I am stuck on this identity \(\displaystyle{\frac{{{2}{{\sin}^{{4}}{x}}+{{\cos}^{{2}}{x}}-{2}{{\cos}^{{4}}{x}}}}{{{3}{{\sin}^{{2}}{x}}-{1}}}}={1}\)

ikramkeyslo4s

ikramkeyslo4s

Answered question

2022-04-01

I am stuck on this identity
2sin4x+cos2x2cos4x3sin2x1=1

Answer & Explanation

Cailyn Hanson

Cailyn Hanson

Beginner2022-04-02Added 11 answers

Let's take this one step at a time. First, move the denominator to the right side. 1)
2sin4(x)+cos2(x)2cos4(x)=3sin2(x)1
Use the identity sin2(x)=1cos2(x) to rewrite sin4(x)=(1cos2(x))2. 2)
2(1cos2(x))2+cos2(x)2cos4(x)=3sin2(x)1
Now foil out 2(1cos2(x))2=2(12cos2(x)+cos4(x)) and distribute 2. 3)
24cos2(x)+2cos4(x)+cos2(x)2cos4(x)=3sin2(x)1
Now there are three things we can do here First we have 2cos4(x) and 2cos4(x) so those will cancel out. Second we have 4cos2(x)+cos2(x)=3cos2(x). And last we can move the 2 from the left side of the equation to the right. 4)
3cos2(x)=3sin2(x)3
Now factor out the 3 and it cancels out 5)
cos2(x)=sin2(x)1
And just rearrange it and we get 6)
1=sin2(x)+cos2(x)

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