Introduction to Calculus/Quiz 1/Answers

If you can pass this quiz, you are ready to take this course

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)