Trigo identities--Pls Help!!

**xeno_vert** · November 28th 2008, 11:33 PM

I beg all of you to pls help me to solve this trigonometric identities equation..the problems are 10, i already solve 5 and obviously, there's a remaining 5 problems..I can't solve them i mean i've been solving those remaining 5 questions for 5 hours and still no success..pls help me...these are the questions...

* those enclosed in parentheses indicated that the number inside is an exponent)

* / = over e.g. cosx/sinx (cosx over sinx)

1.) cos(4)x - sin(4)x = 2cos(2)x - 1

2.) 1 - (sinx / cscx) = cosx / secx

3.) tan x / (1 - tan(2)x) = cosx / (cscx - 2sinx)

4.) secx sinx / (tanx +cotx) = sin(2)x

5.) cos(2)x - sin(2)x = 1 - 2sin(2)x

plss help me...thank you in advance for those who'll helped...

Godbless you all!!!

**great_math** · November 28th 2008, 11:51 PM

Originally Posted by xeno_vert

1.) cos(4)x - sin(4)x = 2cos(2)x - 1

2.) 1 - sinx / cscx = cosx / secx

3.) tan x / 1 - tan(2)x = cosx / cscx - 2sinx

4.) secx sinx / tanx = cotx = sin(2)x

5.) cos(2)x - sin(2)x = 1 - 2sin(2)x

1) $\cos^4x-\sin^4x=2\cos^2x-1$

LHS $=\cos^4x-\sin^4x$

$=(\cos^2x+\sin^2 x)(\cos^2x-\sin^2 x)$

$=\cos^2x-\sin^2x$

$=\cos^2x-(1-\cos^2x)$

$=\cos^2x-1+\cos^2x$

$=2\cos^2x-1=RHS$

2) $1-\frac {\sin x}{\csc x}=\frac{\cos x}{\sec x}$

$LHS=1-\frac {\sin x}{\csc x}$

$=1-\frac{\sin x}{\frac 1{\sin x}}$

$=1-\sin^2 x=\cos^2 x$

$RHS=\frac{\cos x}{\sec x}$

$=\frac{\cos x}{\frac 1{\cos x}}$

$=\cos^2 x=LHS$

**great_math** · November 29th 2008, 12:08 AM

Originally Posted by xeno_vert

3.) tan x / 1 - tan(2)x = cosx / cscx - 2sinx

4.) secx sinx / tanx = cotx = sin(2)x

5.) cos(2)x - sin(2)x = 1 - 2sin(2)x

3) $\frac{\tan x}{1-\tan^2x}=\frac{\cos x}{\csc x-2\sin x}$

$LHS=\frac{\tan x}{1-\tan^2x}$

$=\frac{\frac{\sin x}{\cos x}}{1-\frac{\sin^2x}{\cos^2x}}$

$=\frac{\sin x}{\cos x}\times \frac{\cos^2x}{\cos^2-\sin^2x}$

$=\frac{\sin x\cdot \cos x}{\cos^2x-\sin^2x}$

$RHS=\frac{\cos x}{\csc x-2\sin x}$

$=\frac{\cos x}{\frac{1}{\sin x}-2\sin x}$

$=\frac{\cos x\cdot \sin x}{1-2\sin ^2 x}$

$=\frac{\cos x\cdot \sin x}{\cos^2 - \sin^2 x}=LHS$

4th question seems to be incorrect

5) $\cos^2x - \sin^2x = 1 - 2\sin^2x$

$ RHS=1-2\sin^2x$

$=1-\sin^2x-\sin^2x$

$=\cos^2x-\sin^2x$

PROVED

**xeno_vert** · November 29th 2008, 12:39 AM

3)

I'm just confused how did u get
cos(2)x / cos(2)x - sin(2) x ..
sorry...

**great_math** · November 29th 2008, 12:40 AM

OK so now the edited question
4) $\frac{\sec x\cdot\sin x}{\tan x+\cot x}=\sin^2x $

$LHS=\frac{\sec x\cdot\sin x}{\tan x+\cot x}$

$=\frac{\frac{\sin x}{\cos x}}{\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}}$

$=\frac{\frac{\sin x}{\cos x}}{\frac{\sin^2x+\cos^2x}{\sin x\cdot\cos x}}$

$=\frac{\sin x}{\cos x}\cdot \sin x\cdot\cos x$

$=\sin^2x=RHS$

**great_math** · November 29th 2008, 12:45 AM

Originally Posted by xeno_vert

3)

I'm just confused how did u get
cos(2)x / cos(2)x - sin(2) x ..
sorry...

