Question

Help with trig equations

Answer 1

Write the following expression in terms sinθ or cosθ, and then simplify if possible. cscθ − cotθcos(-θ)

Answer 2

\[\csc(\theta)=\frac{ 1 }{ \sin(\theta) }\]

Answer 3

\[\cot(\theta)=\frac{ \cos(\theta) }{ \sin(\theta) }\]

Answer 4

\[\cos(-x)=\cos(x)\]

Answer 5

These are the identities i know s:

Answer 6

Oh boy. Scared for calc. Good luck :/

Answer 7

you have the correct identities - where are you stuck?

Answer 8

The like simplifying part s:

Answer 9

\[\sin(\theta)+\frac{ \sin(\theta) }{ \cos(\theta) }+ \cos(\theta)\]

Answer 10

not quite - it should be multiplied by cos(theta) not adding on cos(theta)

Answer 11

and you have used cot(theta) incorrectly

Answer 12

and also csc(theta) incorrectly

Answer 13

well I flipped it so we wouldnt have 1/sin

Answer 14

even then i did it wrong cause we would have 1/cos(Theta)

Answer 15

the first step should be to directly substitute in the identities

Answer 16

when you say that is that using the identities? and the equation would look like \[\frac{ 1 }{ \sin(\theta) }-\frac{ \cos(\theta) }{ \sin(\theta) }*\cos(\theta)\] does this look good now?

Answer 17

perfect :)

Answer 18

now you can multiply the two cos(theta)'s on the right-hand term

Answer 19

then you can subtract the numerators because you have a common denominator of sin(theta)

Answer 20

\[\frac{ 1 }{ \sin(\theta) }-\frac{ \cos^2(\theta) }{ \sin(\theta}\]

Answer 21

yup - and carry on to get:\[\frac{ 1 }{ \sin(\theta) }-\frac{ \cos(\theta) }{ \sin(\theta) }*\cos(\theta)=\frac{ 1 }{ \sin(\theta) }-\frac{ \cos^2(\theta) }{ \sin(\theta) }=\frac{1-\cos^2(\theta)}{\sin(\theta)}\]

Answer 22

it cannot simplify any further if I am correct

Answer 23

then you can make use of the identity:\[\cos^2(\theta)+\sin^2(\theta)=1\]

Answer 24

to give you:\[1-\cos^2(\theta)=\sin^2(\theta)\]

Answer 25

use this to simplify further

Answer 26

so add cos^2 to both sides

Answer 27

no - use the final identity I gave to simplify

Answer 28

I am confused how you got the final identity. Where did you get sin^2 from

Answer 29

did you \[\frac{ 1-\cos^2(\theta) }{ \sin(\theta) } * \frac{ \sin(\theta) }{ \sin (\theta) }\]

Answer 30

remember you were left with:\[\frac{1-\cos^2(\theta)}{\sin(\theta)}\tag{1}\]
In here we have \(1-\cos^2(\theta)\) in the numerator and we know that:\[\sin^2(\theta)+\cos^2(\theta)=1\tag{2}\]

Answer 31

oh so you just rearrange it?

Answer 32

yes :)

Answer 33

the identity in (2) can be rearranged to give:\[\sin^2(\theta)=1-\cos^2(\theta)\tag{3}\]

Answer 34

i.e. we can replace \(1-\cos^2(\theta)\) by \(\sin^2(\theta)\) in (1)

Answer 35

this then gives us:\[\frac{1-\cos^2(\theta)}{\sin(\theta)}=\frac{\sin^2(\theta)}{\sin(\theta)}\]and I am sure you know how to simplify this

Answer 36

wouldn't you want to get sin^2 to the other side. I am unsure I am not good with simplifying S:

Answer 37

you can divide the numerator and denominator by \(\sin(\theta)\)

Answer 38

it is equivalent to simplifying the fraction:\[\frac{5^2}{5}\]

Answer 39

\[\frac{5^2}{5}=\frac{5\times\cancel{5}}{\cancel{5}}=5\]

Answer 40

so it would just be sin(theta) then

Answer 41

correct :)

Answer 42

\[\frac{ 1-\cos^2(\theta) }{ \sin(\theta) }=\sin(\theta)\] would be the final then?

Answer 43

yes

Answer 44

Thank you for your assistance :3

Answer 45

yw :)