Question

Prove the following :

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

Accepted Answer

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

Now, $(\sin θ - \cos θ)^{4} =$ ${[(\sin θ - \cos θ)^{2}]}^{2}$

$= {[\sin^{2} θ + \cos^{2} θ - 2 \sin θ \cos θ]}^{2}$

$= [1 - 2 \sin θ \cos θ]^{2}$

$= 1 + 4 \sin^{2} θ \cos^{2} θ - 4 \sin θ \cos θ$

and $(\sin θ + \cos θ)^{2} = \sin^{2} θ + \cos^{2} θ + 2 \sin θ \cos θ$

$= 1 + 2 \sin θ \cos θ$

and $(\sin^{6} θ + \cos^{6} θ) = {(\sin^{2} θ)}^{3} + {(\cos^{2} θ)}^{3}$

$= (\sin^{2} θ + \cos^{2} θ) ({(\sin^{2} θ)}^{2} - \sin^{2} θ \cos^{2} θ + {(\cos^{2} θ)}^{2})$

$= (1) [\sin^{4} θ + \cos^{4} θ - \sin^{2} θ \cos^{2} θ]$

$= [{(\sin^{2} θ + \cos^{2} θ)}^{2} - 2 \sin^{2} θ \cos^{2} θ - \sin^{2} θ \cos^{2} θ]$

$= 1 - 3 \sin^{2} θ \cos^{2} θ$

So, $3 (\sin θ - \cos θ)^{4} + 6 (\sin θ + \cos θ)^{2} + 4 (\sin^{6} θ + \cos^{6} θ)$

$= 3 (1 + 4 \sin^{2} θ \cos^{2} θ - 4 \sin θ \cos θ) + 6 (1 + 2 \sin θ \cos θ) + 4 (1 - 3 \sin^{2} θ \cos^{2} θ)$

$= 3 + 6 + 4$

$= 13$

Hence, proved that $3 (\sin θ - \cos θ)^{4} + 6 (\sin θ + \cos θ)^{2} + 4 (\sin^{6} θ + \cos^{6} θ) = 13$ .

Telangana Board Solutions for Exercise 1: Exercise 6 (a)

Telangana Board Mathematics Solutions for Exercise - Telangana Board Solutions for Exercise 1: Exercise 6 (a)

Questions from Telangana Board Solutions for Exercise 1: Exercise 6 (a) with Hints & Solutions

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