Get the solution to "Prove that3(sin⁡θ−cos⁡θ)4+6(sin⁡θ+cos⁡θ)2+4(sin6θ+cos6θ)−13=0"

Reuben Brennan

Answered question

2022-04-02

Prove that

3 {(\sin θ - \cos θ)}^{4} + 6 {(\sin θ + \cos θ)}^{2} + 4 (\sin^{6} θ + \cos^{6} θ) - 13 = 0

anghoelv1lw

Beginner2022-04-03Added 19 answers

Note that

a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})

Let

s = \sin θ, c = \cos θ

as you did. Since

{(s - c)}^{4} = {(s^{2} + c^{2} - 2 s c)}^{2} = {(1 - 2 s c)}^{2} = 1 - 4 s c + 4 s^{2} c^{2}

{(s + c)}^{2} = s^{2} + c^{2} + 2 s c = 1 + 2 s c

s^{6} + c^{6} = (s^{2} + c^{2}) (s^{4} - s^{2} c^{2} + c^{4})

= s^{4} - s^{2} c^{2} + c^{4}

= s^{4} + 2 s^{2} c^{2} - 3 s^{2} c^{2} + c^{4}

= (s^{4} + 2 s^{2} c^{2} + c^{4}) - 3 s^{2} c^{2}

= {(s^{2} + c^{2})}^{2} - 3 s^{2} c^{2}

= 1 - 3 s^{2} c^{2}

we have

3 {(s - c)}^{4} + 6 {(s + c)}^{2} + 4 (s^{6} + c^{6}) = 3 (1 - 4 s c + 4 s^{2} c^{2}) + 6 (1 + 2 s c) + 4 (1 - 3 s^{2} c^{2})

= (3 + 6 + 4) + (- 12 + 12) s c + (12 - 12) s^{2} c^{2}

= 13

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