Thank you for your time and explanation of trig. identities! I had a mental block in highschool some 30 years ago and now I am learning what I did not learn back then. You have a calm and easy to follow approach to these tutorials. Thank you so much!
Thank you so much! You explain things so well. My teacher spent less than a day going over this along with other lessons as classes are two hours long and its packed with other lessons. Thank you so much, it was very helpful!
Thank you for this video! I'm currently in Advanced Functions and a close friend of mine sent this video in hopes it might help. I'm just writing you to say that it did just that :)
This is like watching a sorcerer do magic! Thank you so much, I've been struggling with this topic and watching your examples helped me understand what I was getting wrong.
Thank you so much. My question wasnt directly answered in this video but hearing your thoufht process while doing it helped me understand how it works and i got it! So helpful!
I am so thankful for this, I have been homeschooling and the course material they have provided is not understandable at some points so I come to youtube for help. I want you to know that it means a lot to me that you upload videos. God bless you!
You're the best! Thanks for your support! I'm sure this isn't necessary to ask you, but please like, share and subscribe to increase the channel's visibility. : )
Hi Ms. Havrot. I think the ratio for pi/6 is wrong because you drew the special triangle wrong (7:14). Thank you so much for the help by the way. You are an amazing teacher :)
I don’t see a mistake. Pi/3 is the 60° angle (bottom) and pi/6 is 30° and is the angle at the top that we cut in two when we split the equilateral triangle in half. Easy to mix up when you see the 6 in the denominator often students think that’s 60° but remember that hits 180°/6 = 30° Thanks for the lovely compliment 😊
hi Mrs Harvot! I was wondering how I could do the very first calculation of the first example (when I switch from radians to degrees), what is the easiest way to do so without a calculator? my class won't be allowed calculators on our quiz and test unfortunately so I need some help with that 🙃 you're an amazing teacher btw, I learn so much from your videos!
Two ways … first of all you know that pi = 180° so you can multiply 13 x 180/12 or slightly easier is to recognize that 13pi/12 is (12/12+1/12)pi and that (12/12)pi is simply pi or 180° so that you just have 180° + pi/12 which is 180° + 15°
16:31 this kinda look like a corelated angle, the cos would become a sin and the angle would be in the 4 quadrant which would make sin positive. Can I also get the answer through that or is this reasoning just conveniently worked out for this question?
Hello, Mrs.Havrot! I just had a question on page 401, question 10, is there actually a possible way to solve it without a calculator? Thank you for your time and effort!
Of course you can! You are given some clues. Sina = 7/25. So use opposite has a length of 7, hypotenuse has a length of 25 so find the adjacent using Pythagorean theorem. Do the same for cosb =5/13. (Find the opposite length) then using the formula for sin(a+b) just plug in the values and away you go!
Hi Ms, For the exact values, I am confused as to how you came up with them. Can you please explain how you got them, especially the 240° for π/4. Thank you in advance.
I'd like to reiterate my question; I understand that they are radian values converted to degrees, but why did you not use π/2 or π? As I noticed a pattern that you did not use the values π/2, π, 3π/2 or 2π , why did you only use the angles in between them? To add on, for π/4, I understand you got 135= 45°x 3= 3π/4 but for 240, 45°x5= 225 and 45° x6= 270, so 240 is an angle between 5π/4 and 6π/4.
Okay now I understand! I wrote 240 on the page for a multiple of 45 degrees and it should have said 225 degrees. You are correct. You can use Pi/2 and Pi ... I’m trying to think of an example where they could be used.
The problem with Pi/2 is that the tan would be undefined. Also Pi plus any of the exact values would simply give you a multiple of an exact value because 180= 45 x 4, 180= 60 x 3 and 180 = 30 x 6
@@mshavrotscanadianuniversit6234 Thank you for your detailed clarification/response Ms. Havrot! Please continue to make lessons on Chapter 8 they are very helpful!
id there is a question like cos(pi-pi/4). That equals 3pi/4 and no exact values add up to 3pi/4. So do I just leave it as it is and draw a graph of cos and show that at pi, cos is 0? we can obviously show what cos is at pi/4 because it's 45 degrees, but what about cos pi? Thank You!
