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Simplify 2cos2xsin2x ?? how to do

i have to use trig identities to solve, confused.

Answer 1

Emery Adkins

You can simplify the expression to #sin(4x)#.

If we let #u=2x#, we can use the sine double angle formula backward:

#color(white)=2sin(2x)# #cos(2x)#

#=2sin(u)# #cos(u)#

#=sin(2u)#

#=sin(2*u)#

#=sin(2*2x)#

#=sin(4x)#

That's as simplified as it gets. Hope this helped!

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Answer 2

Adam Aguirre

To simplify (2\cos(2x)\sin(2x)), you can use the double angle identities for cosine and sine:

[\cos(2x) = 2\cos^2(x) - 1] [\sin(2x) = 2\sin(x)\cos(x)]

Substitute these identities into the expression:

[2(2\cos^2(x) - 1)(2\sin(x)\cos(x))]

Expand and simplify:

[4\cos^2(x) \cdot 2\sin(x)\cos(x) - 2(2\sin(x)\cos(x))]

[8\cos^2(x)\sin(x)\cos(x) - 4\sin(x)\cos(x)]

[4\sin(x)\cos(x)(2\cos^2(x) - 1)]

Now, you can use the Pythagorean identity (\sin^2(x) + \cos^2(x) = 1) to replace (\cos^2(x)) with (1 - \sin^2(x)):

[4\sin(x)\cos(x)(2(1 - \sin^2(x)) - 1)]

[4\sin(x)\cos(x)(2 - 2\sin^2(x) - 1)]

[4\sin(x)\cos(x)(1 - 2\sin^2(x))]

Finally, you can use the double angle identity for sine (\sin(2x) = 2\sin(x)\cos(x)) to replace (2\sin(x)\cos(x)) with (\sin(2x)):

[4\sin(2x)(1 - 2\sin^2(x))]

So, (2\cos(2x)\sin(2x)) simplifies to (4\sin(2x)(1 - 2\sin^2(x))).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.