Another Identities Crisis

**Freaky-Person** · January 13th 2007, 07:08 PM

omg I hate these things...

1. sin^4A + 2cos^2A - cos^4A = 1

2. cos3A = -3cosA+4cos^3A

3. cos^4A = 1/8(3 + 4cos2A + cos4A)

4. (1 - cosA)/sinA = tan(A/2)

5. sin2A = (2tanA)/1 + tan^2A

Well, these are all of them so far, because I haven't looked at them all yet. Trust me I tried to do these by myself, and succeded in some not on here, but I'm not an identities person.

**ThePerfectHacker** · January 13th 2007, 07:16 PM

Originally Posted by Freaky-Person

omg I hate these things...

And think these are fun.

1. sin^4A + 2cos^2A - cos^4A = 1

Write,
$(\sin^4 x-\cos^4 x)+2\cos^2 x$
$(\sin^2 x+\cos^2 x)(\sin^2x-\cos^2 x)+2\cos^2 x$ *
$(1)(\sin^2 x-\cos^2 x)+2\cos^2 x$
$\sin^2 x-\cos^2 x+2\cos^2 x$
$\sin^2 x+\cos^2 x=1$

*)I used the difference of two squares,
$x^4-y^4=(x^2-y^2)(x^2+y^2)$

**ThePerfectHacker** · January 13th 2007, 07:21 PM

Originally Posted by Freaky-Person

2. cos3A = -3cosA+4cos^3A

First, we know that,
$\cos 2x=\cos (x+x)$
Sum of angles formula,
$\cos x\cos x-\sin x\sin x=\cos^2 x-\sin^2 x$
Use Pythagoren Identity,
$\cos^2 x - (1-\cos^2 x)=2\cos^2 x-1$

Thus,
$\cos 3x = \cos (2x+x)$
Use sum of angles formula,
$\cos 2x\cos x-\sin 2x\sin x$
Use double angle for sine, and formula above,
$(2\cos^2 x-1)\cos x-(2\sin x\cos x)\sin x$
$2\cos^3 x-\cos x-2\sin^2 x\cos x$
Use Pythagorean Identity,
$2\cos^3 x-\cos x-2(1-\cos^2 x)\cos x$
$2\cos^3 x - \cos x-2\cos x+2\cos^3 x$
$4\cos^3 x-3\cos x$

**ThePerfectHacker** · January 13th 2007, 07:27 PM

Originally Posted by Freaky-Person

3. cos^4A = 1/8(3 + 4cos2A + cos4A)

You need to know the half-angle identity for cosine,
$\cos^2 \frac{x}{2}= \frac{1+\cos x}{2}$ .
Sometimes it is written as (same thing),
$\cos^2 x=\frac{1+\cos 2x}{2}$ .

Thus,
$\cos^4 x=(\cos^2 x)^2$
Use the identity above,
$\left( \frac{1+\cos 2x}{2} \right)^2$
Open parantheses,
$\frac{1}{4}(1+2\cos 2x+\cos ^2x)$
Use half-angle identity again,
$\frac{1}{4}\left(1+2\cos 2x +\left(\frac{1+\cos 2x}{2} \right) \right)$
Combine fractions inside,
$\frac{1}{4} \left( \frac{3+4\cos 2x+\cos 4x}{2} \right)$
Thus,
$\frac{3+4\cos 2x+\cos 4x}{8}$

**Freaky-Person** · January 13th 2007, 07:27 PM

omg I both hate you and love you. Love you cause I now know some new techniques for how to do these and learned how to in the first place. Hate you because I'm an idiot. XD XP

**Freaky-Person** · January 13th 2007, 07:34 PM

Combine fractions inside,
$\frac{1}{4} \left( \frac{3+4\cos 2x+\cos 4x}{4} \right)$
Thus,
$\frac{3+4\cos 2x+\cos 4x}{8}$ [/QUOTE]

I'm just wondering why it's over 4 on the inside...

**ThePerfectHacker** · January 13th 2007, 07:42 PM

Originally Posted by Freaky-Person

Combine fractions inside,
$\frac{1}{4} \left( \frac{3+4\cos 2x+\cos 4x}{4} \right)$
Thus,
$\frac{3+4\cos 2x+\cos 4x}{8}$

I'm just wondering why it's over 4 on the inside...[/QUOTE]

It was an error, I will fix it.

**ThePerfectHacker** · January 13th 2007, 07:47 PM

Originally Posted by Freaky-Person

4. (1 - cosA)/sinA = tan(A/2)

I like this one.
Work on right hand side.

$\tan \frac{x}{2} = \frac{\sin \frac{x}{2} }{\cos \frac{x}{2} }$
Multiply the numerator and denominator by $\sin \frac{x}{2}$ .
Thus,
$\frac{\sin^2 \frac{x}{2}}{\sin \frac{x}{2} \cos \frac{x}{2}}$
Multiply the numerator and denominator by 2,
$\frac{2\sin^2 x \frac{x}{2}}{2\sin \frac{x}{2} \cos \frac{x}{2}}$ .
Notice the numerator is half-angle Identity.
The denominator is double angle for sine.
Thus,
$\frac{1-\cos x}{\sin x}$

**ThePerfectHacker** · January 13th 2007, 07:52 PM

Originally Posted by Freaky-Person

5. sin2A = (2tanA)/1 + tan^2A

I think you mean,
$\sin 2x = \frac{2\tan x}{1+\tan^2 x}$
Use the Pythagoren Identity,
$\tan^2 x+1=\sec^2 x$
Thus,
$\frac{2\tan x}{\sec^2 x}$
Express as fractions,
$\frac{2\frac{\sin x}{\cos x}}{\frac{1}{\cos^2 x}}$
Simplify,
$\frac{2 \sin x}{\cos x} \cdot \frac{\cos^2 x}{1}$
Kill the terms,
$2\sin x\cos x$
This is the double angle formula for sine,
$\sin 2x$

**Freaky-Person** · January 13th 2007, 07:55 PM

Originally Posted by ThePerfectHacker

Multiply the numerator and denominator by 2,
$\frac{2\sin^2 x \frac{x}{2}}{2\sin \frac{x}{2} \cos \frac{x}{2}}$ .
Notice the numerator is half-angle Identity.
The denominator is double angle for sine.
Thus,
$\frac{1-\cos x}{\sin x}$

what? what happened to the bottom cos(A/2)? or the top cos^2A? what happened?

