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What is root in equation `sin x+cos x` =`sqrt2` `0 <= x<= pi` ` `

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lemong | Student, Undergraduate | (Level 1) Honors

Posted August 5, 2013 at 4:36 PM via web

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What is root in equation `sin x+cos x` =`sqrt2`

`0 <= x<= pi`

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted August 5, 2013 at 5:01 PM (Answer #1)

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You need to solve this trigonometric equation using the following method. You may notice that the coefficient of `cos x` is equal to one, hence, you may replace `tan(pi/4)` for 1, such that:

`sin x + tan(pi/4)*cos x = sqrt2`

You should remember that `tan theta = sin theta/cos theta` , hence, reasoning by analogy, yields:

`sin x + (sin(pi/4))/(cos(pi/4))*cos x = sqrt2`

Bringing the terms to a common denominator yields:

`sin x*(cos(pi/4)) + (sin(pi/4))*cos x = sqrt 2*(cos(pi/4))`

You may  recognize to the left side the expansion of the following trigonometric identity, such that:

`sin theta*cos beta + sin beta*cos theta = sin(theta + beta)`

Reasoning by analogy, yields:

`sin(x + pi/4) = sqrt 2*sqrt2/2`

`sin(x + pi/4) = 1 => x + pi/4 = sin^(-1) 1 => x + pi/4 = pi/2 => x = pi/2 - pi/4 => x = pi/4`

Hence, evaluating the solution to the given equation, over the indicated interval `[0,pi]` , yields `x = pi/4.`

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embizze | High School Teacher | (Level 1) Educator Emeritus

Posted August 5, 2013 at 5:28 PM (Answer #2)

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Solve `sinx+cosx=sqrt(2)` for `0<=x<=pi` :

`sinx+cosx=sqrt(2)`   Square both sides

`sin^2x+2sinxcosx+cos^2x=2`

Use the Pythagorean identity and the double angle identity:

`1+sin2x=2`

`sin2x=1`

`2x=pi/2,2x=(5pi)/2`

`x=pi/4,(5pi)/4` Note that `(5pi)/4` is an extraneous root introduced by squaring both sides. It is not a solution to the original equation.

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The solution on the given interval is `x=pi/4`

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The graph:

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mariloucortez | High School Teacher | (Level 3) Adjunct Educator

Posted August 5, 2013 at 5:54 PM (Answer #3)

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You can solve `sin(x) + cos(x) = sqrt(2)`  by starting to square both sides of the equation so you can get rid of the radical or the square root, it is easier to deal with that way. So, you have:

`(sin(x) + cos(x))^2 = (sqrt(2))^2` 

`sin^2(x) + 2sin(x)cos(x) + cos^2(x)= 2` 

Applying the identity`sin^2(x) + cos^2(x) = 1` , you have the simplified equation:

`1 + 2sin(x)cos(x) = 2` 

Now, combine similar terms by moving 1 to the right side. Remember that moving a term on the other side will change its sign. Or that is the same as subtracting 1 to both sides. Either will give the same result.

`2sin(x)cos(x) = 2 - 1 `

`2sin(x)cos(x) = 1` 

Notice that `2sin(x)cos(x) = sin(2x)` .

Identity of the sin of twice the angle. So replacing the left side with that:

`sin(2x) = 1 `

Take the Arcsine of both sides:

`Asin(sin(2x)) = Asin(1)` 

`2x = pi/2` 

Divide both sides by 2 so that x will be alone on the left side.

`(2x)/2 = pi/(2*2) `

`x = pi/4 `

Hence the answer is pi/4 which indeed in the range of 0<= x <= 2pi.

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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted August 6, 2013 at 1:34 AM (Answer #4)

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`sinx+cosx = sqrt2`

Divide both sides by `sqrt2`

`sinx/sqrt2+cosx/sqrt2 = sqrt2/sqrt2`

We know that;

`1/sqrt2 = cos(pi/4) = sin(pi/4)`

`sinx/sqrt2+cosx/sqrt2 = 1`

`sinxcos(pi/4)+cosxsin(pi/4) = 1`

`sin(x+pi/4) = 1`

`sin(x+pi/4) = sinpi/2`

`x+pi/4 = npi+(-1)^npi/2` where `n in Z`

When n = 0 then `x = pi/4`

When n = 1 then `x = pi/4`

When n = 2 then `x = (9pi)/4`

So the answer is `x = pi/4`.

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