The cos (pi-pi/4) can be calculated using the subtraction formula, however cos(3pi/4) is simply -(sqrt2)/2 (with the denominator rationalized) or - 1/(sqrt2)
Sometimes it takes a second or third time hearing the lesson to really get what's happening. Of course you still need to practice! Please like, share and subscribe to increase the visibility of the channel : )
Hi, thank you so much for these videos, they are of huge help. Can I sak you to please show me how to do question 8.c) on page 401? its "Determine the exact value of each trig ratio" 8.c) is cos ((11pi)/12) Thanks a million!
mshavrot.pbworks.com/w/file/fetch/138233262/cos%20%2811pi%3A12%29.pdf It's sideways! I can't figure out how to turn it but I'm sure you can still read it. : )
Hey, Ms.Havrot. My teacher asked us this question, "Determine the exact value of cos(a-b) given that csca=-5/3, (pi≤a≥3*pi/2) and secb=25/7, (0≤bpi/2)" I can't seem to figure it out I was hoping you could help?
First, change the csca to sina and the secb to cosb. Now that you know the ratios and which quadrants you are in (a in Q3 and b in Q1) you can determine, using Pythagorean theorem the ratios for cosa and sinb. Then all you need to do is use the subtraction formula fir cos (a-b) = cosacosb + sinasinb and plug in your values 😊
Thank you for your time and explanation of trig. identities! I had a mental block in highschool some 30 years ago and now I am learning what I did not learn back then. You have a calm and easy to follow approach to these tutorials. Thank you so much!
You are most welcome! Never too late to learn something new! Thanks for watching and commenting!
Thank you so much! You explain things so well. My teacher spent less than a day going over this along with other lessons as classes are two hours long and its packed with other lessons. Thank you so much, it was very helpful!
It must be very tough for teachers as well as students. I truly sympathize! But, I’m here to fill in the gaps 😊
Thank you for this video! I'm currently in Advanced Functions and a close friend of mine sent this video in hopes it might help. I'm just writing you to say that it did just that :)
That’s great to hear! Hope you have subscribed and shared the love with your friends ❤️
This is like watching a sorcerer do magic! Thank you so much, I've been struggling with this topic and watching your examples helped me understand what I was getting wrong.
How sweet! So happy to be able to help you out 😊
Thank you so much. My question wasnt directly answered in this video but hearing your thoufht process while doing it helped me understand how it works and i got it! So helpful!
Good for you! Happy to have got those wheels spinning 😊
Thank you!! This is exactly the help I needed very thankful for your channel!!!
You are so welcome!
I am so thankful for this, I have been homeschooling and the course material they have provided is not understandable at some points so I come to youtube for help. I want you to know that it means a lot to me that you upload videos. God bless you!
So happy to be of help. Please like, share and subscribe to increase the channel’s visibility.
you are amazing! i recommended all my friends watch your videos
You're the best! Thanks for your support! I'm sure this isn't necessary to ask you, but please like, share and subscribe to increase the channel's visibility. : )
Hi Ms. Havrot. I think the ratio for pi/6 is wrong because you drew the special triangle wrong (7:14). Thank you so much for the help by the way. You are an amazing teacher :)
I don’t see a mistake. Pi/3 is the 60° angle (bottom) and pi/6 is 30° and is the angle at the top that we cut in two when we split the equilateral triangle in half. Easy to mix up when you see the 6 in the denominator often students think that’s 60° but remember that hits 180°/6 = 30°
Thanks for the lovely compliment 😊
hi Mrs Harvot! I was wondering how I could do the very first calculation of the first example (when I switch from radians to degrees), what is the easiest way to do so without a calculator? my class won't be allowed calculators on our quiz and test unfortunately so I need some help with that 🙃 you're an amazing teacher btw, I learn so much from your videos!
Two ways … first of all you know that pi = 180° so you can multiply 13 x 180/12 or slightly easier is to recognize that 13pi/12 is (12/12+1/12)pi and that (12/12)pi is simply pi or 180° so that you just have 180° + pi/12 which is 180° + 15°
16:31 this kinda look like a corelated angle, the cos would become a sin and the angle would be in the 4 quadrant which would make sin positive. Can I also get the answer through that or is this reasoning just conveniently worked out for this question?
thank you! can you explain why you multiplied by 2 root 2 in the second example? Didnt the denominators have the same value
root 2**
@@sanaab2562 You need to rationalize the denominator.
