Relation between Trigonometric Functions

Posted In: Mathematics
  1. $sin\ θ=1/{cosec\ θ}=±√{1-cos^2θ}=±√{1/2(1-cos\ 2θ)}$
    $=2cos^2(θ/2-π/4)-1={2tan{θ/2}}/{1+tan^2{θ/2}}$
  2. $cos\ θ=1/{sec\ θ}=±√{1-sin^2θ}=±√{1/2(1+cos\ 2θ)}$
    $=2cos^2{θ/2}-1={1-tan^2{θ/2}}/{1+tan^2{θ/2}}$
  3. $tan\ θ=1/{cot\ θ}={sin\ θ}/{cos\ θ}=±√{sec^2\ θ -1}$
    $={sin\ 2θ}/{1+cos\ 2θ}={1-cos\ 2θ}/{sin\ 2θ}$
    $=±√{{1-cos\ 2θ}/{1+cos\ 2θ}}={2tan\ {θ/2}}/{1+tan^2{θ/2}}$
  4. $cot\ θ=1/{tan\ θ}={cos\ θ}/{sin\ θ}=±√{cosec^2\ θ -1}$
    $={1+cos\ 2θ}/{cos\ 2θ}={sin\ 2θ}/{1-cos\ 2θ}$
    $=±√{{1+cos\ 2θ}/{1-cos\ 2θ}}={1+tan^2{θ/2}}/{2tan\ {θ/2}}$
  5. $sec\ θ=1/{cos\ θ}=±√{1+tan^2θ}={1+tan^2{θ/2}}/{1-tan^2{θ/2}}$
  6. $cosec\ θ=1/{sin\ θ}=±√{1+cot^2θ}={1+tan^2{θ/2}}/{2tan{θ/2}}$

Post Tags: Trigonometry


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