Question

sin^2(x)cos^2(x)=1/8(1-cos(4x)) can someone plz explain how the trig identies are equal?

Answer 1

Is it \(\sin^2(x)\cos^2(x)=\frac{1}{8}(1-cos(4x))\)?

Answer 2

correct

Answer 3

sin^2xcos^2x = 1/4 (2sinxcosx)^2 = 1/4(sin2x)^2 = 1/4(1-cos^22x) = 1/4 - 1/4cos^22x => 1/4 - 1/8(2cos^22x -1) -1/8 => 1/4 - cos4x/8 -1/8 => 1/8 -cos4x/8 => 1/8(1-cos4x)

Answer 4

Hope Its Correct..

Answer 5

Consider the left side,
\(\sin^2(x) \cos^2(x)\)
Multiply \(\frac{4}{4}\) to the expression, then we get
\(= \frac{4\sin^2(x)\cos^2(x)}{4}\)
\( = \frac{1}{4}(2\sin(x)\cos(x))^2\)
\( = \frac{1}{4}(sin(2x))^2\)
Now, recall \(\cos(2x) = 1-2\sin^2(x)\), that is \(\frac{1-\cos(2x)}{2} = \sin^2(x)\)
Take x=2x and using this identity, we have
\( \frac{1}{4}(sin(2x))^2\)
\(=\frac{1}{4}[\frac{1-\cos(4x)}{2}]\)
\(=\frac{1}{8}(1-\cos(4x))\)

Oops, someone typed faster than me lol

Answer 6

Haha.. No Probs ;)

Answer 7

you guys are life savers. I can get to sleep before midnight tonight

Answer 8

Do you guys know what identity formulas were used in this?