Question

Verify the trigonometric equation by substituting identities to match the right hand side of the equation to the left hand side of the equation:

cot x sec^4 x = cot x + 2 tan x + tan^3 x

Please walk me through this step by step. I don't think I've ever been more lost in math.

Answer 1

Are you sure there are two equal signs in this expression?

Answer 2

Oh sorry, this is the correct one: cot x sec^4 x = cot x + 2 tan x + tan^3 x

Answer 3

that makes a huge difference.

Answer 4

I see you have no fewer than 3 trig functions in this equation.
Can you think of a way to cut that to just 2 trig functions?

Answer 5

Not really. The only thing I know is that you're supposed to start with the right side since it's more complex but after that I don't know what to do.

Answer 6

Need to dig up a table of trig identiies here. Do you have one in front of y ou?

Answer 7

I have the common ones written in my notes.

Answer 8

which one involves (csc x)^2?

Answer 9

1 + cot^2 = csc^2

Answer 10

good. that's correct, accurate.

Answer 11

I see that all terms in your equation involve either tan or cot, except for one: (sec x)^4. Agree or disagree?

Answer 12

Agree

Answer 13

So we might be able to simplify the identity by getting rid of that (sec x)^4. Any ideas on how to do that? How is the sec x function defined?

Answer 14

I'm not really sure. Would we need to do something so that it's like (sec^2)(sec^2) since most of the identities involve sec^2 not sec^4?

Answer 15

That's true. The secant function is defined as 1/ cos x. We don't want to introduce yet another trig function into this inequality, so should not use this.

You are correct in that you can re-write (sec x)^4 as [(sec x)^2]^2. Agreed or not?

Answer 16

Agreed

Answer 17

Try using the identity for (sec x)^2 that we discussed earlier.

Answer 18

I got (1 + tan^2) (1 + tan^2) since 1 + tan^2 = sec^2 and we have sec^4 but I'm not sure how to use the one I wrote earlier for 1 + cot^2 = csc^2

Answer 19

Remember, we don't want to introduce or re-introduce more trig functions here. so (csc x)^2 is out. What is the relationship between (tan x)^2 and (sec x)^2?

Answer 20

1 + tan^2 = sec^2

Answer 21

If you can u se Equation Editor, let's use it. But your latest input is clear. enuf (altho you've left out the argument x.).

Note that we have (sec x)^4 in that equation. How could we use your latest result to eliminate (sec x)^4 from the equation?

Answer 22

this is Equation Editor output:\[\tan ^{2}x+1=\sec^2x\]

Answer 23

Starting with this latest equation, find an equation for \[\sec^4x\]

Answer 24

\[(\tan^2 x + 1)(\tan^2 x + 1) = \sec^4 x\]

Answer 25

I don't think that's right but I'm not sure what to do.

Answer 26

What you have is fine. this identity allows us to eliminate (sec x)^4 from the equation. Why not multiply out\[(\tan^2x+1)^2\]

Answer 27

and replace (sec x)^4 with the result?

Answer 28

So that would make the first side of the equation: \[\cot x (\tan^2 x +1)^2\]

Answer 29

cot x sec^4 x = cot x + 2 tan x + tan^3 x or\[\cot x \sec^4 x = \cot x + 2 \tan x + \tan^3 x\]

Yes, you have the left side correct. Notice that you also have cot x on the right side.

Answer 30

Would it be possible to get rid of the cot x factor?

Answer 31

As you showed me, you now have\[\cot x (\tan^2x + 2 \tan x + 1) \] on the left side, and this is getting to look a lot like the right side.
Note that the cot and tan are reciprocals of one another; can you put that fact to good use?

Answer 32

Is this the right side you have also?
cot x + 2 tan x + tan^3 x\[\cot x + 2 \tan x + \tan^3 x\]

Answer 33

Yes that is the right side. I'm just confused on the left how we went from \[\cot x (\tan^2 x +1)^2 \to \cot x + 2\tan x +1\]

Answer 34

We've gone thru so much detail already, there's bound to be an oversight somewhere. But note that you can and should expand the square of (tan x)^2 + 1).

