Question

(cosA+sinA)
2
+(sinA−cosA)
2
=2​

Answer 1

Here we want to solve:

(cos(A) + sin(A))^2 + ( sin(A) - cos(A))^2 = 2

First, remember the two expansions:

(a + b)^2 = a^2 + 2*a*b + b^2

(a - b)^2 = a^2 - 2*a*b - b^2

Using these relations, we can expand:

(cos(A) + sin(A))^2 = cos(A)^2 + 2*cos(A)*sin(A) + sin(A)^2

( sin(A) - cos(A))^2 = sin(A)^2 - 2*sin(A)*cos(A) + cos(A)^2

Replacing that in our equation, we geT:

(cos(A)^2 + 2*cos(A)*sin(A) + sin(A)^2) + (sin(A)^2 - 2*sin(A)*cos(A) + cos(A)^2) = 2

We can simplify this to get:

2*cos(A)^2 + 2*sin(A)^2 + 2*cos(A)*sin(A) - 2*cos(A)*sin(A) = 2

2*cos(A)^2 + 2*sin(A)^2 = 2

We can take the 2 as a common factor to get:

2*cos(A)^2 + 2*sin(A)^2 = 2*( cos(A)^2 + sin(A)^2) = 2

Now we can divide both sides by 2:

2*( cos(A)^2 + sin(A)^2)/2 = 2/2

cos(A)^2 + sin(A)^2 = 1

And this is a trigonometric property that is true for any value of A.

So the equality:

(cos(A) + sin(A))^2 + ( sin(A) - cos(A))^2 = 2

is true for any real value of A.