Question

Help with a trig identity

Answer 1

\[\sqrt{16-4\sin^2x}\]

Answer 2

how does that equal 4cosx ?

Answer 3

\[\sqrt{16-(4\sin(x))^2} ? \]

Answer 4

yes my bad

Answer 5

\[\sqrt{16-16\sin^2(x)} \\ =\sqrt{16(1-\sin^2(x))}\]

Answer 6

Omg okay thank you i just realized i didn't square root the 4 then factor 16!!!

Answer 7

square the 4
yeah
I figure that is what you meant
because otherwise yeah we wouldn't have been able to get 4cos(x)

Answer 8

Yeah exactly lol thankssssssssss

Answer 9

It is actually equal to 4cos(x) if cos(x)>=0
or equal to -4cos(x) if cos(x)<0

Answer 10

lol right, that's going too into it lol

Answer 11

\[\int\limits(7cosx)/(7cosx)^3\]
Okay so i know it would be 1/49 integral sec^2 but would it be the same result if you left the 1/49 inside the integral ?

Answer 12

@myininaya

Answer 13

@satellite73 ?

Answer 14

Lol somebody ?

Answer 15

you bring constant multiples in or out

Answer 16

would 1/49cos^2 equal 49sec^2 ?

Answer 17

\[\frac{1}{49\cos^2(x)}=\frac{1}{49}\frac{1}{\cos^2(x)}=\frac{1}{49}\sec^2(x)\]

Answer 18

okay thank you!

Answer 19

you're like such a big help today thx lol ;D

Answer 20

lol i will try to be other days

Answer 21

are you good with trig substitutions lol