Question

NEED HELP WITH TRIG ASAP!!! I HAVE TWO HOURS TO COMPLETE http://prntscr.com/4327bl

Answer 1

@zepdrix

Answer 2

Grr I gotta eat my food :d So I might be a little slow with my responses lol.
Which one you stuck on?
All 5?

Answer 3

yes

Answer 4

I have a trig identities sheet, but it takes me forever to figure out one equation. (at the current state, it will be two hours late if I try to work them out alone)

Answer 5

\[\Large\rm \cot x \sec^4x=\cot x(\sec^2x)^2\]

Answer 6

Let's apply our `Pythagorean Identity` for secant.

Answer 7

\[\Large\rm \cot x(\tan^2x+1)^2\]

Answer 8

ya?

Answer 9

I'll bet we can do something with this.

Answer 10

ok

Answer 11

How did you get to (Sec^2x)^2?

Answer 12

\[\Large\rm \sec^4x=(\sec x)^4=\left[(\sec x)^2\right]^2=\left[\sec^2x\right]^2\]

Answer 13

oh, I though you had already solved it... lol

Answer 14

ok continue

Answer 15

The location of the exponent on the trig function makes things a little weird :)

Answer 16

No, we have to match both sides.
So we're maybe.. half way done.

Answer 17

ok

Answer 18

So we'll expand out the outer square,\[\Large\rm \cot x(\tan^2x+1)^2=\cot x(\tan^4x+2\tan^2x+1)\]

Answer 19

And now we're getting very very close.
Does that step make sense?

Answer 20

yes

Answer 21

Here is another identity to keep in mind:\[\Large\rm \cot x \tan x=1\]So do you see what's going to happen when we distribute the cot x?

Answer 22

yup

Answer 23

Thanks for that one. give me a sec to see if I can do the next one and you can eat :)

Answer 24

Yah the second one is nice and easy.
So give it a shot.

Convert to sines and cosines, and use your most basic Pythagorean Identity :)

Answer 25

\[(sinx)(tanx cosx - cotx cosx) = 1-2 \cos^2x\]
Does the "tanx cosx" act as one term or two? (do I distribute to both or just one as a collective term?)

Answer 26

One term :o
I'm not really sure what you're asking though.
Distribute what?

Answer 27

If I distribute "sin x" to the terms inside the parenthesis, do I distribute to "tan ," "cos x," "cotx," etc? Or do I distribute to "tan x cos x"etc?

Answer 28

as in the tan and cos are acting as one term

Answer 29

You distribute to each `term` once.
Example:\[\Large\rm x(ac+b)=acx+bx\]\[\Large\rm x(ac+b)\ne axcx+bx\]

Answer 30

ok

Answer 31

\[(\sin^2x - \cos^2 x) = 1-2\cos^2 x\] That's as far as I got

Answer 32

wait

Answer 33

Ya you're very close.

Answer 34

do I use the power-reducing identities from here

Answer 35

No.
*cough* Pythagorean Identity.

Answer 36

\[\Large\rm \color{orangered}{\sin^2x} - \cos^2 x = 1-2\cos^2 x\]

Answer 37

Gotta deal with that orange term.

Answer 38

but that would make the whole left side -1, wouldn't it?

Answer 39

No.

Answer 40

If I used the "most basic" pythagorean Identity it would make it -1 (atleast in my messed up mind)

Answer 41

Oh you probably setup the Pythagorean Identity backwards then.
It's not,\[\Large\rm \sin^2x\ne \cos^2x-1\]It's this,\[\Large\rm \sin^2x=1-\cos^2x\]

Answer 42

where'd the one come from?

Answer 43

Duuuude.... comeon -_-

Answer 44

\[\Large\rm \sin^2x+\cos^2x=1\]Subtract cos^2x from each side,\[\Large\rm \sin^2x=1-\cos^2x\]

Answer 45

I'm lost

Answer 46

When did we get a one on that side?

Answer 47

and where'd the 2cos^2 x go if we're using that 1?

Answer 48

I'm just showing you the Pythagorean Identity for sine and cosine.
We're going to use it in place of the orange term.

