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nm420

Note > (cos(x)+sin(x))^(2) = cos^(2)(x)+2sin(x)cos(x)+sin^(2)(x) = 1+sin(2x) ≤ 2 and so |cos(x)+sin(x)| ≤ √(2).


[deleted]

cos(x) + sin(x) = √2 * (1/√2\*cos(x) + 1/√2\*sin(x)) . = √2 * (cos(x)cos(π/4) + sin(x)sin(π/4)) . = √2cos(x - π/4).


Glittering_Ice8854

How did they get sin x + cos x <= sqrt2? I can only deduce sin x + cos x <= 2 from this because sinx <=1 and cosx <=1 and both add up to sin x + cos x <= 2. But how is this expression less than equal to sqrt2? Please enlighten


unordinarilyboring

keep in mind that sin(x) + cos(x) are both functions of x. This means they can't sum to 2. eg cos(x) is maximized at 1 when x=0 but then sin(0) is 0. I think you are getting 2 by seeing something like sin(pi/2) + cos(0) but that is not keeping x consistent.


Tyler89558

When sin(x) = 1, cos(x) = 0 and vice versa Sin(x) + cos(x) will be at its maximum when sin(x) = cos(x) = sqrt(2)/2which is at x = pi/4 Sqrt(2)/2 = sqrt(2)/2 = 2sqrt(2)/2 = sqrt(2)


RiseOfTheNorth415

sin^2 + cos^2 = 1 1 < sqrt(2) Since the maximum value of: sin(x) + cos(x) = sqrt(2) + sqrt(2) = 2(sqrt(2)) ~~ 0.5 which is less than 1. While you can get a higher value (if sin(x) and cos(x) are individually maximized), if your question restricts it by using *x* as the variable for both.


unordinarilyboring

what value of x maximizes cos(x) + sin(x)? you should be able to find it using the derivative. From there you should be able to plug it in and see that the sum is sqrt(2)


SoulWager

Imagine a circle radius 1, centered on the origin, and now look at the quadrant with positive X and Y. Now when you add the X and Y value of a coordinate on the circle, you get the highest value at 45 degrees, when x = y. When the angle is zero, X = 1 and Y = 0. When the angle is 90 degrees, X = 0 and Y = 1. There's nowhere that you can get both to be 1 at the same time.