Hey i need some help sin^4 (X) - cos^4 (X) = 1 - 2cos^2(X)
Follow Math Help Forum on Facebook and Google+
Originally Posted by oneway1225 Hey i need some help sin^4 (X) - cos^4 X = 1 - 2cos^2(X) Note that the left hand side is a difference of two squares: $\displaystyle (\sin^2 x - \cos^2 x) (\sin^2 x + \cos^2 x)$.
Hello, Here is a way of solving this: $\displaystyle sin^4(x)-cos^4(x)=[sin^2(x)-cos^2(x)][sin^2(x)+cos^2(x)]$$\displaystyle =sin^2(x)-cos^2(x)=1-cos^2(x)-cos^2(x)=1-2cos^2(x) $ Have a nice day, Hush_Hush.
Originally Posted by Hush_Hush Hello, Here is a way of solving this: $\displaystyle sin^4(x)-cos^4(x)=[sin^2(x)-cos^2(x)][sin^2(x)+cos^2(x)]$$\displaystyle =sin^2(x)-cos^2(x)=1-cos^2(x)-cos^2(x)=1-2cos^2(x) $ Have a nice day, Hush_Hush. how did you go from [sin^2(x)-cos^2(x)][sin^2(x)+cos^2(x)] to sin^2(x)-cos^2(x)
Originally Posted by oneway1225 how did you go from [sin^2(x)-cos^2(x)][sin^2(x)+cos^2(x)] to sin^2(x)-cos^2(x) What is always the value of $\displaystyle \sin^2(x)\, +\, \cos^2(x)$?
Do you need any more help with this problem?
View Tag Cloud