c3 trig

[urwahboy]

ok

 

[Mufaddal]

i have no idea how to even start those questions..i am soo uprepared

 

[Manasa]

m doin gce... Ayesha B urwahboy

 

[Manasa]

i have no idea how to even start those questions..i am soo uprepared

hehe dw ... we wont have such qs tomorow! ... just do past paps and know the rules... gud enuf for A

 

[Ayesha B]

m doin gce... Ayesha B urwahboy

Cool! So... how's your prep now?


And wb C4?

 

[bloom princeton]

hey ppl.. help me out wid this trig

prove:

arctanx+arctan(1/x)=pi/2

arcsinx+arccosx=pi/2

arctanx=pi/2-arccotx

thanks!


http://www.examsolutions.net/maths-revision/core-maths/trigonometry/inverse-functions/example-3.php

 

[Manasa]

help me out with this!
diff y=ln(1+sin2x)

 

[Manasa]

Cool! So... how's your prep now?


And wb C4?

studyd c4 yesterday and few days before.. so just gotta do past papers which ll do day after tomorow
prep is goin fine... m just worryd about the qs where were given an equatn in y and were supposed to find eq of tangent... i get em right but not all of em... :/ have some tips?

 

[Manasa]

?? smth to say lol??

 

[Ayesha B]

help me out with this!
diff y=ln(1+sin2x)

2cos2x/(1 + sin2x)


If they ask you to diff ln|f(x)| then dy/dx = f '(x)/f(x)

 

[Ayesha B]

studyd c4 yesterday and few days before.. so just gotta do past papers which ll do day after tomorow
prep is goin fine... m just worryd about the qs where were given an equatn in y and were supposed to find eq of tangent... i get em right but not all of em... :/ have some tips?

Mhmm those qs are kinda long so my biggest tip would be to take it step by step. Be careful if they're asking for tangent or normal. Then you first have to find dy/dx to find 'm' of the tangent.
Once you get that look for a point (a,b). If it parametric eqns use your t to find both coords.

then set it out m = whatever and (a, b) = whatever.

then y-b = m(x-a) and sub! you're set!

 

[Manasa]

2cos2x/(1 + sin2x)


If they ask you to diff ln|f(x)| then dy/dx = f '(x)/f(x)

Mhmm those qs are kinda long so my biggest tip would be to take it step by step. Be careful if they're asking for tangent or normal. Then you first have to find dy/dx to find 'm' of the tangent.
Once you get that look for a point (a,b). If it parametric eqns use your t to find both coords.

then set it out m = whatever and (a, b) = whatever.

then y-b = m(x-a) and sub! you're set!

tysmm!!!! btw june 13 q8 the last 2 parts... ? can you temme the method?

 

[Manasa]

oh oops sorry i forgot... its actually diff lnsqrt1+sin2x Ayesha B

 

[Ayesha B]

tysmm!!!! btw june 13 q8 the last 2 parts... ? can you temme the method?

I haven't done it sorry! I will be doing it tomorrow morning inshallah so ill let you know

 

[Ayesha B]

oh oops sorry i forgot... its actually diff lnsqrt1+sin2x Ayesha B

that means (1 + sin2x)^1/2

so f '(x) =1/2(1 + sin2x) x 2cos2x = cos2x(1 + sin2x)

so dy/dx wouldbe cos2x(1 + sin2x)/(sqrt(1 + sin2x)) ????

Mhmm.. Idk that's what i would do.!

 

[Ayesha B]

tysmm!!!! btw june 13 q8 the last 2 parts... ? can you temme the method?

You're welcome

 

[Manasa]

that means (1 + sin2x)^1/2

so f '(x) =1/2(1 + sin2x) x 2cos2x = cos2x(1 + sin2x)

so dy/dx wouldbe cos2x(1 + sin2x)/(sqrt(1 + sin2x)) ????

Mhmm.. Idk that's what i would do.!

hehe okay tyy!

 

[urwahboy]

that means (1 + sin2x)^1/2

so f '(x) =1/2(1 + sin2x) x 2cos2x = cos2x(1 + sin2x)

so dy/dx wouldbe cos2x(1 + sin2x)/(sqrt(1 + sin2x)) ????

Mhmm.. Idk that's what i would do.!

lol sorry to interrupt but the answer would be 1/(1+sin2x) u forgot chain rule ... Manasa

 

[Ayesha B]

lol sorry to interrupt but the answer would be 1/(1+sin2x) u forgot chain rule ... Manasa

Lol probably... I'm getting too complicated with simple things :/

 

[Ayesha B]

But waiiiit, if you wanna differentiate lnx isnt the rule

f '(x)/f(x) ??!

Keith Pledger book page 140!!!