Complicated Trigo Identity

**liukawa** · December 11th 2010, 08:48 AM

Hi Guys,

Am Stuck with this few trigo identtes for days !

1) Prove Identity 1-[ (sinx)^6 + (cosx)^6 ] = 3/4(sin2x)^2

2) Prove Identity (sin2x)^2 [ (cotx)^2 - (tanx)^2 ] = 4cos2x

3) If cosec2x + cot2x = cotx , prove that cosec2x + cosec4x + cosec8x = cotx -cot8x

4) Prove that cosx + cos(x+120) + cos (x+240) = 0

5) Prove that sinx[sin(x+120)] + sinx[sin(x-120)] + [sin(x+120)sin(x-120)] = -3/4

6) Prove that [sin(x-120)]^2 + (sinx)^2 + [sin(x+120)]^2 = 3/2

Please help guys thanks alot !

**TheCoffeeMachine** · December 11th 2010, 09:27 AM

Originally Posted by liukawa

1) Prove Identity 1-[ (sinx)^6 + (cosx)^6 ] = 3/4(sin2x)^2

Suppose $a = \sin{x}$ and $b = \cos{x}$ , then we have:

$a^6+b^6 = (a^2+b^2) (a^4+b^4-a^2 b^2) = (a^2+b^2) [(a^2+b^2)^2-2a^2b^2-a^2 b^2]$

............. $= (a^2+b^2) [(a^2+b^2)^2-3a^2 b^2] = 1-3a^2 b^2 = 1-3\left(ab\right)^2$ .

So $1-\left(\sin^6{x}+\cos^6{x}\right) = 1-\left[1-3\left(\sin{x}\cos{x}\right)^2\right]= 3\left(\frac{1}{2}\sin{2x}\right)^2 = \frac{3}{4}\sin^2{2x}.$

**Soroban** · December 11th 2010, 09:43 AM

Hello, liukawa!

Here's the first one . . .

$\text{1) Prove: }\;1-\left(\sin^6\!x + \cos^6\!x\right) \;=\; \frac{3}{4}\sin^2\!2x$

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

. . . . . . . . $=\;1 - \bigg(\sin^4\!x - \sin^2\!x\cos^2\!x + \cos^4\!x\bigg)$

Add and subtract $3\sin^2\!x\cos^2\!x\!:$

. . $1 - \bigg(\sin^4\!x + 2\sin^2\!x\cos^2\!x + \cos^4\!x - 3\sin^2\!x\cos^2\!x\bigg)$

. . $=\; 1 - \bigg(\left[\sin^2\!x + \cos^2\!x\right]^2 - 3\sin^2\!x\cos^2\!x\bigg)$

. . $=\;1 - \left(1 - 3\sin^2\!x\cos^2\!x\right)$

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

. . $=\;3\sin^2\!x\cos^2\!x$

. . $=\;\frac{3}{4}\cdot 4\sin^2\!x\cos^2\!x$

. . $=\;\frac{3}{4}(2\sin x\cos x)^2$

. . $=\;\frac{3}{4}\sin^2\!2x$

Edit: Ah, TheCoffeeMachine beat me to it . . .

**liukawa** · December 11th 2010, 08:51 PM

Thanks guys !

You guys are genius !

Do help out with the rest of the identities also if possible !

thanks alot in advance !

**Archie Meade** · December 12th 2010, 03:33 AM

Originally Posted by liukawa

Hi Guys,

Am Stuck with this few trigo identtes for days !

1) Prove Identity 1-[ (sinx)^6 + (cosx)^6 ] = 3/4(sin2x)^2

2) Prove Identity (sin2x)^2 [ (cotx)^2 - (tanx)^2 ] = 4cos2x

3) If cosec2x + cot2x = cotx , prove that cosec2x + cosec4x + cosec8x = cotx -cot8x

4) Prove that cosx + cos(x+120) + cos (x+240) = 0

5) Prove that sinx[sin(x+120)] + sinx[sin(x-120)] + [sin(x+120)sin(x-120)] = -3/4

6) Prove that [sin(x-120)]^2 + (sinx)^2 + [sin(x+120)]^2 = 3/2

Please help guys thanks alot !

1)

$1-\left[sin^6x+\left(1-sin^2x\right)^3\right]=1-\left[\left(1-2sin^2x+sin^4x\right)\left(1-sin^2x\right)+sin^6x\right]$

$=1-\left[1-sin^2x-2sin^2x+2sin^4x+sin^4x-sin^6x+sin^6x\right]=-3sin^4x+3sin^2x$

$=3sin^2x\left(1-sin^2x\right)=3\left(sinxcosx\right)^2=3\left(\fra c{sin2x}{2}\right)^2$

2)

$cos2x=cos^2x-sin^2x$

$sin2x=2sinxcosx$

$\left(sin2x\right)^2\left[\frac{cos^2x}{sin^2x}-\frac{sin^2x}{cos^2x}\right]=4sin^2xcos^2x\left[\frac{cos^2x}{sin^2x}-\frac{sin^2x}{cos^2x}\right]=4\left[cos^4x-sin^4x\right]$

$=4\left[cos^2x-sin^2x\right]\left[cos^2x+sin^2x\right]$

**liukawa** · December 13th 2010, 05:46 AM

hi guys.

please help with the other identities if possible thanks !

