this really helps and makes things a little clearer how would i go about i = sin (5t^3 + 6t -3 ) would this be i = cos(5t^3 + 6t -3) then (15t^2 + 6 ) ?? not sure i did that correctly
Not to get started but in further questions such as simplifying answers or finding stationary points that involve trig. functions you may have to know some of the basic identities and solving equations.
ExamSolutions There are quite a number of videos on your site regarding trigonometry,i actually don't know which video would be specific to help me to learn some of the basic identities and solving equations. Can you send me a link please?
+eyeman I was thinking the exact same thing, he might've made a silly mistake or then again we could be wrong too, although since 3/7cosx is the same as (3/1)/(1/7cosx) I'd say that we're probably right.
+eyeman just think of it as 3÷ (7*secx) and secx is 1/cosx so 3÷(7*1/cosx) which is the same as 3÷(7/cosx) and if we change the division sign to multiplication sign we should take the reciprocal of 7/cosx which becomes 3*(cosx/7) which becomes 3cosx/7.
this really helps and makes things a little clearer how would i go about i = sin (5t^3 + 6t -3 ) would this be i = cos(5t^3 + 6t -3) then (15t^2 + 6 ) ?? not sure i did that correctly
For these series of videos,do i need to understand Basic Trigonometry to be able to integrate/differentiate Sin,Cos and Tan?
Not to get started but in further questions such as simplifying answers or finding stationary points that involve trig. functions you may have to know some of the basic identities and solving equations.
ExamSolutions Do you have any videos that you can recommend to me to help me to learn some of the basic identities and solving equations?
Bayezid Yes, have you been on my site and looked under trigonometry?
ExamSolutions There are quite a number of videos on your site regarding trigonometry,i actually don't know which video would be specific to help me to learn some of the basic identities and solving equations.
Can you send me a link please?
Bayezid heck out the trigonometry section on here www.examsolutions.net/maths-revision/syllabuses/Edexcel/period-1/C3/module.php
A great lesson, explaining so clearly and helpfully.
What is sec^2?
+Azad Kaya sec(x) is 1 / cos(x) so sec^2(x) is just this all squared.
What do you do if you have y=5sin(3x^2)
Use the chain rule. where t= 3x^2 so y=sin(t).
Then dy/dt = cos(t) = cos(6x) and dt/dx = 6x .
Since dy/dx = (dy/dt ) (dt/dx) = 6x sin(3x^2)
30x cos(3x^2)
confused to why it becomes 3/7cosx instead of 21cosx
you multiplied cosx with the 3 but the 7 stayed as a denominator, help!
+eyeman I was thinking the exact same thing, he might've made a silly mistake or then again we could be wrong too, although since 3/7cosx is the same as (3/1)/(1/7cosx) I'd say that we're probably right.
+Ahoora Saadat I am correct, there is no mistake. It is -3(1/7)[1/sec(x)] which is -3(1/7)cos(x) which when differentiated is (3/7)sin(x)
+eyeman just think of it as 3÷ (7*secx) and secx is 1/cosx so 3÷(7*1/cosx) which is the same as 3÷(7/cosx) and if we change the division sign to multiplication sign we should take the reciprocal of 7/cosx which becomes 3*(cosx/7) which becomes 3cosx/7.
Sorry, are you sure that in second equation the answer will be 5sec^2x ? I managed to get 5cosec^2x
does this follow the quotient rule?