"Simplification of cos4(x)+sin4(x) (sin⁡x)4+(cos⁡x)4=(1−cos⁡2x)24+(1+cos⁡2x)24 =1−2cos⁡2x+(cos⁡2x)2+1+2cos⁡2x+(cos⁡2x)24 =1+(cos⁡2x)22 its correct?" solved and explained

jelentetvq

Answered question

2022-01-28

Simplification of

\cos^{4} (x) + \sin^{4} (x)

{(\sin x)}^{4} + {(\cos x)}^{4} = \frac{{(1 - \cos 2 x)}^{2}}{4} + \frac{{(1 + \cos 2 x)}^{2}}{4}

= \frac{1 - 2 \cos 2 x + {(\cos 2 x)}^{2} + 1 + 2 \cos 2 x + {(\cos 2 x)}^{2}}{4}

= \frac{1 + {(\cos 2 x)}^{2}}{2}

its correct?

Alfred Mueller

Beginner2022-01-29Added 10 answers

A faster way:

\sin^{4} x + \cos^{4} x = {(\sin^{2} x + \cos^{2} x)}^{2} - 2 \sin^{2} x \cos^{2} x = 1 - 2 \sin^{2} x \cos^{2} x

= 1 - \frac{1}{2} {(2 \sin x \cos x)}^{2} = 1 - \frac{1}{2} \sin^{2} 2 x

= 1 - \frac{1}{2} \frac{1 - \cos 4 x}{2} = \frac{3 + \cos 4 x}{4}

Aiden Cooper

Beginner2022-01-30Added 14 answers

{(\frac{e^{i x} - e^{- i x}}{2 i})}^{4} + {(\frac{e^{i x} + e^{- i x}}{2})}^{4} = \frac{1}{2^{4}} ({(e^{i x} - e^{- i x})}^{4} + {(e^{i x} + e^{- i x})}^{4})

but

{(a + b)}^{4} + {(a - b)}^{4} = 2 (a^{4} + 6 a^{2} b^{2} + b^{4})

hence

\sin^{4} x + \cos^{4} x = \frac{1}{2^{3}} (2 \cos (4 x) + 6) = \frac{1}{4} (\cos (4 x) + 3)

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