can someone help with verifying a trig function?

- anonymous

can someone help with verifying a trig function?

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- jamiebookeater

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- anonymous

try us? lols
won't know till you post the question

- anonymous

\[(\sin \theta-\cos \theta)^{2}=1-2\sin^{2} \theta \cot \theta , Verify?\]

- anonymous

i'm going to skip the theta sign so here goes
sin^2 - 2sin*cos + cos^2 = 1 - 2sin^2(cos/sin)
sin^2 - cos^2 = 1
1 - 2 sin*cos = 1 - 2sin(cos
they are equal
make sense?

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## More answers

- anonymous

thank you but just give me a second to make sure i got it

- anonymous

sure thing

- anonymous

how did you get to sin^2 - 2sin*cos + cos^2 = 1 - 2sin^2(cos/sin)

- anonymous

where did the 2sin come from specifically

- anonymous

i foiled the left side of the equation
and tangent is equal to sin/cos
so cot is equal to cos/sin

- anonymous

(a - b)^2
a^2 -2ab + b^2
it's foil

- anonymous

does that make sense?

- anonymous

i think im having trouble with the foiling

- anonymous

umm
do you know how to foil?
(a - b)(a - b)
a^2 - ab - ab + b^2
(sin - cos)(sin - cos)
sin^2 - sin*cos - sin*cos + cos^2
sin^2 - 2sin*cos + cos^2

- anonymous

okay thank you for yor help i see what i was doing wrong, i was foiling wrong, little rusty on the algebra

- anonymous

no problem

- anonymous

hey you have time for another?

- anonymous

sure

- anonymous

Sec^4-tan^4, what trig function is it equal to?

- anonymous

sec is 1/cos

- anonymous

1/cos^4 - sin^4/ cos^4
(1 - sin^4) / cos^4
cos^2 + sin^2 = 1
so cos^2 = 1 - sin^2
[(1 - sin^2)(1 + sin^2)]/ cos^4
(1 + sin^2)/ cos^2

- anonymous

i'm not sure if you can get more simplified than that..

- anonymous

sin^2 = 1 - cos^2
(1 + 1 - cos^2)/ cos^2
(2 - cos^2) / cos^2
2sec^2 - 1
there that's one trig function

- anonymous

the choices are
A) Sec^2
B) csc
C) tan
D) Sec^2+tan^2
E) sin^2
F) sin*tan

- anonymous

the answer is D
where i stopped first
(1 + sin^2)/ cos^2
if you divide that out
you'll get sec^2 + tan^2

- anonymous

ohh man thanks that one was killin me lol just for my own knowledge where do you start first when you see a problem like that? like what goes through your mind first?

- anonymous

i hate doing trig functions
all i can say is something that you've probably heard 100s of times
practice
but seriously, the more you practice, the easier it is
key hints
cos^2 + sin^2 = 1 <-- super big and useful
tan = sin/cos
there are also the rarer ones
half angle formulas

- anonymous

just move the functions around until you see something familiar and you'll see things to cancel out
your object to make it as simple as possible

- anonymous

ok cool keep those in mind i have one question, how did you go from [(1 - sin^2)(1 + sin^2)]/ cos^4 to (1 + sin^2)/ cos^2

- anonymous

i wrote before that line that
cos^2 = 1 - sin^2
which i got from sin^2 + cos^2 = 1 see? very useful
since those are equal, i canceled from top and bottom
does that make sense?

- anonymous

yes i see it now, and i see how you were left with the answer. your helping me out big time lol

- anonymous

lol awesome ^_^

- anonymous

alright ive got another one for you ive been trying it for a while but keep hittin dead ends, got time?

- anonymous

go for it

- anonymous

1/2(1+sin/cos+cos/1+sin) , what trig function is equal?
Sin
cos
tan
csc
sec
cot

- anonymous

cos/1? you sure, caz that's just cos

- anonymous

yup cos/(1+sin)

- anonymous

oh i see you need to use parenthesis rocco, otherwise there's a lot of miss understandings

- anonymous

ohh im sorry

- anonymous

oh and the first part is (1+sin)/cos

- anonymous

1/2(1+tan+cos+cot))

- anonymous

T-T

- anonymous

i feel quite useless....

- anonymous

give me some time, i'm going to write this out, sorry

- anonymous

no problem, and ure not useless dont worry about that lol

- anonymous

rocco
i have no luck for this one :x
sorry!
repost this one

- anonymous

ok no prob, found another one, wanna take a crack at it, i thnk i got half of it?

- anonymous

rofl sure

- anonymous

ok its csc - cot = 1/ (csc + cot) verify

- anonymous

(csc - cot)(csc + cot) = 1
csc^2 - cot^2 = 1
^-- that's another rule you should remember

- anonymous

ohh but can you cross the = sign when verifying?

- anonymous

i'm not sure i understand what you mean?

- anonymous

somewhere i heard that when you verify the = becomes like a wall that you cant cross and only work on either side to make them equal, maybe im just understanding your work wrong

- anonymous

i don't think i've heard that rule
2 + 4 = 8 - 2
if i add 2 on to both sides (crossing?)
i get 6 = 6
do you mean crossing like this?

- anonymous

yea like that

- anonymous

it's perfectly fine
as i just shown up there LOL
6 is equal to 6
so it's fine
cheers ^_^

- anonymous

ohh i see, haha thanks youve been a big help!!!

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