Introduction to Calculus/Quiz 1/Answers

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  1. Evaluate \tan(\theta) \,\ in terms of \sin(\theta) \,\
    \tan(\theta)=\sin(\theta)/\sqrt(1-(\sin^2(\theta)) \,\
    Shyam (T/C)
  2. If \csc(\theta)=1/x, \,\ then what does x \,\ equal?
    x=\sin(\theta) \,\ where x = [-1, 1] \,\
    Shyam (T/C)
  3. Prove \tan^2(\theta)+1=\sec^2(\theta) \,\ using \,\ \sin^2(\theta)+ \cos^2 (\theta)=1
    \sin^2(\theta)+ cos^2(\theta)=1 \,\
    divide both sides by \cos^2(\theta)=>\sin^2(\theta)/cos^2(\theta)+1=1/cos^2(\theta) \,\
    =>\tan^2(\theta)+1=sec^2(\theta) \,\
    Shyam (T/C)
  4. \cos(A+B)=\cos(A)\cos(B)-\sin(A)\sin(B) \,\
    • Find the double angle idenities for the cosine function using the above rule.
      replace B \,\ by A=>\cos(A+A)=\cos(A)\cos(A)-\sin(A)\sin(A) \,\
      =>\cos(2A)=\cos^2(A)-\sin^2(A) \,\
    Shyam (T/C)
    • Find the half angle idenities from the double angle idenities.
      =>\cos(2A)=\cos^2(A)-\sin^2(A) \,\
      replace A \,\ by A/2=>\cos(A)=2\cos^2(A/2)-1 \,\ using \sin^2(A)+ \cos^2(A)=1 \,\
    Shyam (T/C)
    • Find the value of \,\ \cos^2(\theta) without exponents using the above rules
      =>\cos(2\theta)=2\cos^2(\theta)-1 \,\
      =>\cos^2(\theta)=(1+\cos(2\theta))/2 \,\
    Shyam (T/C)
    • (Challenge) Find the value of \,\ \cos^3(\theta) without exponents
      \cos(3\theta)=cos(\theta+2\theta) \,\
      =>\cos(3\theta)=cos(\theta)\cos(2\theta)-sin(\theta)\sin(2\theta) \,\
      =>\cos(3\theta)=cos(\theta)(2\cos^2(\theta)-1)-sin(\theta)(2\sin(\theta)\cos(\theta)) \,\ using \cos(2\theta)=2\cos^2(\theta)-1 \,\ and \sin(2\theta)=2\sin(\theta)\cos(\theta) \,\
      =>\cos(3\theta)=cos(\theta)((2\cos^2(\theta)-1)-2sin^2(\theta)) \,\
      =>\cos(3\theta)=cos(\theta)(4\cos^2(\theta)-3) \,\ using \,\ \sin^2(\theta)+ \cos^2(\theta)=1
      =>4\cos^3(\theta)=cos(3\theta)+3cos(\theta) \,\
      =>\cos^3(\theta)=(cos(3\theta)+3cos(\theta))/4 \,\
    Shyam (T/C) 19:42, 18 November 2006 (UTC)
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