How do you simplify #sin^2x sin^2x#?

Question

20377 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-07T15:33:43

#sin^4 x#

#sin^2x.sin^2 x = sin^4 x#

Answer 2 · 2016-05-07T15:46:29

#cos^4x^2#-2#cos^2x#+1=0

the formula is #sin^2x#+#cos^2x#=1

#sin^2x# =1-#cos^2x#

Substitute this instead of #sin^2x# #sin^2x#

The equation is now like this:

(1-#cos^2x#)(1-#cos^2x#)=0

Multiply out the brackets:

1-#cos^2x#-#cos^2x#+#cos^4x^2#

Group like terms and put them in the right order:

#cos^4x^2#-2#cos^2x#+1=0