Recent content by kenshaw93

thank you that really helped. this is what i have managed to do with your help: d/dx(1+3sec^2(y))= 6sec^3y.siny.dy/dx d/dx(dy/dx)^2=6sec^3y.siny.dy/dx i canceled the dy/dx from the RHS and LHS to get: d2y/dx2= 6 sec^2y tany thats a great step forward but the answer needed is 3sec^2y tany...

i also tried finding dy/dx of siny=2sinx and i got 2cos(X)/cos(y), then i squared that to get (dy/dx)^2 and it matched the given one ie. 1+3sec^2(y) which i also found it to be equal to (cos^2y + 3)/cos^2y. so: dy/dx=2cos(X)/cos(y) (dy/dx)^2=1+3sec^2(y)=(cos^2y + 3)/cos^2y.

i meant right not write*-

sorry i didn't write it, i thought it would be useless but i tried differentiating 1+3sec^2(y) and all i got was 3tany(dy/dx)... if that's write then i don't know how to continue

Homework Statement If siny=2sinx and (dy/dx)^2=1+3sec^2(y) show that: by differentiating 1+3sec^2(y) with respect to x, d^2y/dx^2=3sec^2(y)tan(y) Homework Equations The Attempt at a Solution

Homework Statement Without using tables(calculators) find the numerical value of Sin[Pi/8]^2 - Cos[3 (Pi/8)]^4 Homework Equations The Attempt at a Solution I tried changing it to: 1-cos[pi/8]^2 - cos[3pi/8]^4 but have no idea where to go... its really got me scratching my...

haha indeed i am, still young i guess, just scratching the surface of mathematics. The subject really gets you thinking

Ok i found out how to work it out. i first multiply by 2 throughout. then i used the factor formulae until i end up with the terms : 2sin5A sin4A / 2sin4Acos5A which is identicle to sin5A/cos5A which is identicle to tan5A...PROVED! The satisfaction of proving an identity or equation is...

Hi I've got this problem which has really been bothering me. How are you supposed to prove that: (Sin[A] Sin[2 A] + Sin[3 A] Sin[6 A])/(Sin[A] Cos[2 A] + Sin[3 A] Cos[6 A]) is identicle to tan[5A]. I am almost sure that I've got to use the factor formulae, but I've had no luck. Maybe...