Question

Quick question regarding simplification of trig functions

Answer 1

Can I simplify this further
\[\cos^2(5 \ln|x|) + \sin^2(5\ln|x|)\]

Answer 2

@Vincent-Lyon.Fr

Answer 3

think Pythagorean identities

Answer 4

\[\cos^2(x)+\sin^2(x)=?\]
for any x....

Answer 5

what is the ?

Answer 6

I know that cos^2 x + sin^2 x = 1 but in this case its 5 ln|x|

Answer 7

great you got it 1

Answer 8

it doesn't matter what x is...

Answer 9

cos^2(x)+sin^2(x)=1 for any x

Answer 10

are you sure?

Answer 11

cos^2(kitty)+sin^2(kitty)=1

Answer 12

cos^2(frozen yogurt)+sin^2(frozen yogurt)=1

Answer 13

lol how about this then

\[4x \cos (5\ln|x|)\sin(5\ln|x|) = 2x \sin(10\ln|x|)\]

Answer 14

is that right?

Answer 15

\[2 \sin(u)\cos(u)=\sin(2u) \\ \text{ where } u=5\ln|x|\]

Answer 16

yes so looks good

Answer 17

I'm not doing any substitution I am directly plugging it in?. Trig identities work for any variable inside the function?

Answer 18

yep on their domain

Answer 19

and sin and cos have domain all real numbers

Answer 20

what do you mean you aren't doing any sub?

Answer 21

aren't you pluggin into the identities to get what you got?

Answer 22

Alright so this is the question. Prove that y1 = 1 and y2 = e^4x are linearly independent using the Wronskian determination

Answer 23

I did the algebra and I'm left out with
\[\frac{1}{2x} \cos^{2}(\frac{1}{2}\ln|x|)+\frac{1}{2}\sin^{2}(\frac{1}{2}\ln|x|)\]

Answer 24

I'm sorry that is 1/2x before sin^2

Answer 25

\[\frac{1}{2x}[1] \neq 0\]
for some any value of x including zero

Answer 26

\[\left[\begin{matrix}1 & e^{4x} \\ (1)' & (e^{4x})'\end{matrix}\right] \\ =1(e^{4x})'-e^{4x}(1)' \\ =4e^{4x}\]
isn't this the wronskian determinant
or am I thinking something else?

Answer 27

I am sorry I gave you the wrong functions
y1 = cos(1/2 ln x) and y2 = sin(1/2 ln x)

Answer 28

What you did was right, but I want to make sure if my answer is right for the following functions

Answer 29

\[\cos(\frac{1}{2} \ln(x)) \cdot \frac{1}{2x} \cos(\frac{1}{2}\ln(x))-\sin(\frac{1}{2}\ln(x)) \cdot [-\frac{1}{2x}\sin( \frac{1}{2} \ln(x))] \\ \frac{1}{2x} [ \cos^2(\frac{1}{2} \ln(x))+\sin^2(\frac{1}{2} \ln(x))]=\frac{1}{2x}[1]=\frac{1}{2x}\]
you did great

Answer 30

Thanks. That was really helpful :)

Answer 31

Where you from?

Answer 32

from a planet far far a way named america

Answer 33

ok united states is where i'm from

Answer 34

Which state, I'm currently residing in Alabama

Answer 35

stalker.-.