difficult trig identity help plz!

**finalfantasy** · November 2nd 2007, 07:37 AM

I am having a lot of trouble proving these trig identities! If anyone can help me out, that would be fantastic!

2 csc 2x = sec x csc x
2 cot 2x = cot x - tan x

and ...

cos^6x + sin^6x = 1 - 3 sin^2x + 3 sin^4x ( question editted!!! )

THANKS

**ticbol** · November 2nd 2007, 10:11 AM

2 csc 2x = sec x csc x
LHS = 2 csc 2x
= 2 / sin 2x
= 2 / 2 sin x cos x
= 1 / sin x cos x
= csc x sec x
= RHS

Thefore, proven.

2 cot 2x = cot x - tan x
LHS = 2 cot 2x
= 2 / tan 2x
= 2 / [2 tan x / 1 - tan^2 x]
= 2[1 -tan^2 x] / 2 tan x
= 1 - tan^2 x / tan x
= cot x -tan x
= RHS

Therefore, proven.

and ...

cos6x + sin6x = 1 - 3 sin2x + 3 sin4x

Let us check if that is true when x = 30 degrees,
cos(6*30deg) +sin(6*30deg) =? 1 -3sin(2*30deg) +3sin(4*30deg)
-1 +0 =? 1 -3(0.866) +3(0.866)
-1 =? 1
No.
So, cos6x + sin6x = 1 - 3 sin2x + 3 sin4x is not an identity.

**Soroban** · November 2nd 2007, 01:10 PM

Hello, finalfantasy!

Another approach . . .

$1)\;\;2\csc 2x \:= \:\sec x \csc x$

We have: . $\sec x\csc x \;=\;\frac{1}{\cos x}\cdot\frac{1}{\sin x} \;=\;\frac{2}{2\sin x\cos x} \;=\;\frac{2}{\sin2x} \;=\;2\csc2x$

$2)\;\;2\cot 2x \:= \:\cot x - \tan x$

We have: . $\cot x - \tan x \;=\;\frac{1}{\tan x} - \tan x \;=\;\frac{1-\tan^2\!x}{\tan x} \;=\;\frac{2(1-\tan^2\!x)}{2\tan x}$

. . $= \;\frac{2}{\frac{2\tan x}{1-\tan^2\!x}} \;=\;\frac{2}{\tan2x} \;=\;2\cot2x<br /> <br />$

**finalfantasy** · November 3rd 2007, 08:31 AM

Hey guys! I editted the last question, can you solve it now? THANKS SO MUCH!

@ Soroban: how does 1 / cos x * 1 / sin x give you 2 / 2sinxcosx

1 times 1 is 1 right? So where does the 2 come in?

@ticbol:

how do you get this step highlighted in bold?

LHS = 2 csc 2x
= 2 / sin 2x
= 2 / 2 sin x cos x
= 1 / sin x cos x
= csc x sec x
= RHS

what happens to the first 2 in 2 csc 2x?

**topsquark** · November 3rd 2007, 08:40 AM

Originally Posted by finalfantasy

cos^6x + sin^6x = 1 - 3 sin^2x + 3 sin^4x

Recall that $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$

Here I am setting $a = cos^2(x)$ and $b = sin^2(x)$

Thus
$(cos^2(x) + sin^2(x))(cos^4(x) - cos^2(x)sin^2(x) + sin^4(x)) =? ~ 1 - 3sin^2(x) + 3sin^4(x)$

$cos^4(x) - cos^2(x)sin^2(x) + sin^4(x) =? ~ 1 - 3sin^2(x) + 3sin^4(x)$

$(1 - sin^2(x))^2 - (1 - sin^2(x))sin^2(x) + sin^4(x) =? ~ 1 - 3sin^2(x) + 3sin^4(x)$

Now expand out the LHS and simplify.

-Dan

**topsquark** · November 3rd 2007, 08:41 AM

Originally Posted by finalfantasy

Hey guys! I editted the last question, can you solve it now? THANKS SO MUCH!

@ Soroban: how does 1 / cos x * 1 / sin x give you 2 / 2sinxcosx

1 times 1 is 1 right? So where does the 2 come in?

@ticbol:

how do you get this step highlighted in bold?

LHS = 2 csc 2x
= 2 / sin 2x
= 2 / 2 sin x cos x
= 1 / sin x cos x
= csc x sec x
= RHS

what happens to the first 2 in 2 csc 2x?

He noted that $2sin(x)cos(x) = sin(2x)$ . He only had a $sin(x)cos(x)$ , so he multiplied the numerator and denominator by 2 (essentially multiplying the fraction by 1.)

You can get the same result by noting that $sin(x)cos(x) = \frac{1}{2}sin(2x)$ , putting this into the denominator and then simplifying the fraction. Since many people really don't like complex fractions, I think Soroban chose the best course.

-Dan

**Soroban** · November 3rd 2007, 09:51 AM

Hello, finalfantasy!

$\cos^6\!x + \sin^6\!x \;= \;1 - 3\sin^2\!x + 3\sin^4\!x$

On the left side, use: . $a^3 + b^3 \:=\:(a+b)(a^2-ab +b^2)$

$\text{We have: }\;\cos^6\!x +\sin^6\!x \:=\:\underbrace{(\cos^2\!x + \sin^2\!x)}_{\text{This is 1}}(\cos^4\!x - \cos^2\!x\sin^2\!x + \sin^4\!x)$

. . . $= \;\cos^4\!x - \cos^2\!x\sin^2\!x\,{\color{blue} + \,3\cos^2\!x\sin^2\!x} + \sin^4\!x\,{\color{blue} -\, 3\cos^2\!x\sin^2\!x}$

. . . $= \;\underbrace{\cos^4\!x + 2\cos^2\!x\sin^2\!x + \sin^4\!x} - 3\cos^2\!x\sin^2\!x$

. . . $= \qquad\underbrace{(\cos^2\!x + \sin^2\!x)^2}_{\text{This is 1}} - 3\cos^2\!\sin^2\!x$

We have: . $1 - 3\cos^2\!x\sin^2\!x \;=\;1 - 3(1-\sin^2\!x)\sin^2\!x\;=\; 1 - 3\sin^2\!x + 3\sin^4\!x$

**ticbol** · November 3rd 2007, 11:21 AM

Originally Posted by finalfantasy

Hey guys! I editted the last question, can you solve it now? THANKS SO MUCH!

@ Soroban: how does 1 / cos x * 1 / sin x give you 2 / 2sinxcosx

1 times 1 is 1 right? So where does the 2 come in?

@ticbol:

how do you get this step highlighted in bold?

LHS = 2 csc 2x
= 2 / sin 2x
= 2 / 2 sin x cos x
= 1 / sin x cos x
= csc x sec x
= RHS

what happens to the first 2 in 2 csc 2x?

Huh? How would I answer that?

2csc(2x)
= 2[1/sin(2x)
= 2 / sin(2x)

sin(2x) = 2sin(x)cos(x)

So,
= 2 / 2sin(x)cos(x)

Thread: difficult trig identity help plz!

LinkBack

Thread Tools

Display

difficult trig identity help plz!

Similar Math Help Forum Discussions

Difficult-ish Trig Identity

Difficult Trig Identity Question

Difficult Trig Proof

Very Difficult Trig Identity Cot(2X) - Cot(3X) = 1/2SecXCsc(3X)

difficult identity

Search Tags