Prove this trigonometric identity (1 Viewer)

[phoenix159]

cos⁶x + sin⁶x = 1/4 (1 + 3 cos² 2x)



Let cos (x) = c and sin (x) = s

LHS = c⁶ + s⁶
= (c² + s²)(c² - c⁴s⁴ + s²)
= 1 X (c² - c⁴s⁴ + s²)
= c² - c⁴s⁴ + s²

sin²x = 1 - cos²x

LHS = (c² - c⁴[1 - c⁴] + [1 - c²])
= c² - c⁴ + c⁸ + 1 - c²
= c⁸ - c⁴ + 1

RHS = 1/4 (1 + 3 cos² 2x)

cos 2x = cos²x - sin²x = cos²x - (1 - cos²x) = 2cos²x - 1

RHS = 1/4 (1 + 3 cos²2x)
= 1/4 (1 + 3 [2 cos²x - 1]²)
= 1/4 (1 + 3 [2 cos²x - 1]²)
= 1/4 (1 + 3 [2 c - 1]²)
= 1/4 (1 + 3 [4 c² - 4 c + 1])
= 1/4 (1 + 12 c² - 12 c + 3)
= 1/4 (4 + 12 c² - 12 c)
= 3 c² - 4c + 1

 

[Sy123]

cos⁶x + sin⁶x = 1/4 (1 + 3 cos 2x)

Are you sure there is no typo?

x=pi/2 does not yield equality

 

[braintic]

I believe it is: cos⁶x + sin⁶x = 1/4 (1 + 3 cos^2 2x)

 

[phoenix159]

Thank you!

 

[phoenix159]

Cos is meant to be cos^2 --> cos6x + sin6x = 1/4 (1 + 3 cos^2 2x)

 

[braintic]

See attachment: View attachment Identity.pdf

 

[phoenix159]

Thanks heaps!

 

[pheelx3]

heh, so who are you, Fort St?
(how did you think of our exam?)

 

[jxballistic]

bro that question so hectic in da fort st examinimification