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

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