Question

Please help

Answer 1

Prove the identity
\[\frac{ \sin x - \cos x }{ \sec x - \csc x} = \frac{ \sin 2 x }{ 2 }\]

Answer 2

more trig?

Answer 3

Yes! Its so confusing. As soon as I think I understand it, I look at another problem and its back to square one

Answer 4

yeah this one is annoying because of the compound fraction and also a trick
but it is not impossible
it is easier to do the algebra with letters instead of sines and cosines
lets replace sine by b and cosine by a

Answer 5

I tried using reciprocal identities in the denominator as I have in other problems but it didnt work out

Answer 6

\[\frac{ \sin( x) - \cos (x) }{ \sec (x) - \csc (x)} = \frac{ \sin (2 x) }{ 2 }\] lets start instead with
\[\frac{b-a}{\frac{1}{a}-\frac{1}{b}}\]and simplify that compound fraction

Answer 7

is it clear how i got that? i replaced \(\sin(x)\) by \(b\) and \(\cos(x)\) by \(a\)

Answer 8

this part has nothing to do with trig, it is just algebra is all, that is why it is easier to work with letters than it is to work with the words "sine" and "cosine"

Answer 9

yeah, its much easier to see using a and b

Answer 10

I work with the denominator first, correct?

Answer 11

ok then there are a couple ways to simplify this
one is to do just what you said

Answer 12

that should get you
\[\large \frac{b-a}{\frac{b-a}{ab}}\]

Answer 13

perfect, thats what i got

Answer 14

now can i multiply by the reciprocal of the denominator?

Answer 15

yes, exactly

Answer 16

okay, i tried these steps before (without substituting for a and b) and it got really confusing. Would it be \[\frac{ (b-a) (ab) }{ b-a }\] and then cancel b-a to equal ab?

Answer 17

yes

Answer 18

is it clear now that you are done basically?

Answer 19

somewhat, how does that equal sin2x/2 though?

Answer 20

putting back the sine and cosine we get \[ab=\cos(x)\sin(x)\] on the left

Answer 21

on the right you use the identity
\[\sin(2x)=2\sin(x)\cos(x)\] so
\[\frac{\sin(2x)}{2}=\frac{2\sin(x)\cos(x)}{2}=\sin(x)\cos(x)\] so the left and right are equal

Answer 22

okay, I think I understand this a lot more now.

Answer 23

good
this one was kind of tricky but mostly just algebra
usually there is mostly algebra
that is why the \(a,b\) trick is a good idea

Answer 24

I never thought to do that, thank you. can i use that trick in simplifying
\[\frac{ 1+\sin2x+\cos2x }{ cosx }\]

Answer 25

i would write out the double angel formula for both terms in the numerator first
that would probably be the best start here

Answer 26

matter of fact, this one is much easier than the last one
you should get
\[\frac{1+2\sin(x)\cos(x)+2\cos^2(x)-1}{\cos(x)}\] then the ones go and you can divide each term by \(\cos(x)\)

Answer 27

For the double angle formula of cos i have written down:
cos2x=cos^2x-sin^2x
2cos^2x-1
1-2sin^2x
why did you use 2 cos^2x -1 in place of the others?

Answer 28

because i guessed that that would be the easiest one to allow you to simplify
wasn't really much of a guess, since that is the only one with only cosine in it

Answer 29

that is almost always the one you want to use anyway
it says if you know \(\cos(x)\) then you can easily find \(\cos(2x)\) without finding the sine

Answer 30

okay, that makes sense. I got 2(sinx + cosx)

Answer 31

me too

Answer 32

You're so much help, I really appreciate this

Answer 33

no problem
you have a lot of trig it seems

Answer 34

yeah, im studying for a chapter test so im trying to cover everything

Answer 35

good luck on the test

Answer 36

thank you,would it be a bother to ask you how to solve this in radians? \[\cos^{-1} \frac{ \sqrt{3} }{ 2 }\]

Answer 37

not at all

Answer 38

this is asking if you know a number (or if you prefer and angle) between \(0\) and \(\pi\) where the cosine of that angle is \(\frac{\sqrt3}{2}\)
it should spring to mind, but if it does not, find where the first coordinate on the unit circle is \(\frac{\sqrt3}{2}\) and that will be your angle

Answer 39

this is a pretty useful cheat sheet
not that you are allowed one in the test
look at the unit circle on the last page
find where you have the point \((\frac{\sqrt3}{2},\frac{1}{2})\) that angle is the one you are looking for

Answer 40

let me know when you see the answer, i will check it

Answer 41

Wow that sheet has everything. Is it pi/6 ?

