Question

Need help with precalculus will give owl bucks

Answer 1

@jim_thompson5910 i think ill be fine with that one for the test im allwoed to use the formula sheet i hope the formula isnt hard when i use heron

Answer 2

if you move RHS to LHS and then combine second and third term, magic will happen

Answer 3

Take the left side of the expression and factor out sin^2 x.

sin^4 x + sin^2 x * cos^2 x =

sin^2 x ( ? + ? )

Answer 4

cos^2-1=-sin^2

Answer 5

im confused with both

Answer 6

is thsi the one i have to GCF

Answer 7

it might help to think of \(\Large \sin^4(x)\) as \(\Large \sin^2(x)*\sin^2(x)\)

Answer 8

>>if you move RHS to LHS and then combine INCORRECT PROCEDURE

That will work only if you already know the expression is an identity in which case we would not be doing this problem.

@caozeyuan

Answer 9

Oh ok thank you so i would have to factor out sin 4x?

Answer 10

\[\Large \sin^4(x) + \sin^2(x)\cos^2(x) = \sin^2(x)\]

\[\Large \sin^2(x)*\sin^2(x) + \sin^2(x)*\cos^2(x) = \sin^2(x)\]

\[\Large {\color{red}{\sin^2(x)}}*\sin^2(x) + {\color{red}{\sin^2(x)}}*\cos^2(x) = \sin^2(x)\]


GCD in red

Answer 11

whats gcd

Answer 12

GCD or GCF, sorry

Answer 13

\[\sin ^{4}+\sin ^{2}\cos ^{2}-\sin ^{2}=\sin4+\sin ^{2}(\cos ^{2}-1)=\sin ^{4}+\sin ^{2}*-\sin ^{2=s}\]

Answer 14

GCD = greatest common divisor
GCF = greatest common factor

Answer 15

oh ok why not that second sin2x

Answer 16

idk any of this stuff sorry

Answer 17

because we can only factor out one copy of sin^2

\[\Large {\color{red}{\sin^2(x)}}*\sin^2(x) + {\color{red}{\sin^2(x)}}*\cos^2(x) = \sin^2(x)\]

\[\Large {\color{red}{\sin^2(x)}}\left(\sin^2(x) + \cos^2(x)\right) = \sin^2(x)\]

Answer 18

oh ok thank you

Answer 19

Ok, so I should phrase it as this: subtract sin^2 from LHS, prove the expression is 0, so the LHS=sin^2 is proven

Answer 20

then you'll use the pythagorean identity

Answer 21

So what happeens after we fctored it out and found gcf? Also wats LHS

Answer 22

Left hand side

Answer 23

LHS means left hand side
RHS means right hand side

Answer 24

oh ok loll dang i didnt know that

Answer 25

pythagorean identity

\[\Large \sin^2(x) + \cos^2(x) = 1\]

this equation is true for all real numbers x

Answer 26

ok

Answer 27

the biggest trick here is remember sin^2+cos^2=1, other than that just simple algebra

Answer 28

there are other identites?

Answer 29

OK well this is most important

Answer 30

like tan and sec and csc stuff, used to do those a lot in high school now completely forgotten

Answer 31

yes there are many identities

http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf

some teachers allow you to have an identity sheet for the test

Answer 32

lol, yeah i have the formula sheet i will probably get is tommorow

Answer 33

So i have 1 what do i do

Answer 34

replace the "sin^2 + cos^2" portion with "1"

Answer 35

then you use the idea that

1 times any number = same number

Answer 36

ok si sin2*sin2+1=sin2

Answer 37

1*x = x*1 = x

Answer 38

wait did we take out those gcf?

Answer 39

\[\Large \sin^4(x) + \sin^2(x)\cos^2(x) = \sin^2(x)\]

\[\Large \sin^2(x)*\sin^2(x) + \sin^2(x)*\cos^2(x) = \sin^2(x)\]

\[\Large {\color{red}{\sin^2(x)}}*\sin^2(x) + {\color{red}{\sin^2(x)}}*\cos^2(x) = \sin^2(x)\]

\[\Large {\color{red}{\sin^2(x)}}\left(\sin^2(x) + \cos^2(x)\right) = \sin^2(x)\]

\[\Large \sin^2(x)*\left(1\right) = \sin^2(x)\]

\[\Large \sin^2(x) = \sin^2(x) \ \ \ {\color{green}{\checkmark}}\]


So the original equation given in the problem is definitely an identity and has been confirmed to be an identity.

Answer 40

Hello!

Answer 41

Ok awesome! I was just confused on one part , where did the other sin go in part 4

Answer 42

Hi disney

Answer 43

I factored it out

a*b + a*c = a*(b+c)

Answer 44

ooo, ok thank! Everyone who helped me i will send owl bucks

Answer 45

In this case,

a = sin^2
b = sin^2
c = cos^2