Question

Help me verify these identities?

Answer 1

\[1) \frac{ \cos x }{ 1 + \sin x } + \frac{ 1 + \sin x }{ \cos x } = 2 \sec x\]

Answer 2

\[2) \cot (x-\frac{ \pi }{ 2 }) = -\tan x\]

Answer 3

I know they are easy, I kindof just want to make sure I understand...

Answer 4

@Mertsj

Answer 5

@whpalmer4

Answer 6

for the first one, I suggest a substitution: \[a = \sin x, b = \cos x\]Leave the right side alone. Put the left side over a common denominator and simplify, then substitute back and simplify some more.

Answer 7

I find that a bit easier, but your mileage may vary.

Answer 8

Haha. Car reference. :P So I would multiply them both by b/b?

Answer 9

\[\frac{b}{1+a}+\frac{1+a}{b}\]multiply left fraction by \(\frac{b}{b}\) and right fraction by \(\frac{1+a}{1+a}\) and combine

Answer 10

Yup, just saw that. My bad.

Answer 11

So you get.. \[\frac{ \cos^2x +1 + 2\sin x + \sin^2x }{ \cos x+ \cos x \sin x }\]

Answer 12

Yeah, but I wouldn't have distributed the denominator just yet. Now, what can we do with \(\sin^2x+\cos^2x\)?

Answer 13

Okay, so we should leave it:\[\frac{ \cos^2 x +1 +2 \sin x +\sin^2 }{ \cos x(1 + \sin x) }\]
And-->sin^2(x) + cos^2(x) = 1

Answer 14

right, so after we do that, we have...

Answer 15

Now we have\[\frac{ 2 + 2\sin x }{ \cos x(1+\sin x) }\]

Answer 16

Right. Can you see any common factors in the numerator? After you factor the numerator, can you simplify the fraction further?

Answer 17

So you factor the numerator, and get (2(1+sin x))/(cos x(1+sin x)). Then you can cancel the 1+sin x, and you get 2/cos x. Right?

Answer 18

Yes. Now what's the definition of \(\sec x\)?

Answer 19

1/cos x, so 2/cos would be 2sec x. :D

Answer 20

winner winner chicken dinner!

Answer 21

On to the other one?

Answer 22

Fantastic. Just had some fried chicken, too. :) Now about the second one. Can't you just say you know it works because cot((pi/2)-x)=tan x, so if you multiply everything by -1, then you get what we have here. It's just the opposite, right?

Answer 23

well, you could, I suppose, if you remembered that identity, which I never do :-)

break it down: cot = 1/tan = 1/(sin/cos) = cos/sin

what about cos[x-pi/2]? what does that equal?
what about sin[x-pi/2]? what does that equal?
Substitute those into cos/sin and see what you get.

Answer 24

Okay, wait. cot=1/tan. tan=1/(sin/cos).. I'm confused..

Answer 25

Okay, just looked at my identities. I'm caught up now. Well, halfway. cos[x-pi/2] is tan. That's the way I went first. But what the heck do you do with the (x-pi/2) after you change cot to cos/sin???

Answer 26

\[\cot x = \frac{1}{\tan x}\]\[\tan x = \frac{\sin x}{\cos x}\]\[\cot x = \frac{1}{\frac{\sin x}{\cos x}} = \frac{\cos x}{\sin x}\]Agreed?

Answer 27

Got that part. :p

Answer 28

\[\cos (x-\frac{\pi}{2})\ne \tan x\]or any other argument to tan!

Answer 29

Huh??

Answer 30

what is cos 0? what is cos pi/2? what is cos pi? what is cos 3pi/2?

Answer 31

cos 0 = 1, .9996, .9984, .9966.

Answer 32

Here are graphs of cos x and tan x. Does it look like one of them is just a shifted version of the other? That's what subtracting or adding to the argument of a function does — it shifts the graph right or left...

Answer 33

Do what? One goes vertical, and one doesn't..

Answer 34

\[\cos 0=1\]\[\cos \frac{\pi}2 = 0\]\[\cos \pi = -1\]\[\cos \frac{3\pi}{2} = 0\]

Answer 35

the curve in blue is cos x. the curve in purple is tan x. you said that " cos[x-pi/2] is tan."

I'm trying to demonstrate that it is not :-)

Answer 36

Oh, oops. I meant cos(pi/2-x).

Answer 37

That's not any better! Here's a graph of \(\cos x\) (in blue) vs \(\cos (x-\frac{\pi}2)\) (purple). Suggesting anything to you?

Answer 38

Here's a hint. Notice how similar this graph of \(\cos x\) and \(\sin x\) is :-)

Answer 39

Well, it's an identity. cos((pi/2)-x)=tan x
Now I'm lost..

