Homework Help

`sinx/sqrt 2 = cos^2x`

user profile pic

schlule | Student, Grade 11 | eNotes Newbie

Posted May 16, 2013 at 12:13 AM via web

dislike 1 like

`sinx/sqrt 2 = cos^2x`

3 Answers | Add Yours

user profile pic

crmhaske | College Teacher | (Level 3) Associate Educator

Posted May 16, 2013 at 4:39 AM (Answer #4)

dislike 1 like

Correction:

`u_1=-1` and` u2=0.5`

Therefore:

`0.5=cos^2x_2-gtx_2= pi/4`

The second root still does not exist for cosine, so we can convert and end up with the same result as oldnick

`-1=cos^2x_1=1-sin^2x_1`

`-2=-sin^2x_1`

`sinx_1=sqrt(2)`

You will find that if you try to calculate `arcsinsqrt2`  in your calculator you will get an error.  That is because it does not exist.  The values that exist for arcsin fall between -1 and 1, and `sqrt(2)>1`

The value of `sin((3pi)/4)` is `sqrt(2)/2` ; therefore, that cannot be the solution to `arcsinsqrt2`  .  Rather, it is a multplie of the one existing root `pi/4` .

Also, observe in the graph of the function that the graph does not cross the x-axis at sqrt(2), further supporting that the root does not exist. Rather, all roots of this function are multiples of the 1 existing root, pi/4, as it is an infinitely repeating function.

``

user profile pic

crmhaske | College Teacher | (Level 3) Associate Educator

Posted May 16, 2013 at 12:36 AM (Answer #1)

dislike 0 like

Presumably you are wanting to solve for x?

`sinx/sqrt2=cos^2x`

We know that:

`sin^2x=1-cos^2x`

Therefore if we square both sides of equation we wish to solve:

`(sin^2x)/2=cos^4x`

Now substitute the identity above:

`1-cos^2x=2cos^4x`

Rearranging we have the equation:

`2cos^4x+cos^2x-1=0`

In order to solve for the roots of this equation we will simplify by setting `u=cos^2x`

`2u^2+u-1=0`

`u^2+(1/2)u-1/2=0`

Two numbers that muliply to -0.5 and add to 0.5 are 1 and -0.5.  Therefore:

u1=1 and u2=-0.5

Now we can solve for x1 and x2:

`1=cos^2x_1 -> x_1=0`

`-0.5=cos^2x_2`

As you cannot take the square root of a negative number, x2 does not exist.

Sources:

user profile pic

oldnick | (Level 1) Valedictorian

Posted May 16, 2013 at 12:48 AM (Answer #3)

dislike 0 like

WROOOOOOOOOONG CMHASKE!

Root exists, for `x=pi/4 +kpi`  and `x=3/4 +kpi`

Indeed  `sinx=sqrt(2)/2`

and `sinx/sqrt(2)= sqrt(2)/2sqrt(2)=1/2=cos^2x=(sqrt(2)/2)^2)=1/2` !!!!!!!!!!!!!

WROOOOOOOOONNNNNNNNNNNNNNNNNNNNNNGGGGGGG!

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes