Use a counter example to show that cos(x+y)= cosx +cosy is not an identity
Let x=π/4, y=π/4,
cos(x+y)
=cos(π/4+π/4)
=cos(π/2)
=0
cos(x)+cos(y)
=cos(π/4)+cos(π/4)
=√2/2+√2/2
=√2
Since √2 ≠ 0,
cos(x+y) ≠ cosx +cosy
Sure, let's take x = 0 and y = π/2.
According to the given expression,
cos(0 + π/2) = cos(0) + cos(π/2)
However,
cos(π/2) = 0 and cos(0) = 1
Therefore, the expression becomes:
cos(π/2) = 1 + 0
But it is clear that the left side of the equation (cos(π/2)) is not equal to the right side (1 + 0).
So, cos(x+y) = cosx + cosy is not an identity. It's more like a "sidentity."
To disprove the equation cos(x+y) = cosx + cosy as an identity, we need to find specific values for x and y that make the equation invalid.
Let's choose x = 0 and y = π/2.
Plugging these values into the equation, we have:
cos(0 + π/2) = cos(0) + cos(π/2)
cos(π/2) = 1 + 0
0 = 1
Since 0 does not equal 1, we have shown that the equation cos(x+y) = cosx + cosy is not an identity.
To find a counter example, we need to find values of angles x and y that satisfy the equation cos(x+y) = cos(x) + cos(y).
Let's try x = 0 and y = π/2:
cos(x+y) = cos(0 + π/2) = cos(π/2) = 0
cos(x) + cos(y) = cos(0) + cos(π/2) = 1 + 0 = 1
Since 0 ≠ 1, this counter example shows that the equation cos(x+y) = cos(x) + cos(y) is not an identity.