Answer 40

Did you keep the receipt for the book where you looked up that identity? You may want to get a refund!

Answer 41

It's online.. http://www.sosmath.com/trig/Trig5/trig5/trig5.html
Under Co-function Identities... :/

Answer 42

My calculator graphs them as the same.. :(

Answer 43

What is \(\tan \dfrac{\pi}4=\)

What is \(\cos (\dfrac{\pi}{2}-\dfrac{\pi}{4})= \cos\dfrac{\pi}4=\)

Answer 44

Do what? :/ Are you sure I'm not confusing you? The pi/2 comes first..

Answer 45

Impossible. Absolutely impossible. \(f(x + a)\) produces an exact copy of \(f(x)\) shifted along the x-axis. The shape of the graph of cos x and tan x do not look even remotely similar.

Answer 46

You misread the trig identity page. There are *3* identities on the same line!

Answer 47

Look, let x = pi/4. Tell me what tan pi/4 is. Then tell me what cos (pi/2-x) is.

Answer 48

I just found what hapened. Somewhere along the line we strayed from the problem I was given and got confused. My problem doesn't even have cosine... it's cotangent.. -__-

Answer 49

No, we didn't stray. cot = cos/sin

Answer 50

I know, but I was saying, or trying to, that cot((pi/2)-x)=tan.

Answer 51

no, you said cos!

Answer 52

"cos[x-pi/2] is tan."

Answer 53

I know. I messed up because I got confused with the cos stuff.. I sowwy. :'(

Answer 54

anyhow, the fact that you're arguing with me about what the graph of cos looks like vs tan means that you definitely need to do it my way, breaking it down to the elementary functions :-)

Answer 55

so, go back to where I said "that's not any better", and look at those posts and graphs again, please.

Answer 56

I didn't know that we were talking about two totally different things, or else I wouldn't have been arguing. I know cos and tan aren't related. :p

Answer 57

then you need to read carefully!

Answer 58

I do. Too much going on at one time. My baby is in my ribs.

Answer 59

when in doubt, if arguing with someone who is quite insistent that you are wrong, recheck your work before saying further :-)

Answer 60

saying anything further, that is...

Answer 61

a rule I am quite religious about following!

Answer 62

Yes, sir. Let's continue. So now we have cos/sin((pi/2)-x), right?

Answer 63

If you weren't a male, I would say just wait until you get pregnant. I have trouble remembering what day it is, let alone mathematics.

Answer 64

Here's a little table:
\[\begin{array}{ccc}
\theta & \cos \theta & \sin \theta\\\hline\\ 0 & 1 & 0 \\
\frac{\pi }{6} & \frac{\sqrt{3}}{2} & \frac{1}{2} \\
\frac{\pi }{3} & \frac{1}{2} & \frac{\sqrt{3}}{2} \\
\frac{\pi }{2} & 0 & 1 \\
\frac{2 \pi }{3} & -\frac{1}{2} & \frac{\sqrt{3}}{2} \\
\frac{5 \pi }{6} & -\frac{\sqrt{3}}{2} & \frac{1}{2} \\
\pi & -1 & 0 \\
\frac{7 \pi }{6} & -\frac{\sqrt{3}}{2} & -\frac{1}{2} \\
\frac{4 \pi }{3} & -\frac{1}{2} & -\frac{\sqrt{3}}{2} \\
\frac{3 \pi }{2} & 0 & -1 \\
\frac{5 \pi }{3} & \frac{1}{2} & -\frac{\sqrt{3}}{2} \\
\frac{11 \pi }{6} & \frac{\sqrt{3}}{2} & -\frac{1}{2} \\
2 \pi & 1 & 0 \\
\end{array}\]

Answer 65

Notice how the values of \(\sin \theta\) follow 3 steps behind the values of \(\cos \theta\)?

Answer 66

That's because \[\sin\theta = \cos(\theta - \frac{\pi}2)\]

Answer 67

I have a 10-year-old kid, and have observed this problem at close range, thanks :-)

Answer 68

Well, I'm also a teenager.. So there's that. I was already spastic. Did I spell that right? Anyways! Are you saying to change the sine from our cos/sin to cos(x-(pi/2))??

Answer 69

Similarly, if we put a - sign in front of the values, \[\sin(\theta -\frac{\pi}{2}) = -\cos\theta\]

Answer 70

Oh, gosh.. I'm lost. What does our equation look like?

Answer 71

So that gives us \[\cot(x-\frac{\pi}{2}) = \frac{\cos(x-\frac{\pi}{2})}{\sin(x-\frac{\pi}{2})} = \frac{\sin x } {-\cos x } = -\tan x\]

Answer 72

A thing of beauty, ain't it? :-)

Answer 73

The second step is where you lost me.. :/ Do both of them have it just because cot did?

