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If sin θ + cos θ = √2, then what is sin6 θ + cos6 θ + 6 sin2 θ cos2 θ equal to?
1. \(\frac{1}{{4}}\)
2. \(\frac{3}{{4}}\)
3. 1
4. \(\frac{7}{{4}}\)

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Correct Answer - Option 4 : \(\frac{7}{{4}}\)

Given                       

sin θ + cos θ = √2

Formula used :

sin2 θ + cos2 θ = 1 

(x + y)3 = x3 + y3 + 3.x.y(x + y) 

Calculations :

(cos2 θ + sin2 θ)3 = cos6 θ + sin6 θ + 3.sin2 θ.cos2 θ(cos2 θ + sin2 θ) 

⇒ cos6 θ + sin6 θ + 3.sin2 θ.cos2 θ(cos2 θ + sin2 θ) = 13 = 1 

⇒  sin6 θ + cos6 θ + 3 sin2 θ cos2 θ = 1

Now, 

sin θ + cos θ = √2 

Squarring both the sides 

Sin2 θ + cos2 θ + 2 sin θ cos θ = 2 

⇒ 2 sin θ cos θ = 1 

⇒ sin θ cos θ = 1/2 

⇒ sin2 θ cos2 θ = 1/4 

⇒ 3 sin2 θ cos2 θ  = 3/4 

Now,

sin6 θ + cos6 θ + 6 sin2 θ cos2 θ = sin6 θ + cos6 θ + 3 sin2 θ cos2 θ + 3 sin2 θ cos2 θ  

⇒ 1 + (3/4) 

⇒ 7/4 

∴ Option 4 is the correct answer.

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