Question

Please Help!!:D
Rewrite the expression as a trigonometric function of a single angle measure.

Answer 1

sec = 1/cos

Answer 2

Is he correct @YanaSidlinskiy

Answer 3

Please stop. \[\frac{ \tan4\Theta +\tan2\Theta }{ 1-\tan 4\Theta \tan2\Theta } \]

Answer 4

Hate these things like crazy.

Answer 5

u telling me to stop @YanaSidlinskiy

Answer 6

No, Maxwell:)

Answer 7

@paki

Answer 8

oh... stay away form trig.... :(

Answer 9

I really need help. This is one of my questions on my assignment.

Answer 10

@SolomonZelman please help here...

Answer 11

\(\LARGE\color{blue}{ \frac{\tan 4θ+\tan 2θ}{1-\tan 4θ\tan 2θ}\large{=\tan(4θ+2θ)=\tan(6θ)} }\)

Answer 12

Like this, you want it, or everything in terms `tanθ` ?

Answer 13

warning, in terms of `tanθ` will give a very awkward answer... so don't recommend .

Answer 14

That's what I was thinking. No. I don't need it in terms of tan. I've tried it and I know what it's like. I agree. It's definitely crazy:)

Answer 15

Thanx sooooooooooooooo much:)

Answer 16

Anytime :) btw, my grade for these was 98 :)

Answer 17

OMG!!!!!!!!!!!!!!!!!!!!!!!!!!!!! I am soooooooo mad. Wanna teach me?

Answer 18

What do you want to know?

Answer 19

How to solve these stupid things:)

Answer 20

Like examples.

Answer 21

Well, I just remember the rules... I can only say that it is common sense after you know the identities.

One tip for verifying trigonometric equations though, is to translate everything into sines and cosines (usually).

Answer 22

Ok, Well, then tell me the rules:) Hopefully I'll be smart like ya:)

Answer 23

\(\huge\color{blue}{ \tan (a+b)=\frac{\tan(a)+\tan(b)}{1-\tan(a)\tan(b)} }\)


\(\huge\color{blue}{ \tan (a-b)=\frac{\tan(a)-\tan(b)}{1+\tan(a)\tan(b)} }\)

Answer 24

\(\LARGE\color{blue}{ \sin (a±b)= \sin(a)\cos(b)± \sin(b)\cos(a) }\)

\(\LARGE\color{blue}{ \cos (a±b)= \cos(a)\cos(b)∓ \sin(b)\sin(a) }\)

Answer 25

\(\LARGE\color{blue}{ \sin^2x+\cos^2x=1 }\)

Answer 26

The rest can be derived from these.

Answer 27

Well besides the half angle ones. You will always find the online.

Answer 28

I have a quick question. On the first rule like tan(a+b) and tan(a-b) will they always be in that form?

Answer 29

Yes even if you have imaginary/complex angles.

Answer 30

Wow! Ok. I think I can write a whole book about you! explaining better than a book:) For real! Filled with informationxD!!!

Answer 31

Can we do an example of each?

Answer 32

Anyone explains better than a book, books don't explain they are there to confuse you. Well, beside English and History books.

Answer 33

besides

Answer 34

I agree. Definitely!:) *they are there to confuse*

Answer 35

And..Can we do an example of each if you don't mind?

Answer 36

Okay, but you will need to do it. And the questions won't be telling which rule to use.
Ready ?

Answer 37

Umm.. Ok. I guess so. You have to check it though too. Ok?

Answer 38

Verify the rule below.
\(\normalsize\color{blue}{ \cos(2θ)=\cos^2θ - \sin^2θ }\)

Answer 39

Ok. Is it this?
\(\normalsize\color{blue}{ \cos (a±b)= \cos(a)\cos(b)∓ \sin(b)\sin(a) }\)

Answer 40

yes, so the rule is \(\normalsize\color{blue}{ \cos(a+b)=\cos(a) \cos(b) - \sin(b) \sin(a) }\)

Answer 41

use it:)

Answer 42

Ok. Give me a sec. See what I can get:)

Answer 43

sure

Answer 44

I dunno:( I don't know where to plug them in.

Answer 45

Wait..Wait.. I *think*

Answer 46

Hint: `2θ = θ + θ `

Answer 47

I'm just curious..Is the answer going to be 0?

Answer 48

You just need to show that
`cos(2θ)= cos²θ - sin²θ`

Answer 49

Umm. Ok. well..
cos\[\cos(0\pm0) = \cos(0)^{2}\cos(0)^{2}\sin(2)^{2}\sin(0)\]
I really don't know:) lol. At least I gave it a try:) Show me:)

Answer 50

\(\normalsize\color{blue}{ \cos(2θ)=\cos(θ+θ)=\cos(θ)\cos(θ)- \sin(θ)\sin(θ)= }\)
\(\normalsize\color{blue}{ \cos^2θ- \sin^2θ. }\)

Answer 51

Makes sense ?

Answer 52

Yes. Better than mine!!!!:D Wow!! Some more? Are you tired?

Answer 53

I can give you one more question.
Show that \(\normalsize\color{blue}{ \cos~90 = 0. }\)

Answer 54

Post each step as you go,

Answer 55

Will I use this formula?
\(\normalsize{ \sin^2x+\cos^2x=1 }\)

Answer 56

No

Answer 57

90 = 45 + 45 (hint)

Answer 58

Will it be the same formula as we used for the other question?

Answer 59

Yes....
and you will agree that sin(45)=soc(45) right ?
Now, do the proof

Answer 60

sin(45)=cos(45) I meant

Answer 61

Ok. let me try.

Answer 62

sure, but on your way, do not forget to prove that sin(45)=cos(45), and find their values, and to plug them in.

Answer 63

That's a bummer. For real.

Answer 64

Do i need to show that cos 90=0 first?

Answer 65

showing that cos(90) will not be possible before you show that cos(45)=sin(45)

Answer 66

showing that cos(90)=0 will not be... (I meant)

Answer 67

Ok. Here's what I got:
\[\cos(90)=\cos(45+45)=\cos(45)\cos(45)-\sin(45)\sin(45)=\cos ^{45} -\sin ^{45}\]
Am I wrong? Ughh. You're going too hard one meh.xD

Answer 68

I think I messed up, for real. And probably BIG time too.

Answer 69

Well, it is equal to `cos²(45) - sin²(45)`

Answer 70

Now, show that sin(45) = cos(45)

Draw a 45-45-90 and the proportions of it

Answer 71

So it was only there \(\huge{\uparrow}\) that I messed up?

Answer 72

yes, you wrote 45's as powers, but 45s are angles and they are to the second power each.

Answer 73

Whew! I seriously am startin to understand it now:) And I'm not sure how I would do that.

Answer 74

Draw a 45-45-90 triangle, and write the length of each side in terms of S.

Answer 75

Does it matter where I put the numbers?

Answer 76

And what do you mean by "terms of S?" Like Sin?

Answer 77

Put the numbers next to the sides, because they are to lavel the side lengths.

No, "S" is just a variable, can be "x", "a" "b" or any other variable.

Answer 78

They are to LABEL the side lengths (sorry for the typo)

Answer 79

??