Simplifying sin^2 2θ/(1+cos^2 2θ) as a Function of sin θ

[UrbanXrisis]

1. Express $$\frac{sin^2 2 \theta}{1+cos^2 2 \theta}$$ as a function of $$sin \theta$$

here's what I did:
$$= \frac{4 sin^2 \theta cos^2 \theta}{1+(2cos^2 \theta -1)^2}$$
$$= \frac{4 sin^2 \theta cos^2 \theta}{1+(4cos^4 \theta - 4 cos^2 \theta + 1)}$$
$$= \frac{2sin^2 \theta cos^2 \theta}{2cos^2 \theta - 2cos^2 \theta +1}$$

is this correct? can I simplify it more?

2. Write $$sin(a+b)sin(a-b)$$ as a function of double angles

I used the sum and product formulas to simplify the equation but I did not use the double angle formulas. I'm not quite sure what the question is asking.

Here's what I did:
$$= .5 [cos(a+b-a+b)-cos(a+b+a-b)]$$
$$= .5 [cos(2b)-cos(2a)]$$

not sure where the double anges come in. any ideas?

3. simplify $$\sqrt{2-2cos4 \Theta}$$

i can make it become 1-cos4x but I don't know how to simplify it further because of the cos4. no clue on this one, any help would be appreciated.

 

[TD]

1. Express $$\frac{sin^2 2 \theta}{1+cos^2 2 \theta}$$ as a function of $$sin \theta$$
here's what I did:
$$= \frac{4 sin^2 \theta cos^2 \theta}{1+(2cos^2 \theta -1)^2}$$
$$= \frac{4 sin^2 \theta cos^2 \theta}{1+(4cos^4 \theta - 4 cos^2 \theta + 1)}$$
$$= \frac{2sin^2 \theta cos^2 \theta}{2cos^2 \theta - 2cos^2 \theta +1}$$
is this correct? can I simplify it more?

Well it's not written as a function of only sin(x), so it's not correct yet.
Although there may be ways to simplify, you can reduce the sin²2x in function of only sinx like this:

$$\sin ^2 2x = 4\sin ^2 x\cos ^2 x = 4\sin ^2 x\left( {1 - \sin ^2 x} \right)$$

In the denominator, there's a cos²2x but that's equal to 1-sin²2x, giving you the case above again.

2. Write $$sin(a+b)sin(a-b)$$ as a function of double angles
I used the sum and product formulas to simplify the equation but I did not use the double angle formulas. I'm not quite sure what the question is asking.
Here's what I did:
$$= .5 [cos(a+b-a+b)-cos(a+b+a-b)]$$
$$= .5 [cos(2b)-cos(2a)]$$
not sure where the double anges come in. any ideas?

I'm not sure but perhaps this was the point, you have now rewritten it as a function of the double angles 2a and 2b.

3. simplify $$\sqrt{2-2cos4 \Theta}$$
i can make it become 1-cos4x but I don't know how to simplify it further because of the cos4. no clue on this one, any help would be appreciated.

There are 3 version of the double-angle formule for the cosine, choose the one which makes the constant disappear:

$$\sqrt {2 - 2\cos 4x} = \sqrt {2 - 2\left( {1 - 2\sin ^2 2x} \right)}$$

 

[maverick6664]

1.
$$\frac {\sin^2 2 \theta} {2- \sin^2 2\theta} = \frac 2 {2 - \sin^2 2\theta} - 1$$
and simplify the denominator using $$\sin 2\theta = 2 \sin \theta \cos \theta$$ and $$\cos^2 \theta = 1 - \sin^2 \theta$$...? Still complicated.

2. looks correct.

3. $$\sqrt{ 2( 1 - \cos 4\theta)} = \sqrt {2( 1 - \cos^2 2\theta + \sin^2 2\theta)} = 2 | \sin 2\theta} | = 4 | \sin \theta \cos \theta }|$$

hmmm...I have to go to see a class day of the kindergarten of my daughter next month... :)

 

[captainvyom_akash]

Your two problems can be solved easily-------------Akash

Sin(a-b)Sin(a+b)
=Sin(a^2) -Sin(b^2) /////////////////////you can check this easily. it is a formula
=[1-cos(2a)-{1-Cos2b)]/2
=[1-cos(2a)-1+cos(2b)]/2
=[cos(2a)-cos(2b)]/2 ////////////////expressed in double angle in cos





(2-2cos4A)^½=(2(1-cos4A))^½ ////////////A IS ANGLE THETA
=(2(2Sin^2A))^½
=(4Sin^2A)^½
=2Sin^2A)------------------------------Soved

Akash
akash_413@sify.com
captainvyom_akash@yahoo.com