Introduction to Calculus/Quiz 1/Answers

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

(\theta)=1$$
 * 1) Evaluate $$\tan(\theta) \,\ $$ in terms of $$\sin(\theta) \,\ $$
 * $$\tan(\theta)=\sin(\theta)/\sqrt(1-(\sin^2(\theta)) \,\ $$
 * Shyam (T/C)
 * 1) If $$\csc(\theta)=1/x, \,\ $$ then what does $$x \,\ $$ equal?
 * $$x=\sin(\theta) \,\ $$ where $$x = [-1, 1] \,\ $$
 * Shyam (T/C)
 * 1) Prove $$\tan^2(\theta)+1=\sec^2(\theta) \,\ $$using $$ \,\ \sin^2(\theta)+ \cos^2
 * $$\sin^2(\theta)+ cos^2(\theta)=1 \,\ $$
 * 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)
 * 1) $$\cos(A+B)=\cos(A)\cos(B)-\sin(A)\sin(B) \,\ $$
 * 2) * 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)
 * 1) * 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)
 * 1) * 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)
 * 1) * (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)