Question

Inverse Trigo (Diff Calculus)

Answer 1

For each of the trigo formulas for cos2x, deduce by differentiation the trigo formula of sin2x.

Answer 2

HELP. i dont understand the problem..

Answer 3

hmm..
you know what is the trigo formula of cos(2x) ?

Answer 4

yes

Answer 5

its cos^2x - sin^2 x

Answer 6

good
so now.. differentiate on both sides

Answer 7

okay, wait

Answer 8

s it.. 2 cosx (-sinx) - 2sinx (cosx)??

Answer 9

so its -2cosxsinx - 2sinxcosx

Answer 10

yea.. don't forget the left hand side

Answer 11

your equation was \[\cos(2x) = \cos^{2}x - \sin^2x\]

Answer 12

ok, so.. whats next? =)

Answer 13

i told you :P.. differentiate on both sides.. then simplify!

Answer 14

hmm. -2cosx (sinx+sinx)???

Answer 15

ehh.. wrong :-/

Answer 16

you dunno \[\frac{d}{dx}\cos(2x) = ?\]

shame on you.. SHAME ON YOU :P..

Answer 17

ok, im sorry

Answer 18

alright try again!

Answer 19

-2 Sin[2 x]

Answer 20

?

Answer 21

hey.. its -2sin2x

Answer 22

yes yes.. that is correct.. now simply fy.. u need to find what

sin(2x) is right?

Answer 23

yes..

Answer 24

what do you mean *now simply fy*?

Answer 25

sin2x is sin(2x) = 2 sin x cos x

Answer 26

'sin2x is sin(2x) = 2 sin x cos x"
u have to derive this darling..

you differentiated an Equation remember

your equation was cos(2x) = cos^2 x - sin^ x
so when u differentiate that equation what do you get madam? (write both sides of that equation)

Answer 27

sin2x is 2cos2x

Answer 28

i said.. the entire equation.. left hand side AND right side

Answer 29

-2sin2x = -2cosxsinx - 2sinxcosx

Answer 30

\[\cos(2x) = \cos^2x - \sin^2x\]
differentiating on both sides
\[-2\sin(2x) =2 sinxcosx - 2sinxcox\]

Answer 31

yes exactly .. finally phew :D
now simplify that expression.. and you get the formula for sin(2x)

Answer 32

yay its 2sinxcosx

Answer 33

yup yup congrats :P

Answer 34

thank you =) hahaha finally shame on me

Answer 35

last question?

Answer 36

\[arc \sin \frac{ x }{ a } + \frac{ \sqrt{a^2-x^2} }{ x }\]

Answer 37

first derivative

Answer 38

arc sin x/a is 1/ sq rt a^2 - x^2

Answer 39

yes correct

Answer 40

and the answer should be - sq rt a^2 - x^2 / x^2

Answer 41

no idea deriving the answer

Answer 42

because u differentiated only the arc sin part, what about the other part?