Simplifying (2asin@cos@)^2+a^2(cos^2@-sin^2@)^2=a^2

[m6drifter]

I am stuck trying to figure out a part of my problem. The object of the problem is to simplify one side to equal the other side. My problem is:

(2asinθcosθ)^2 + a^2 (cos^2θ - sin^2θ)^2 = a^2

I've simplified the second part of a^2 (cos^2θ - sin^2θ)^2:
--> a^2 (cos^2θ - sin^2θ) (cos^2θ - sin^2θ)
--> a^2 (1 - 2sin^2θ) (1 - 2sin^2θ)
--> a^2 (1 - 4sin^2θ + 4sin^2θ)
--> a^2 (1)
--> a^2

So I'm curious how I simplify:
(2asinθcosθ)


Any help would be appreciated. Let me know if I did the other part incorrectly, but it looks correct to me.

 

[Deleted member 4993]

Use

sin(2*x) = 2 * sin(x) * cos(x)

and

cos(2*x) = {cos(x)}^2 - {sin(x)}^2

Then factor out a^2 and further use

{cos(x)}^2 + {sin(x)}^2 = 1

 

[galactus]

\(\displaystyle \L\\(2a\cdot{sin{\theta}cos{\theta})^{2}}+a^{2}(cos^{2}{\theta}-sin^{2}{\theta})^{2}\)

You could do it this way:

\(\displaystyle \L\\2a\cdot{sin{\theta}cos{\theta}}=a\cdot{sin2{\theta}}\)

and

\(\displaystyle \L\\cos^{2}{\theta}-sin^{2}{\theta}=cos2{\theta}\)

Therefore, you have:

\(\displaystyle \L\\a^{2}sin^{2}2{\theta}+a^{2}cos^{2}2{\theta}\)

Factor out a^2 and get:

\(\displaystyle \L\\a^{2}(\underbrace{sin^{2}2{\theta}+cos^{2}2{\theta}}_{\text{this equals 1}})\)

So, you have \(\displaystyle \L\\a^{2}\)


EDIT: Oops, SK beat me to it. Oh well, hopefully, you see now it's a matter of applying the correct identities.

 

[m6drifter]

Wow, thanks guys. Apparently I was making it harder than it should have been.

 

[soroban]

Hello, m6drifter!

Your approach should have worked, but you made an error . . .

Show that: \(\displaystyle \:\left(2a\sin\theta\cos\theta\right)^2\,+\,a^2\left(\cos^2\theta\,-\,\sin^2\theta\right)^2 \;= \;a^2\)

We have: \(\displaystyle \:4a^2\sin^2\theta\cos^2\theta \,+\,a^2\left(\cos^4\theta\,-\,2\cos^2\theta\sin^2\theta\,+\,\sin^4\theta)\\right)\)

. . . . . .\(\displaystyle = \;a^2\left(4\sin^2\theta\cos^2\theta\,+\,\cos^4\theta\,-\,2\sin^2\theta\cos^2\theta\,+\,\sin^4\theta\right)\)

. . . . . .\(\displaystyle = \;a^2\left(\cos^4\theta\,+\,2\cos^2\theta\sin^2\theta\,+\,\sin^4\theta\right)\)

. . . . . .\(\displaystyle = \;a^2\underbrace{\left(\cos^2\theta\,+\,\sin^2\theta\right)}_{\text{This is 1}}^2\)

. . . . . .\(\displaystyle = \;a^2\)