$1-\frac {\sin^2 x}{\cos^2x}$

$=\frac{\cos^2x-\sin^2x}{\cos^2x}$

So the fraction becomes

$\frac{\frac{\sin x}{\cos x}}{\frac{\cos^2x-\sin^2x}{\cos^2x}}$

Got it?

**xeno_vert** · November 29th 2008, 12:47 AM

sorry..I'm confused again..

Regarding Question 3:

where did u get the cos(2)x??? (the denominator of cosx . sinx / cos(2) - sin(2)x = LHS

**great_math** · November 29th 2008, 12:48 AM

Originally Posted by xeno_vert

sorry..I'm confused again..

where did u get the cos(2)x??? (the denominator of cosx . sinx / cos(2) - sin(2)x = LHS

Check the 5th question keeping in mind the denominator...

**xeno_vert** · November 29th 2008, 01:07 AM

i'm terribly sorry for i ask a lot of questions..i'm really sorry..i'm really analyzing it but i need to be guided sometimes..i'm really sorry.. im confused again with this one...

Regarding Question 5:

how did u express secx sinx to sinx / cosx?
sinx /cosx in terms of quotient relation is equivalent to tanx

another things is that secx sinx in terms of product relation is equivalent to tanx (the original form is sinx secx, i just interchange them)

tnx for the reminder...huhuhuh i'm really sorry...

**great_math** · November 29th 2008, 01:22 AM

Originally Posted by xeno_vert

i'm terribly sorry for i ask a lot of questions..i'm really sorry..i'm really analyzing it but i need to be guided sometimes..i'm really sorry.. im confused again with this one...

how did u express secx sinx to sinx / cosx?
sinx /cosx in terms of quotient relation is equivalent to tanx

another things is that secx sinx in terms of product relation is equivalent to tanx (the original form is sinx secx, i just interchange them)

tnx for the reminder...huhuhuh i'm really sorry...

Sorry I messed up Latex.. You can check it now..

$ \sec x=\frac 1{\cos x}$

$\sec x\cdot\sin x=\frac{\sin x}{\cos x}=\tan x $

**xeno_vert** · November 29th 2008, 01:29 AM

Originally Posted by great_math

Sorry I messed up Latex.. You can check it now..

$ \sec x=\frac 1{\cos x}$

$\sec x\cdot\sin x=\frac{\sin x}{\cos x}=\tan x $

will it changed the equation?

**xeno_vert** · November 29th 2008, 01:35 AM

So the fraction becomes

im sorry...the question supposed to be is that where does the cos(2)x / cos(2) - sin(2)x was derived?

**great_math** · November 29th 2008, 01:42 AM

Originally Posted by xeno_vert

will it changed the equation?

No..We will land again where were are...

Originally Posted by xeno_vert

So the fraction becomes

im sorry...the question supposed to be is that where does the cos(2)x / cos(2) - sin(2)x was derived?

well what is $\frac{\frac{1}{15}}{\frac{1}{5}}$ ?

that is same as $\frac{1}{15}\div\frac{1}{5}$

$=\frac{1}{15}\times 5$

$=\frac 1{3}$

Do you now get from where cos(2)x / cos(2) - sin(2)x was derived?

**xeno_vert** · November 29th 2008, 02:32 AM

Regarding the last part of the solution to Question 5:

im sorry again..i'll just asked what happened to cosx?

i min if that's the form then i guess it wud be sin(2)x / cos(2)x

**mr fantastic** · November 29th 2008, 02:56 AM

Originally Posted by xeno_vert

im sorry again..i'll just asked what happened to cosx?

i min if that's the form then i guess it wud be sin(2)x / cos(2)x

Read it more carefully. The cos x's cancel.

This whole thread is a headache because of the number of questions originally posted and the number of follow-up questions that have been asked.

Next time: Do not ask more than one question in your original post.

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