**Unknown008** · December 13th 2010, 06:15 AM

3) Oh well...

$cosec 2x + \cot 2x = \cot x$

Then, we also have:

$cosec 4x + \cot 4x = \cot 2x$

$cosec 8x + \cot 8x = \cot 4x$

Combining those three, we get:

$cosec 2x + cosec 4x + cosec 8x + \cot 2x + \cot 4x + \cot 8x = \cot x + \cot 2x + \cot 4x$

Can you simplify this?

**Soroban** · December 13th 2010, 05:05 PM

Hello, liukawa!

There's always some fool who will do all these problems for you.

Okay, it's me . . .

$\text{4) Prove: }\:\cos x + \cos(x+120) + \cos (x+240) \:=\: 0$

$\cos x + \bigg[\cos x\cos120 - \sin x\sin120\bigg] + \bigg[\cos x\cos240 - \sin x\sin240\bigg]$

. . $=\;\cos x + \bigg[\cos x\left(\text{-}\frac{1}{2}\right) - \sin x\left(\frac{\sqrt{3}}{2}\right)\bigg] + \bigg[\cos x\left(\text{-}\frac{1}{2}\right) - \sin x\left(\text{-}\frac{\sqrt{3}}{2}\right)\bigg]$

. . $=\;\cos x - \frac{1}{2}\cos x - \frac{\sqrt{3}}{2}\sin x - \frac{1}{2}\cos x + \frac{\sqrt{3}}{2}\sin x$

. . $=\quad 0$

$\text{5) Prove that:}$
. . $\sin x[\sin(x\!+\!120)] + \sin x[\sin(x\!-\!120)] + [\sin(x\!+\!120)\sin(x\!-\!120)] \:=\: -\frac{3}{4}$

$\sin x\bigg[\sin x\cos120 + \cos x\sin120\bigg] + \sin x\bigg[\sin x\cos120 - \cos x\sin120 \bigg]$

. . . . . . $+ \bigg[\sin x\cos120 - \cos x\sin120\bigg]\,\bigg[\sin x\cos120 + \cos x\sin120\bigg]$

. . $=\;\sin x\bigg[\sin x\left(\text{-}\frac{1}{2}\right) + \cos x\left(\frac{\sqrt{3}}{2}\right)\bigg] + \sin x\bigg[\sin x\left(\text{-}\frac{1}{2}\right) - \cos x\left(\frac{\sqrt{3}}{2}\right)\bigg]$

. . . . . . $+ \bigg[\sin x\left(\text{-}\frac{1}{2}\right) + \cos x\left(\frac{\sqrt{3}}{2}\right)\bigg]\,\bigg[\sin x\left(\text{-}\frac{1}{2}\right) - \cos x\left(\frac{\sqrt{3}}{2}\right)\bigg]$

. . $=\;\text{-}\frac{1}{2}\sin^2\!x + \frac{\sqrt{3}}{2}\sin x\cos x - \frac{1}{2}\sin^2\!x - \frac{\sqrt{3}}{2}\sin x\cos x + \frac{1}{4}\sin^2x - \frac{3}{4}\cos^2\!x$

. . $=\;\text{-}\frac{3}{4}\sin^2\!x - \frac{3}{4}\cos^2\!x \;\;=\;\;-\frac{3}{4}\underbrace{(\sin^2\!x + \cos^2\!x)}_{\text{This is 1}} \;\;=\;\;-\frac{3}{4}$

$\text{6) Prove: }\:\sin^2(x-120) + \sin^2\!x+ \sin^2(x+120) \:=\: \frac{3}{2}$

$\bigg[\sin x\cos120 - \cos x\sin120\bigg]^2 + \sin^2\!x + \bigg[\sin x\cos120 + \cos x\sin120\bigg]^2$

. . $=\;\sin^2\1x\cos^2120 - 2\sin x\cos x\sin120\cos120 + \cos^2\!x\sin^2\!120 + \sin^2\!x$

. . . . . . $+ \sin^2\!x\cos^2\!120 + 2\sin x\cos x\sin120\cos120 + \cos^2\!x\sin^2\!120$

. . $=\;2\sin^2\!x\cos^2\!120 + 2\cos^2\!x\sin^2\!120 + \sin^2\!x$

. . $=\;2\sin^2\!x\left(\text{-}\frac{1}{2}\right)^2 + 2\cos^2\!x\left(\frac{\sqrt{3}}{2}\right)^2 + \sin^2\1x$

. . $=\;\frac{1}{2}\sin^2\!x + \frac{3}{2}\cos^2\!x + \sin^2\!x \;=\;\frac{3}{2}\sin^2\!x + \frac{3}{2}\cos^2\!x$

. . $=\;\frac{3}{2}\left(\sin^2\!x + \cos^2\1x\right) \;=\;\frac{3}{2}$

**liukawa** · December 13th 2010, 08:58 PM

Originally Posted by Soroban

Hello, liukawa!

