(cosx - sinx)^2 + (cosx + sinx)^2 = 2
Iam in this step:
2cos^2(x) + 2sin^2(x) = 2
How can I make the equation on the right equal 2
oh, please. What is the first and most fundamental trig identity you ever learned?
cos^2 + sin^2 = 1
right? Time for the duhh! and head slap.
Thank You
To make the equation on the right side equal to 2, you can simplify the terms on the left side.
First, distribute the squared terms:
(cosx - sinx)^2 = cos^2(x) - 2cos(x)sin(x) + sin^2(x)
(cosx + sinx)^2 = cos^2(x) + 2cos(x)sin(x) + sin^2(x)
Now, substitute these values back into the original equation:
(cos^2(x) - 2cos(x)sin(x) + sin^2(x)) + (cos^2(x) + 2cos(x)sin(x) + sin^2(x)) = 2
Next, simplify by combining like terms:
2cos^2(x) + 2sin^2(x) - 2cos(x)sin(x) + 2cos^2(x) + 2sin^2(x) + 2cos(x)sin(x) = 2
Notice that the -2cos(x)sin(x) and +2cos(x)sin(x) terms cancel each other out. This simplifies the equation to:
2cos^2(x) + 2sin^2(x) + 2cos^2(x) + 2sin^2(x) = 2
Now, combine like terms again:
4cos^2(x) + 4sin^2(x) = 2
Finally, divide both sides of the equation by 4 to isolate the terms:
cos^2(x) + sin^2(x) = 2/4
Since cos^2(x) + sin^2(x) equals 1 (by the Pythagorean identity), the equation becomes:
1 = 2/4
Simplifying further:
1 = 1/2
Therefore, the equation is not true.