https://phys.libretexts.org/Courses/Georgia_State_University/GSU-TM-Physics_II_(2212)/02%3A_Math_Review/2.11%3A_Vectors/2.11.01%3A_Review_of_Trigonometry
      \(\begin{align} \cos (α+β) &= \cos α \cos β −\sin α \sin β \\ \cos (α−β) &= \cos α \cos β+\sin α \sin β \\ \sin (α+β) &= \sin α \cos β+\cos α \sin β \\ \sin (α−β) &= \sin α \cos β−\cos α \sin β \\ \ta...\(\begin{align} \cos (α+β) &= \cos α \cos β −\sin α \sin β \\ \cos (α−β) &= \cos α \cos β+\sin α \sin β \\ \sin (α+β) &= \sin α \cos β+\cos α \sin β \\ \sin (α−β) &= \sin α \cos β−\cos α \sin β \\ \tan (α+β) &= \frac{\tan α+\tan β}{1−\tan α \tan β} \\ \tan (α−β) &= \frac{\tan α− \tan β}{1+\tan α \tan β} \end{align}\)
    • 2.9.1: Review of Trigonometry
      https://phys.libretexts.org/Courses/Georgia_State_University/GSU-TM-Introductory_Physics_II_(1112)/02%3A_Math_Review/2.09%3A_Vectors/2.9.01%3A_Review_of_Trigonometry
      \(\begin{align} \cos (α+β) &= \cos α \cos β −\sin α \sin β \\ \cos (α−β) &= \cos α \cos β+\sin α \sin β \\ \sin (α+β) &= \sin α \cos β+\cos α \sin β \\ \sin (α−β) &= \sin α \cos β−\cos α \sin β \\ \ta...\(\begin{align} \cos (α+β) &= \cos α \cos β −\sin α \sin β \\ \cos (α−β) &= \cos α \cos β+\sin α \sin β \\ \sin (α+β) &= \sin α \cos β+\cos α \sin β \\ \sin (α−β) &= \sin α \cos β−\cos α \sin β \\ \tan (α+β) &= \frac{\tan α+\tan β}{1−\tan α \tan β} \\ \tan (α−β) &= \frac{\tan α− \tan β}{1+\tan α \tan β} \end{align}\)
    • 2.1: Review of Trigonometry
      https://phys.libretexts.org/Courses/Georgia_State_University/GSU-TM-Physics_I_(2211)/02%3A_Vectors_and_Math_Review_Topics/2.01%3A_Review_of_Trigonometry
      \(\begin{align} \cos (α+β) &= \cos α \cos β −\sin α \sin β \\ \cos (α−β) &= \cos α \cos β+\sin α \sin β \\ \sin (α+β) &= \sin α \cos β+\cos α \sin β \\ \sin (α−β) &= \sin α \cos β−\cos α \sin β \\ \ta...\(\begin{align} \cos (α+β) &= \cos α \cos β −\sin α \sin β \\ \cos (α−β) &= \cos α \cos β+\sin α \sin β \\ \sin (α+β) &= \sin α \cos β+\cos α \sin β \\ \sin (α−β) &= \sin α \cos β−\cos α \sin β \\ \tan (α+β) &= \frac{\tan α+\tan β}{1−\tan α \tan β} \\ \tan (α−β) &= \frac{\tan α− \tan β}{1+\tan α \tan β} \end{align}\)

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