Answer 35

I think you've learned a lot here and made some progress towards proving the identity. At this point I ask you which would benefit y ou more, to complete this proof or to move on to another problem.

Answer 36

I can try to move on to another problem I suppose and maybe that one will be easier and I can come back to this one.

Answer 37

I've copied this one down and am gtoing to work on it a bit more. If you'll promise to come back and review what we've done here, then I'd suggest you move on to another proof. Each of us has only so much time, and so it's essential to put every min. to good use.

Answer 38

Okay. I found another one on my paper that looks slightly easier. I can go try to work on that for now.

Answer 39

Hey, I think I've proven the one we were working on. I'll share that with you later.
Post the new problem now...Ask a (new) question, not adding on to this already long discussion.

Answer 40

Okay, the new problem I have is:\[1 + \sec^2 x \sin^2 x = \sec^2 x\]

Answer 41

Definitely have finished the proof.

Answer 42

That's great. Thank you for taking the time to help me.

Answer 43

And you're supposed to prove this new identity?

Answer 44

Yes

Answer 45

1 + \sec^2 x \sin^2 x = \sec^2 x\[1 + \sec^2 x \sin^2 x = \sec^2 x\]

Answer 46

the tan and sec functions pair up in an identity, as do the cot and csc.

Answer 47

Also, the reciprocal of the cos is the sec function.
Could you use either or both of these facts to begin simplifying the given equation?

Answer 48

Note: the sine and cosine are most familiar to most of us, so y ou might benefit from eliminating the sec function in favor of its reciprocal, the cos.

Answer 49

Simply take the original equation and substitute \[\frac{ 1 }{ \cos^2 x }\]

Answer 50

for (sec x)^2.

Answer 51

just on the left side.

Answer 52

\[1 + 1/\cos^2x + \sin^2x\]

Answer 53

Al, don't we haver multiplication in the left-hand term? You've introduced a 2nd addition.

Answer 54

Oh my bad \[1+1/\cos^2x \sin^2x\]

Answer 55

Basically right; need parentheses around the reciprocal of (cos x)^2. But anyway.

What you have now is [1 / (cos x)^2\*[sin x]^2. OK? Can you simplify that?

Answer 56

\[1+\frac{ \sin^2x }{ \cos^2x }=?\]

Answer 57

Leave the 1 alone. The 2nd term is easily simplified.

Answer 58

I'm struggling to find the identity to use. Would it be tan x = sin x/ cos x

Answer 59

yes, preciselyl.

Answer 60

So that means 1 + tan^2 = sec^2?

Answer 61

Yes. is that true or not?

Answer 62

True

Answer 63

So that's the end of that proof?

Answer 64

Then you've proven the identity.

Answer 65

yes.

Answer 66

Now, regarding the previous problem:

We have \[\cot x(\tan^4x+2\tan^2x + 1) = \cot x + 2\tan x + \tan^3x.\]

Answer 67

convert that cot x to 1 / tan x and then multiply everything within parentheses by 1 / tan x.

Answer 68

\[2\tan x/\tan x + \tan^3 x /\tan x = \]

Answer 69

Compare that to the right side. See how close you're getting?

Answer 70

Where you should have 2 (tan x)^2, you have written 2 tan x. fix that please.

Answer 71

\[2 (\tan x)^2/\tan x + \tan^3/ \tan x\]

Answer 72

Simplify the left side, then again compare the left and right sides of y our equation.

Answer 73

\[2 \tan x + \tan^2 x\]

Answer 74

Aside from some very minor mistakes you've succeeded in proving this identiy. I encourage you to review our discussion back to what we typed in 8 minutes ago, and make the changes necessary to validate your proof.

typing 2 tan x instead of 2 (tan x)^2 was one example.

I so much appreciate your involving y ourself in this work as much as you have.

Answer 75

As much as I'd like to continue, I need to shift my focus elsewhere. Hope to have the privilege of working wi th you again soon. :)

Answer 76

I understand. Thank you for your help and spending so much time with me on this. I understood the other proof we completed and hopefully I will be able to figure this one out eventually.

Answer 77

Cool. It's worth the effort! See you, Alex! Bye.