If (cos q + sin q) = 31/25, then what will be the value of cos² q?

Given: 

(cos q + sin q) = 31/25

Concept used:

Sin²θ + Cos²θ = 1

Calculation:

(cos q + sin q) = 31/25      -----(i)

⇒ (cos q + sin q)² = (31/25)²

⇒ cos² q + sin² q + 2cosqsinq = 961/625

⇒ 1 + 2cosqsinq = 961/625

⇒ 2cosqsinq = 961/625 - 1

⇒ 2cosqsinq = 336/625

Now,

(cos q - sin q)2 = cos2 q + sin2 q - 2cosqsinq

⇒ (cos q - sin q)2 = 1 - 336/625

⇒ (cos q - sin q)2 = 289/625

⇒ (cos q - sin q) = 17/25      ----(ii)

From eq (i) and eq (ii) we get,

2cos q = 48/25

⇒ cos q = 24/25

So, cos2 q = (24/25)²

⇒ 576/625

∴ The required answer is 576/625.

Shortcut Trick

Concept Used:

Sinθ = Opposite/ Hypotenuse(O/H)

Cosθ = Adjacent/ Hypotenuse(A/H)

Calculation:

According to the question, 

⇒ cos q + sin q = 31/25

⇒ A/H + O/H = 31/25

⇒ (A + O)/H = 31/25

So we get, H as 25 and (A + O) = 31

Now we know a triplet as 7, 24, and 25

This satisfies the given condition,

Using the values, we get,  

Cos q = A/H = 24/25

So, cos2 q = (24/25)2

⇒ 576/625

∴ The required answer is 576/625.