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what is the value for sinx+cosx=?

+1 vote
asked Feb 14, 2013 in TRIGONOMETRY by venkatesh Rookie

1 Answer

+1 vote

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 = 22/2 = 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 = -22/2 = -2

sin(x) + cos(x) = -2.

Therefore sin(x) + cos(x) = 2 or -2.

answered Feb 14, 2013 by britally Apprentice

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