What is the solution for the equation sinx=cosx?

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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.

tg x -1 =0

We'll move the free term in the right side of the equal sign, like this:

tg x = 1

x= arctg 1 + k*pi

x = pi/4 + k*pi

pi/4=45 degrees.

We could also make the note that if the ratio sin x / cos x=1, it means that the terms from the ratio are equal

=> sin x = cosx!

If sin x = cos x => the angles of the triangle are equal, too, so, in a right angle triangle, the angles could only be = 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

2*x=180-90

x=90/2

**x=45 degrees**

sinx = cosx , given. square both sides and get

(sinx)^2 = ( cosx)^2 =1-(sinx)^2x , as (sinx)^2+cosx)^2 = 1 is a well known identiy .

Rearrange to get, 2(sinx)^2 = 1

Therefore, sinx = +1/sqrt2 or -1/sqrt2

**When sinx =1/sqrt2 ,x=45degree** or

**when sinx -1/sqrt2, ****x=180+45 = 225 degree.**

*********

Check:

cos45 = 1/sqrt2 =sin45 and

co225 = -1/sqrt2 =sign225

sinX=cosX

sinX=cos(Pi/2-X)

so X=pi/4(4n+1)..................n belongs to Real numbers

eg:-

n=0 you get X=45degrees

n=1 you get X=225degrees

put n=-1,-2,-3,....2,3,4,............you got perfact answers

Square both sides of the original equation to get

`sin^2x=cos^2x`

From the identity `cos^2x=1-(sin^2x)`

we get `sin^2x=1-(sin^2x)`

Rearrange to get `2(sin^2x)=1`

`sin^2x=1/2`

`sinx=1/sqrt(2) or sinx=-1/sqrt(2)`

When `sinx=1/sqrt(2)` ,**`x=45^@ or pi/4 rad` **

When `sin=-1/sqrt(2)` , `x=225^@ or 5pi/4 rad`

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