Answer 74

It's gorgeous. I love math. I just have to catch on first. Normally I'm not THIS slow.. Geez.

Answer 75

\[\cot u = \frac{\cos u}{\sin u}\]so if \(u = x-\frac{\pi}2\) that's what you get...

Answer 76

truth be told, I never remember the shifting identities like I used here, I have to figure them out each time by making a little table for myself :-)

Answer 77

don't use them quite often enough to keep them fresh and reliable.

Answer 78

But that cos on the top. When you do the x first: (x-(pi/2)), it makes the answer negative.. Right?

Answer 79

cos (x - pi/2) = sin x

Here's a table:
\[\begin{array}{ccc}
x & \cos(x-\frac{\pi}{2}) & \sin x \\
\hline \\
0 & 0 & 0 \\
\frac{\pi }{6} & \frac{1}{2} & \frac{1}{2} \\
\frac{\pi }{3} & \frac{\sqrt{3}}{2} & \frac{\sqrt{3}}{2} \\
\frac{\pi }{2} & 1 & 1 \\
\frac{2 \pi }{3} & \frac{\sqrt{3}}{2} & \frac{\sqrt{3}}{2} \\
\frac{5 \pi }{6} & \frac{1}{2} & \frac{1}{2} \\
\pi & 0 & 0 \\
\frac{7 \pi }{6} & -\frac{1}{2} & -\frac{1}{2} \\
\frac{4 \pi }{3} & -\frac{\sqrt{3}}{2} & -\frac{\sqrt{3}}{2} \\
\frac{3 \pi }{2} & -1 & -1 \\
\frac{5 \pi }{3} & -\frac{\sqrt{3}}{2} & -\frac{\sqrt{3}}{2} \\
\frac{11 \pi }{6} & -\frac{1}{2} & -\frac{1}{2} \\
2 \pi & 0 & 0 \\
\end{array}\]

Answer 80

looks pretty identical, right? :-)

Answer 81

Similarly:

\[\begin{array}{ccc}
x & \sin (x-\frac{\pi}2) & \cos x\\
\hline\\
0 & -1 & 1 \\
\frac{\pi }{6} & -\frac{\sqrt{3}}{2} & \frac{\sqrt{3}}{2} \\
\frac{\pi }{3} & -\frac{1}{2} & \frac{1}{2} \\
\frac{\pi }{2} & 0 & 0 \\
\frac{2 \pi }{3} & \frac{1}{2} & -\frac{1}{2} \\
\frac{5 \pi }{6} & \frac{\sqrt{3}}{2} & -\frac{\sqrt{3}}{2} \\
\pi & 1 & -1 \\
\frac{7 \pi }{6} & \frac{\sqrt{3}}{2} & -\frac{\sqrt{3}}{2} \\
\frac{4 \pi }{3} & \frac{1}{2} & -\frac{1}{2} \\
\frac{3 \pi }{2} & 0 & 0 \\
\frac{5 \pi }{3} & -\frac{1}{2} & \frac{1}{2} \\
\frac{11 \pi }{6} & -\frac{\sqrt{3}}{2} & \frac{\sqrt{3}}{2} \\
2 \pi & -1 & 1 \\
\end{array}\]

The signs are opposite, so that's why we picked up a - sign in the denominator

Answer 82

Oop, okay. Just tried it. For some reason, I didn't think that would turn out the same... Weird.. Okay, so I have to explain it to get full credit. Let me see if I can explain it properly. So we start with cot(x-(pi/2))=-tan x. We know that cot=cos/sin, so then we get (cos(x-(pi/2)))/(sin(x-(pi/2))). Then from our identities, we know that cos(x-(pi/2))=sin x, and sin(x-(pi/2))=-cosx, so we are left with sin x/-cos x. sin/cos is the same as tan, and it was negative, so it is equal to -tan x. Right?

Answer 83

Can you explain it to me a little dumber why the sine was negative and the cosine wasn't. Like, in a way I would remember that if I didn't have that graph..

Answer 84

Did I do that right?

Answer 85

exactly!

Answer 86

well, cosine leads sine around the unit circle. at x = 0, cos x = 1, sin x = 0. At x = pi/2, cos x = 0, sin x = 1. At x = pi, cos x = -1, sin x = 0, at x = 3pi/2, cos x = 0, sin x = -1. and at x = 2pi, we've come full circle.

Answer 87

Subtracting pi/2 from x before feeding it into the cos or sin machine shifts the graph by x = pi/2, and when we do that, we either get the other function, or - the other function.

Answer 88

I'm proof you don't need to remember the exact relationship, you just need to remember your unit circle values and then you can figure it out again...

Answer 89

Okay, I got it now I think. Thank you, oh so very much. You have been fantastic. :P God bless you and your family. :)

Answer 90

You're welcome!