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trigonometry problemwhen sin x-cosx=0? when sin is maxim, cos is minim, so the...

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nena1993 | Student, Grade 11 | (Level 2) Honors

Posted February 9, 2011 at 2:44 AM via web

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trigonometry problem

when sin x-cosx=0?

when sin is maxim, cos is minim, so the difference is -1 or 1 but is not zero.

my tutor said that's not true! why?

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted February 9, 2011 at 4:12 AM (Answer #2)

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Your tutor is correct! There are not just 2 possibilities for the angle x.

It is obvious that the values for sin x and cos x have to be equal in order to cancel the expression sin x - cos x.

sin x - cos x=0

We could divide the equation by (cos x) and the result will be:

sin x /cos x - 1 =0

But we know that the ratio between sinx and cos x determine the tangent function.

tan x -1 =0

tan x = 1

x= arctan 1 + k*pi

x = pi/4 + k*pi

pi/4=45 degrees.

We could also notice that if the ratio sin x / cos x=1, it means that the numerator and denominator are equal.

=> sin x = cosx!

If sin x = cos x => the x values  are equal, too, so, in a right angle triangle, the angles could only be of 45 degrees (the conclusion is based on fact that in a triangle, the sum of angles is 180 degrees, and one of them is 90 degrees and the other 2 are equals, so

90 + 2*x=180



x=45 degrees

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted February 9, 2011 at 5:43 AM (Answer #3)

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You want the value of x for which sin x - cos x = 0. Why are you taking the cases where cos x = 1, sin x = 0, etc.

Instead, solve for x.

sin x - cos x = 0

=> sin x = cos x

=> sin x / cos x = 1

=> tan x = 1

=> x = arc tan 1

=> x = pi/4 + n*pi

So the solution of the equation is x = pi/4 + n*pi

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