what is the value for sinx+cosx=?

Answer 1

Let f(x) = sin(x) + cos(x)

Apply derivative each side.

f'(x) = (d/dx)[sin(x) + cos(x)]

f'(x) = [sin(x)]' + [cos(x)]'

Recall: derivative of sin(x) is cos(x) and derivative of cos(x) is -sin(x)

So, f'(x) = cos(x) - sin(x)

Now, you make f'(x) = 0.

0 = cos(x) - sin(x)

Add sin(x) to each side.

sin(x) = cos(x)

Divide each side by cos(x).

sin(x)/cos(x) = 1

tan(x) = 1

x = arc tan(1)⇒x = π/4

Therefore x = {π/4+nπ} where n is an integer.

If x = π/4 then f(x) = sin(π/4) + cos(π/4) ⇒ f(x) = 1/√2 + 1/√2

f(x) = 2/√2 = √2√2/√2 = √2

sin(x) + cos(x) = √2.

If x = 5π/4 then f(x) = sin(5π/4) + cos(5π/4) ⇒ f(x) = -1/√2 - 1/√2

f(x) = -2/√2 = -√2√2/√2 = -√2

sin(x) + cos(x) = -√2.

Therefore sin(x) + cos(x) = √2 or -√2.