@@mshavrotscanadianuniversit6234 thank you !
You’re amazing thank you so much for these videos
Glad you like them! ❤️
Hello, Mrs.Havrot! I just had a question on page 401, question 10, is there actually a possible way to solve it without a calculator? Thank you for your time and effort!
Of course you can! You are given some clues. Sina = 7/25. So use opposite has a length of 7, hypotenuse has a length of 25 so find the adjacent using Pythagorean theorem. Do the same for cosb =5/13. (Find the opposite length) then using the formula for sin(a+b) just plug in the values and away you go!
@@mshavrotscanadianuniversit6234 thank you so much! that clears everything up!
Hi Ms,
For the exact values, I am confused as to how you came up with them. Can you please explain how you got them, especially the 240° for π/4. Thank you in advance.
I'd like to reiterate my question; I understand that they are radian values converted to degrees, but why did you not use π/2 or π? As I noticed a pattern that you did not use the values π/2, π, 3π/2 or 2π , why did you only use the angles in between them? To add on, for π/4, I understand you got 135= 45°x 3= 3π/4 but for 240, 45°x5= 225 and 45° x6= 270, so 240 is an angle between 5π/4 and 6π/4.
I am not sure what you are asking. Could you refer to a time on the video for me? 240 degrees is just 4pi/3 (Pi/3=60 degrees so 4pi/3 = 240 degrees)
Okay now I understand! I wrote 240 on the page for a multiple of 45 degrees and it should have said 225 degrees. You are correct. You can use Pi/2 and Pi ... I’m trying to think of an example where they could be used.
The problem with Pi/2 is that the tan would be undefined. Also Pi plus any of the exact values would simply give you a multiple of an exact value because 180= 45 x 4, 180= 60 x 3 and 180 = 30 x 6
@@mshavrotscanadianuniversit6234 Thank you for your detailed clarification/response Ms. Havrot! Please continue to make lessons on Chapter 8 they are very helpful!
At 7:01 isn’t cos (pie over 4) = square root of 2 over 2)?
^Nvm they mean the same thing
id there is a question like cos(pi-pi/4). That equals 3pi/4 and no exact values add up to 3pi/4. So do I just leave it as it is and draw a graph of cos and show that at pi, cos is 0? we can obviously show what cos is at pi/4 because it's 45 degrees, but what about cos pi? Thank You!
The cos (pi-pi/4) can be calculated using the subtraction formula, however cos(3pi/4) is simply -(sqrt2)/2 (with the denominator rationalized) or - 1/(sqrt2)
Ok Thanks!
Thanks this helped bare still
Sometimes it takes a second or third time hearing the lesson to really get what's happening. Of course you still need to practice! Please like, share and subscribe to increase the visibility of the channel : )
Hi, thank you so much for these videos, they are of huge help. Can I sak you to please show me how to do question 8.c) on page 401? its "Determine the exact value of each trig ratio" 8.c) is cos ((11pi)/12) Thanks a million!
Sure can. I’ll do it, then send a link to the pbwiki site
mshavrot.pbworks.com/w/file/fetch/138233262/cos%20%2811pi%3A12%29.pdf It's sideways! I can't figure out how to turn it but I'm sure you can still read it. : )
Ms Havrot's Canadian University Math Prerequisites thank you very much, you have been most helpful :)
Hey, Ms.Havrot. My teacher asked us this question, "Determine the exact value of cos(a-b) given that csca=-5/3, (pi≤a≥3*pi/2) and secb=25/7, (0≤bpi/2)" I can't seem to figure it out I was hoping you could help?
First, change the csca to sina and the secb to cosb. Now that you know the ratios and which quadrants you are in (a in Q3 and b in Q1) you can determine, using Pythagorean theorem the ratios for cosa and sinb. Then all you need to do is use the subtraction formula fir cos (a-b) = cosacosb + sinasinb and plug in your values 😊
Absolutely !
is there a video of yours that I can watch that will help me with simplifying these numbers
th-cam.com/video/bn0_s7Ug0UY/w-d-xo.html
Operations with radicals
Grateful for you
Awww thanks! ❤️