For proving \frac{16}{\cos 4x+7}=\frac{1}{\sin^4 x+\cos^2 x}+\frac{1}{\sin^2 x+\cos^4 x} I tried to use

Jamarion Kerr

Jamarion Kerr

Answered question

2022-01-23

For proving
16cos4x+7=1sin4x+cos2x+1sin2x+cos4x
I tried to use that:
sin4x+cos4x=sin4x+2sin2xcos2x+cos4x2sin2xcos2x
=(sin2x+cos2x)22sin2xcos2x
=12122sinxcosx2
=112sin2(2x)
=1121cos4x2
=34+14cos4x

Answer & Explanation

Prince Huang

Prince Huang

Beginner2022-01-24Added 15 answers

sin4x+cos2x=(1cos2x)24+1+cos2x2=3+cos22x4=3+1+cos4x24=7+cos4x8
sin2x+cos4x=sin4(π2x)+cos2(π2x)=7+cos4(π2x)8=7+cos4x8
Karlie Wilkins

Karlie Wilkins

Beginner2022-01-25Added 7 answers

Note that
sin4x+cos2x=sin2x(1cos2x)+cos2x
=sin2x+cos2x(1sin2x)=sin2x+cos4x
Hence, according to your work,
2(sin4x+cos2x)=(sin4x+cos2x)+(sin2x+cos4x)
=sin4x+cos4x+1=7+cos4x4

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