# Question #b7e2c

##### 1 Answer

You would like to prove that

#2(sin^6 Theta + cos^6 Theta) - 3(sin^4 Theta + cos^4 Theta) + 1 = 0#

This can be proven by starting at the left side and transforming until we have the right side, so in this case,

I will use the following formulas and identites:

[1]

#" "sin^2 Theta + cos^2 Theta = 1 #

[2]# " "(a - b)^2 = a^2 - 2ab + b^2#

[3]# " "(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3#

Let's start with the proof:

#2(sin^6 Theta + cos^6 Theta) - 3(sin^4 Theta + cos^4 Theta) + 1#

#= 2([sin^2 Theta]^3 + cos^6 Theta) - 3([sin^2 Theta]^2 + cos^4 Theta) + 1#

... use [1 ]

#= 2([1 - cos^2 Theta]^3 + cos^6 Theta) - 3([1 - cos^2 Theta]^2 + cos^4 Theta) + 1#

... use the binomial formulas [2] and [3] ....

#= 2([1 - 3 cos^2 Theta + 3 cos^4 Theta - cos^6 Theta] + cos^6 Theta) - 3([1 - 2 cos^2 Theta + cos^4 Theta] + cos^4 Theta) + 1#

#= 2(1 - 3 cos^2 Theta + 3 cos^4 Theta cancel( - cos^6 Theta + cos^6 Theta)) - 3(1 - 2 cos^2 Theta + 2 cos^4 Theta) + 1#

#= 2 - 6 cos^2 Theta + 6 cos^4 Theta - 3 + 6 cos^2 Theta - 6 cos^4 Theta + 1#

#= 2 cancel(color(green)(- 6 cos^2 Theta)) cancel(color(blue)(+ 6 cos^4 Theta)) - 3 cancel(color(green)(+ 6 cos^2 Theta)) cancel(color(blue)(- 6 cos^4 Theta)) + 1#

#= 2 - 3 + 1#

#= 0#

q.e.d.