**ticbol** · January 13th 2007, 07:57 PM

Originally Posted by Freaky-Person

omg I hate these things...

1. sin^4A + 2cos^2A - cos^4A = 1

2. cos3A = -3cosA+4cos^3A

3. cos^4A = 1/8(3 + 4cos2A + cos4A)

4. (1 - cosA)/sinA = tan(A/2)

5. sin2A = (2tanA)/1 + tan^2A

Well, these are all of them so far, because I haven't looked at them all yet. Trust me I tried to do these by myself, and succeded in some not on here, but I'm not an identities person.

If you can only post about one or two problems per posting, then these are inviting problems. More than two are causing loss of interest.
These 5 problems, you should have separated them into 5 , 4, or 3 postings.

So I try the first two here. If you can re-post the next 3 in 3 or 2 separate postings, then I'd try to solve them also.

--------------------
1) sin^4A + 2cos^2A - cos^4A = 1

LHS =
= sin^4(A) +2cos^2(A) -cos^4(A)
Rearranging,
= [sin^4(A) -cos^4(A)] +2cos^2(A)
= [(sin^2(A) +cos^2(A))*(sin^2(A) -cos^2(A))] +2cos^2(A)
= [(1)*(sin^2(A) -cos^2(A))] +2cos^2(A)
= sin^2(A) -cos^2(A) +2cos^2(A)
= sin^2(A) +cos^2(A)
= 1
= RHS

Therefore, proven.

Note: We used,
----sin^2(A) +cos^2(A) = 1
----a^2 -b^2 = (a+b)(a-b)

-----------------------------------------------
2) cos3A = -3cosA+4cos^3A

LHS =
= cos(3A)
= cos(2A +A)
= cos(2A)cosA -sin(2A)sinA
= [2cos^2(A) -1]cosA -[2sinAcosA]sinA
= [2cos^3(A) -cosA] -2sin^2(A)cosA
= 2cos^3(A) -cosA -2cosA[1 -cos^2(A)]
= 2cos^3(A) -cosA -2cosA +2cos^3(A)
= -3cosA +4cos^3(A)
= RHS

Therefore, proven.

We used:
----cos(A+B) = cosAcosB -sinAsinB
----cos(2A) = 2cos^2(A) -1
----sin(2A) = 2sinAcosA
----sin^2(A) = 1 -cos^2(A)

**ThePerfectHacker** · January 13th 2007, 07:59 PM

Originally Posted by Freaky-Person

what? what happened to the bottom cos(A/2)? or the top cos^2A? what happened?

You should try some time to figure this thy self. I am not going to be there when you are going to be taking an exam.

You have,
$\sin^2 \frac{x}{2}=\frac{1-\cos x}{2}$
This is the half-angle identity.
Multiply by 2 both sides,
$2\sin^2 \frac{x}{2}=1-\cos x$

For the second part you have,
$\sin x = \sin 2\left(\frac{x}{2} \right)$
Double angle formula for sine,
$2\sin \frac{x}{2} \cos \frac{x}{2}$

**Freaky-Person** · January 13th 2007, 08:06 PM

Originally Posted by ThePerfectHacker

You should try some time to figure this thy self. I am not going to be there when you are going to be taking an exam.

You have,
$\sin^2 \frac{x}{2}=\frac{1-\cos x}{2}$
This is the half-angle identity.
Multiply by 2 both sides,
$2\sin^2 \frac{x}{2}=1-\cos x$

For the second part you have,
$\sin x = \sin 2\left(\frac{x}{2} \right)$
Double angle formula for sine,
$2\sin \frac{x}{2} \cos \frac{x}{2}$

I'm sorry but I just don't have that identity, and it's not on the exam i'm just doing this because I couldn't get into 11 AP so I'm going for 12 AP...

**ThePerfectHacker** · January 14th 2007, 07:34 AM

Originally Posted by Freaky-Person

I'm sorry but I just don't have that identity, and it's not on the exam i'm just doing this because I couldn't get into 11 AP so I'm going for 12 AP...

So let me write the identity.

$\cos x= \cos 2\cdot \frac{x}{2}$ .
Using the double angle for cosine and sine.

$\cos x =2\cos^2 \frac{x}{2} -1$
Thus,
$\frac{1+\cos x}{2}=\cos^2 \frac{x}{2}$
And now for sine,
$\cos x = 1-2\sin^2 \frac{x}{2}$
Thus,
$\frac{1-\cos x}{2} = \sin^2 \frac{x}{2}$

I think you made a mistake these identities must be on you exam. They appear in all textbooks I seen.

**Freaky-Person** · January 14th 2007, 12:04 PM

I think you made a mistake these identities must be on you exam. They appear in all textbooks I seen.

They're not. We have some identites but not these ones. All we have is 8 of the easiest ones. Anything multiplied or divided by two, or even associated with the angles is in the AP or Pre-calculus course. This is Canada for you...

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