Graphs of Reciprocal Trigonometric Functions

This lesson shows how to graph the reciprocal trigonometric functions (y = cscx, y = secx and y = cotx) using the y = sinx, y = cosx and y = tanx functions. This lesson was created for the MHF4U Advanced Functions course in the province of Ontario, Canada

[dragonflame]

you're amazing! you're teaching me in a few minutes what my teacher couldnt teach me in weeks! i have a test tomorrow, and you saved my life. I THANK

[taylaplasket]

You helped me gain conceptual understanding of the topic. We need more teachers like you. Thank you!!

[AlRichards]

@chroncile In order to evaluate cos(5pi/12) I used the cos (x+y) identity. I wanted to break down the 5pi/12 into a sum of 2 angles that add to 5pi/12 and are both special angles on the unit circle (to get exact values). If you look at the angles in the first quadrant: pi/6, pi/4, pi/3, .. and change all of them so they have denominators of 12, they become 2pi/12, 4pi/12, 3pi/12 ... Looking for 2 angles that add to 5pi/12 we see that 2pi/12 and 3pi/12 do so we simplify & use pi/6+pi/4.

[jennamarcus]

i was having panic attacks cause i have so many tests and i don't understand this stuff but you just helped me a lot! thank you so much! your extra effort to help us out really makes a difference.

[AlRichards]

@zacmiley12 It is actually cos(TT/6 + TT/4). I used the trig identity cos(A + B) = cosAcosB - sinAsinB with A = TT/6 and B = TT/4. Therefore, cos(TT/6 + TT/4) = cos(TT/6)cos(TT/4) - sin(TT/6)sin(TT/4) = .... This is done so we can find the exact values for sec(5TT/12).

[AlRichards]

@PFRC2010 You can have a root in the denominator. Many people rationalize the denominator, so there isn't a root there, but it isn't always necessary. Let R(x) mean the root of x.
6000/([R(6) - R(2)] we'd rationalize the denominator by multiplying (top & bottom) by the conjugate of R(6) - R(2), or R(6)+R(2). So = {6000R(6)-6000R(2)}/(6-2)} = {6000R(6) - 6000R(2)}/4} = 1500R(6) - 1500R(2), which is a bit more complex than the orginial expression.

[emmacxoxo]

omg this literally saved my life, i have a test tomorrow and my teacher is terrible thank u soooo much this was amazing ! <3

[yashbatra]

This is an extremely high-quality video for 2010! 😊

[mandyw]

i dont really understand any of these examples but i really like your voice. (u sound like a teacher that really care about their student .)
(im watching it with my sister, she needed help)

[AlRichards]

Are you refering ot the fairly complex fraction in the last example? If so, the only reason that was as complicated as it was because the question asked for the "exact value". I don't know of a simpler way to do that at this point.

[AlRichards]

Which example are you refering to? There are three different graphs drawn in the video with vertical asymptotes.

[JenaSherry]

You rock! You've just made my daughter's life MUCH easier. :) Bravo!! Thanks! 

[garfieldcatgc]

for sec, cosec and cot can we generate some sort of compound angle formula so that we can avoid having awkward fractions like we had in the video. It is just that when i have non-c alc exam it is often better to simplfy expressions when possible and avoid making error

[AlRichards]

I don't believe I said 1/infinity = 0. However, 1 divided by a very large number approaches a value of 0. Consider 1/100 = 0.01, 1/1000 = 0.001, = Notice that the larger the number divided into 1, the smaller the result and the closer it gets to 0.
At pi/2 the tangent function approaches an infnite value, although nothing can actually equal infinity. So at pi/2 1 is divided by a very, very large number which will essentially be equal to 0.

[AlRichards]

Perhaps a better way to evaluate cot(pi/2) is using a right triangle. If the angle is pi/2 then the adjacent side has a value of 0, so cot(pi/2) = adj/opp = 0/opp = 0.

[kavinlingaraju]

Thank you so much for these! You've just made my studying a lotttt easier!

[AlRichards]

Hey, thanks. I really appreciate the comment.

[flrtboy]

You are awesome...I just need a review on them.

[SuperMilliee]

very helpful! thx! im gonna ace it for tmw's test

[PFRC]

I thought you can't have square roots in the denominator of your answer, is that true?