Answer 42

yes

Answer 43

since \(\cos(\frac{\pi}{6})=\frac{\sqrt3}{2}\) you have
\[\cos^{-1}(\frac{\sqrt3}{2})=\frac{\pi}{6}\]

Answer 44

so if, for example. it was sin^-1 (sqrt3/2) it would be pi/3? because the sine of pi/3 is sqrt3/2?

Answer 45

yes

Answer 46

but be careful when you do this

Answer 47

for example \(\sin(\frac{2\pi}{3})=\frac{\sqrt3}{2}\) as well
you have to know which one to pick

Answer 48

for arcsine and arctangent the angle must be between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\)
for arccosine it must be between \(0\) and \(\pi\)

Answer 49

okay, ill remember that. for something like \[\sin^{-1} (\sin(\frac{ \pi }{ 5 }))\] i know the arcsine and sine cancel leaving pi/5, but how would i make the angle in range?

Answer 50

yes this one would because
\[-\frac{\pi}{2}\leq \frac{\pi}{5}\leq \frac{\pi}{2}\]

Answer 51

it is already in the range in other words
but
\[\sin^{-1}(\sin(\frac{3\pi}{2}))\neq \frac{3\pi}{2}\] because \(\frac{3\pi}{2}>\frac{\pi}{2}\)

Answer 52

okay, so as long as the fraction falls within the range for the designated trig function, i can leave it as the final answer. In this case, pi/5. correct?

Answer 53

yes

Answer 54

but if its something like \[\tan^{-1} (\tan(\frac{ 4\pi }{ 5}))\] i would need to get 4pi/5 between -pi/2 and pi/2. could i subtract by pi? \[\frac{ 4\pi }{ 5 } - \frac{ 5\pi}{ 5 }=\frac{ -\pi }{ 5 }\]

Answer 55

yes, exactly

Answer 56

that is because tangent it periodic with period \(\pi\)
you would have to do something a bit different for
\[\sin^{-1}(\sin(\frac{3\pi}{2}))\]

Answer 57

Okay, this part is fairly easy but i get confused when they use 2 different trig functions
\[\cos^{-1} (\tan(\frac{ -3\pi }{ 4 }))\]

Answer 58

you could find the tangent of \(-\frac{3\pi}{4}\) and then take the arccosine of that number

Answer 59

using my calculator or the unit circle?

Answer 60

you better use the unit circle for these at least at the beginning

Answer 61

do you know how to find \(\tan(-\frac{3\pi}{4})\) using the unit circle?

Answer 62

would it be\[-(\frac{\frac{ -\sqrt2 }{ 2 }}{ \frac{ \sqrt2 }{ 2 } })\]

Answer 63

better known as \(1\)

Answer 64

i mean of course that is a first step, but you have the same number top and bottom, so it is just \(1\)

Answer 65

then find \(\cos^{-1}(1)\) to finish the problem

Answer 66

0

Answer 67

that wasnt so bad

Answer 68

no not too bad

Answer 69

what if its 5pi/4 and needs to fit in the restrictions of arcsine?

Answer 70

i.e. \[\sin^{-1}(\sin(\frac{5\pi}{4}))=-\frac{\pi}{4}\]

Answer 71

on the other hand you could say
\[\sin(\frac{5\pi}{4})=-\frac{\sqrt2}{2}\]and \[\sin^{-1}(-\frac{\sqrt2}{2})=-\frac{\pi}{4}\]

Answer 72

I dont understand how it is -pi/4, wouldnt arcsin(-sqrt2/20= 7pi/4?