Is { cos (a) + sin (b)} * { cos (a) + sin (b)}=1+ 2*cos (a)*cos (b)
To determine if the expression {cos(a) + sin(b)} * {cos(a) + sin(b)} is equal to 1 + 2*cos(a)*cos(b), we can simplify both sides of the equation and compare them.
First, let's expand the left-hand side of the equation:
{cos(a) + sin(b)} * {cos(a) + sin(b)} = cos(a) * cos(a) + cos(a) * sin(b) + sin(b) * cos(a) + sin(b) * sin(b)
= cos^2(a) + 2 * cos(a) * sin(b) + sin^2(b)
Now, let's simplify the right-hand side of the equation:
1 + 2 * cos(a) * cos(b)
Since we have cos^2(a) appearing in the left-hand side, we can rewrite it using the identity cos^2(x) = 1 - sin^2(x), giving us:
1 - sin^2(a) + 2 * cos(a) * sin(b) + sin^2(b)
Now, we can see that the expression on the left-hand side is equal to the expression on the right-hand side. Therefore, {cos(a) + sin(b)} * {cos(a) + sin(b)} = 1 + 2*cos(a)*cos(b).