There's always some fool who will do all these problems for you.

Okay, it's me . . .

$\cos x + \bigg[\cos x\cos120 - \sin x\sin120\bigg] + \bigg[\cos x\cos240 - \sin x\sin240\bigg]$

. . $=\;\cos x + \bigg[\cos x\left(\text{-}\frac{1}{2}\right) - \sin x\left(\frac{\sqrt{3}}{2}\right)\bigg] + \bigg[\cos x\left(\text{-}\frac{1}{2}\right) - \sin x\left(\text{-}\frac{\sqrt{3}}{2}\right)\bigg]$

. . $=\;\cos x - \frac{1}{2}\cos x - \frac{\sqrt{3}}{2}\sin x - \frac{1}{2}\cos x + \frac{\sqrt{3}}{2}\sin x$

. . $=\quad 0$

$\sin x\bigg[\sin x\cos120 + \cos x\sin120\bigg] + \sin x\bigg[\sin x\cos120 - \cos x\sin120 \bigg]$

. . . . . . $+ \bigg[\sin x\cos120 - \cos x\sin120\bigg]\,\bigg[\sin x\cos120 + \cos x\sin120\bigg]$

. . $=\;\sin x\bigg[\sin x\left(\text{-}\frac{1}{2}\right) + \cos x\left(\frac{\sqrt{3}}{2}\right)\bigg] + \sin x\bigg[\sin x\left(\text{-}\frac{1}{2}\right) - \cos x\left(\frac{\sqrt{3}}{2}\right)\bigg]$

. . . . . . $+ \bigg[\sin x\left(\text{-}\frac{1}{2}\right) + \cos x\left(\frac{\sqrt{3}}{2}\right)\bigg]\,\bigg[\sin x\left(\text{-}\frac{1}{2}\right) - \cos x\left(\frac{\sqrt{3}}{2}\right)\bigg]$

. . $=\;\text{-}\frac{1}{2}\sin^2\!x + \frac{\sqrt{3}}{2}\sin x\cos x - \frac{1}{2}\sin^2\!x - \frac{\sqrt{3}}{2}\sin x\cos x + \frac{1}{4}\sin^2x - \frac{3}{4}\cos^2\!x$

. . $=\;\text{-}\frac{3}{4}\sin^2\!x - \frac{3}{4}\cos^2\!x \;\;=\;\;-\frac{3}{4}\underbrace{(\sin^2\!x + \cos^2\!x)}_{\text{This is 1}} \;\;=\;\;-\frac{3}{4}$

$\bigg[\sin x\cos120 - \cos x\sin120\bigg]^2 + \sin^2\!x + \bigg[\sin x\cos120 + \cos x\sin120\bigg]^2$

. . $=\;\sin^2\1x\cos^2120 - 2\sin x\cos x\sin120\cos120 + \cos^2\!x\sin^2\!120 + \sin^2\!x$

. . . . . . $+ \sin^2\!x\cos^2\!120 + 2\sin x\cos x\sin120\cos120 + \cos^2\!x\sin^2\!120$

. . $=\;2\sin^2\!x\cos^2\!120 + 2\cos^2\!x\sin^2\!120 + \sin^2\!x$

. . $=\;2\sin^2\!x\left(\text{-}\frac{1}{2}\right)^2 + 2\cos^2\!x\left(\frac{\sqrt{3}}{2}\right)^2 + \sin^2\1x$

. . $=\;\frac{1}{2}\sin^2\!x + \frac{3}{2}\cos^2\!x + \sin^2\!x \;=\;\frac{3}{2}\sin^2\!x + \frac{3}{2}\cos^2\!x$

. . $=\;\frac{3}{2}\left(\sin^2\!x + \cos^2\1x\right) \;=\;\frac{3}{2}$

Thanks dude !

nah you are not an idiot.

I am the idiot =P

You are a genius dude !

Thanks alot !

Appreciate your effort and time spent thanks !

**liukawa** · December 13th 2010, 08:59 PM

Originally Posted by Unknown008

3) Oh well...

$cosec 2x + \cot 2x = \cot x$

Then, we also have:

$cosec 4x + \cot 4x = \cot 2x$

$cosec 8x + \cot 8x = \cot 4x$

Combining those three, we get:

$cosec 2x + cosec 4x + cosec 8x + \cot 2x + \cot 4x + \cot 8x = \cot x + \cot 2x + \cot 4x$

Can you simplify this?

Yup .

I can carry on from here.

Thanks for your enlightenment !

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