Double angle problem.

[brentwoodbc]

First I'll say hi, my name is Ryan,
My question is.
Use the identity for sin2x and an identity for cos2x to verify that sin^2(2x)+cos^2(2x)=1

I know sin^2x +cos^2x=1
but when I use the formula for double angles like where sin2x=2sinxcosx with a sin^2 I am pretty lost.

Thanks for any help you may have.

 

[stapel]

Use the identity for sin2x and an identity for cos2x to verify that sin^2(2x)+cos^2(2x)=1

...when I use the formula for double angles like where sin2x=2sinxcosx with a sin^2 I am pretty lost.

Well, post what you've got so far, starting from where you plugged "2sin(x)cos(x)" in for "sin(2x)" and "cos^2(x) - sin^2(x)" in for "cos(2x)". We'll be glad to help you get un-stuck! :wink:

 

[galactus]

Use the identity \(\displaystyle cos(2x)=cos^{2}(x)-sin^{2}(x)\) along with \(\displaystyle sin(2x)=2sin(x)cos(x)\)

It should fall into place.

 

[brentwoodbc]

I know this is probley wrong but.

sin^2(2x)+cos^2(2x)=1
2sin^2(x)cos^2(x)+cos^2(x)-sin^2(x) This is what is where I dont know what to do.

 

[galactus]

You did not square them.

\(\displaystyle (2sin(x)cos(x))^{2}+(cos^{2}(x)-sin^{2}(x))^{2}\)

 

[brentwoodbc]

You did not square them.

\(\displaystyle (2sin(x)cos(x))^{2}+(cos^{2}(x)-sin^{2}(x))^{2}\)

I pretty much get it now though its just the second bracket that is confusing me.
Thank you.

 

[stapel]

To learn how to multiply polynomials, try here. Then multiply out the square just as you'd multiply out (x - y)[sup:3501576p]2[/sup:3501576p]. :wink:

 

[celiajean]

Hi. I think I may be posting in the wrong spot, although i dont know where to find where to post a 'new' post. I have 2 problems. One is: tan^2(2theta)-3=0 I dont know whether i should add three first...take the square root of both sides and then replace the left hand side with the double angle equation for tangent...I need the general formula for the solutions. The second problem is similar 2) cos(2theta) + 5cos(theta)=-3 i can replace the cos(2theta) with another equation like 2cos^2(theta)-1 but then factoring? quadratic? I'm lost.

 

[Deleted member 4993]

Hi. I think I may be posting in the wrong spot, although i dont know where to find where to post a 'new' post. I have 2 problems.

One is: tan^2(2theta)-3=0

I dont know whether i should add three first --- No ... isolate tan(2?) by adding 3 to both sides ? find the value 2? by inverting (what should be value of 2? so that tan(2?) = ?3 ?)

...take the square root of both sides and then replace the left hand side with the double angle equation for tangent...I need the general formula for the solutions. The second problem is similar 2) cos(2theta) + 5cos(theta)=-3

i can replace the cos(2theta) with another equation like 2cos^2(theta)-1 but then factoring? quadratic? Yes ..

I'm lost.

 

[celiajean]

to isolate the tan (2theta) i have to minus three...but then what about the tan^2...original problem is tan^2(2theta)-3...so i have tan^2(2theta)=3 so do i take the square root of both sides and have sqrt(3)=tan(2theta)? then replace tan(2 theta) with something else? I'm sorry for my ignorance in this subject. :S