Answer 49

OH FROM THE IDENTITY

Answer 50

ok then, now I'm fllowing

Answer 51

following*

Answer 52

\[\Large\rm \color{orangered}{\sin^2x} - \cos^2 x = 1-2\cos^2 x\]So then just plug in the identity:\[\Large\rm \color{orangered}{1-\cos^2x} - \cos^2 x = 1-2\cos^2 x\]and simplify.

Answer 53

ok.

Answer 54

Third problem should also be a piece of cake.
Convert your secant to cosines to start.

Answer 55

one sec, writing down #2

Answer 56

Using reciprocal identities?

Answer 57

yes

Answer 58

would it be \[(\frac{ 1 }{ \cos x })^2 or \frac{ 1 }{ \cos^2 x0 }\]

Answer 59

don't mind that 0...

Answer 60

Both of those are exactly the same.
The second one is more simplified though. That's what we want to use.

Answer 61

I'm stuck at \[1 + ((\sin^2 x)/(\cos^2 x)) = 1/\cos^2 x\]

Answer 62

No no no. Don't apply the Reciprocal Identity to the right side.
When doing these problems we always LEAVE one of the sides alone.

Answer 63

ok

Answer 64

You need to recall another identity from there.\[\Large\rm \frac{\sin^2x}{\cos^2x}=\left(\frac{\sin x}{\cos x}\right)^2\]\[\Large\rm \frac{\sin x}{\cos x}=?\]

Answer 65

tan x

Answer 66

\[\Large\rm 1+(\tan x)^2=\sec^2x\]Ok good, I guess that gets us to here.

Answer 67

alright. next one!

Answer 68

any tips for the 4th one?

Answer 69

This next one is actually kind of difficult.
You have to remember how to multiply conjugates, and you need to be very careful with the multiplication.

Answer 70

ok

Answer 71

Remember conjugates?\[\Large\rm (a-b)(a+b)=a^2-b^2\]How this applies to trig:\[\Large\rm (1-\cos x)(1+\cos x)=1-\cos^2x\]And from there we can apply our Pythagorean Identity to simplify it to a single term.

Answer 72

So what we're doing is, starting out by looking for a common denominator.

Answer 73

yup

Answer 74

\[\large\rm \frac{\sin x}{1-\cos x}\color{royalblue}{\left(\frac{1+\cos x}{1+\cos x}\right)}+\frac{\sin x}{1+\cos x}\color{orangered}{\left(\frac{1-\cos x}{1-\cos x}\right)}=2 \csc x\]

Answer 75

ok

Answer 76

DO NOT multiply out the numerators.
Leave the brackets in the tops.

Multiply out the denominators.

Answer 77

done, got sin^2x for both

Answer 78

\[\Large\rm \frac{\sin x(1+\cos x)}{\sin^2x}+\frac{\sin x(1-\cos x)}{\sin^2x}=2\csc x\]Ok good.

Answer 79

Hmmm what next..?

Answer 80

can we make it so that the equation becomes \[\frac{ 1 + \cos x }{ \sin x } + \frac{ 1-\cos x }{ \sin x }\] ?

Answer 81

\[\Large\rm \frac{ 1 + \cos x }{ \sin x } + \frac{ 1-\cos x }{ \sin x }=2 \csc x\]Canceling out some sines? Ok good.

Answer 82

They have the same denominator, so hmmm what next? :d

Answer 83

combine

Answer 84

sooooo \[\frac{ 2 }{ \sin x }\]

Answer 85

k good.

Answer 86

\[\Large\rm \frac{2}{\sin x}=2\frac{1}{\sin x}\]

Answer 87

Hmmmm what from here...

Answer 88

2 csc x = 2 csc x

Answer 89

yay team

Answer 90

lol. Last one for now...

Answer 91

Ahhh that last one is so easy.. come on you can do it >.<\[\Large\rm -\tan^2x+\color{orangered}{\sec^2x}=1\]

Answer 92

lol. this one is easy

Answer 93

alright, only 5 more to go: http://prntscr.com/433evv

Answer 94

uploading them 2 at a time

Answer 95

Mmm k I need a math break >.<

Answer 96

I have 30 mins

Answer 97

zep?

Answer 98

http://prntscr.com/433gw2