Tan(theta)-sin(theta)/sin(cube)(theta) When limit theta tends to zero.
asked Oct 8, 2016 in Calculus Answers by anonymous

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:

To avoid this verification in future, please log in or register.

1 Answer

Writing t for theta, tan(t)=sin(t)sec(t),

so if t≠0 we can write: (tan(t)-sin(t))/(sin(t))^3=(sec(t)-1)/(sin(t))^2.

The expansion of cos(t)=1-t^2/2 when t is small, and sin(t)=t.

Also, sec(t)=1/cos(t)=(1-t^2/2)^-1=1+t^2/2 when t is small. Therefore sec(t)-1=t^2/2 And (sin(t))^2=t^2.

So the quotient approaches t^2/2t^2=1/2 as t approaches zero.

The limit is therefore 1/2.

answered Oct 8, 2016 by Rod Top Rated User (425,260 points)

Related questions

Welcome to MathHomeworkAnswers.org, where students, teachers and math enthusiasts can ask and answer any math question. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry, fractions, solving expression, simplifying expressions and more. Get answers to math questions. Help is always 100% free!
77,982 questions
81,731 answers
60,935 users