sin(x) + cos(x)= 1/2

• Jun 20th 2011, 08:31 AM
neilofbodom
sin(x) + cos(x)= 1/2
Id like to know the solution to

sin(x) + cos(x)= 1/2

rather than the answer, i'd like to know how to work it out since i got a wrong answer.

thanks,
neilofbodom
• Jun 20th 2011, 08:39 AM
Also sprach Zarathustra
Re: sin(x) + cos(x)= 1/2
Quote:

Originally Posted by neilofbodom
Id like to know the solution to

sin(x) + cos(x)= 1/2

rather than the answer, i'd like to know how to work it out since i got a wrong answer.

thanks,
neilofbodom

sin(x) + cos(x)= 1/2

(sin(x) + cos(x))^2= (1/2)^2

sin(2x)=(1/4)-1

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• Jun 20th 2011, 08:40 AM
Prove It
Re: sin(x) + cos(x)= 1/2
First thing you will need to do is graph the function, to see how many solutions you are expecting. You need to square both sides of the function to solve this equation, and squaring could bring in extraneous solutions.

\displaystyle \displaystyle \begin{align*} \sin{x} + \cos{x} &= \frac{1}{2} \\ (\sin{x} + \cos{x})^2 &= \left(\frac{1}{2}\right)^2 \\ \sin^2{x} + \cos^2{x} + 2\sin{x}\cos{x} &= \frac{1}{4} \\ 1 + \sin{2x} &= \frac{1}{4} \\ \sin{2x} &= -\frac{3}{4} \\ 2x &= \left\{ \pi + \arcsin{\left(\frac{3}{4}\right)}, 2\pi - \arcsin{\left(\frac{3}{4}\right)}\right\} + 2\pi n\textrm{ where }n \in \mathbf{Z} \\ x &= \left\{\frac{\pi}{2} + \frac{1}{2}\arcsin{\left(\frac{3}{4}\right)} , \pi - \frac{1}{2}\arcsin{\left(\frac{3}{4}\right)} \right\} + \pi n\end{align*}

Now disregard any extraneous solutions.
• Jun 20th 2011, 10:15 AM
Re: sin(x) + cos(x)= 1/2
Quote:

Originally Posted by neilofbodom
Id like to know the solution to

sin(x) + cos(x)= 1/2

rather than the answer, i'd like to know how to work it out since i got a wrong answer.

thanks,
neilofbodom

This has no solution in the 3rd quadrant
as sin(x) and cos(x) are negative there.

We can examine whether there is a solution in the 1st quadrant, where both are positive.

$\displaystyle f(x)=sinx+cosx\Rightarrow\ f'(x)=cosx-sinx$

The turning point occurs when sin(x) = cos(x)

$\displaystyle \Rightarrow\ x=\frac{\pi}{4}$

$\displaystyle f''(x)=-sinx-cosx$

hence the first quadrant contains a maximum as $\displaystyle f''(x)<0,\;\;\;x=\frac{\pi}{4}$

More significantly, the tangent slope is always decreasing in the first quadrant,
hence the minimum values in the first quadrant occur either side of the maximum at $\displaystyle x=0,\;\;\;x=\frac{\pi}{2}$

where f(x) =1 in both cases.

Hence, there is no solution in the first quadrant either.

There are certainly solutions in the 2nd and 4th quadrants, since sin(x) and cos(x)
have opposite sign there.

We can use the fact that $\displaystyle cosx=sin\left(\frac{\pi}{2}-x\right)$

coupled with the identity

$\displaystyle sinA+sinB=2sin\frac{A+B}{2}cos\frac{A-B}{2}$

to obtain

$\displaystyle sinx+cosx=sinx+sin\left(\frac{\pi}{2}-x\right)=2sin\frac{\pi}{4}cos\left(x-\frac{\pi}{4}\right)$

$\displaystyle \Rightarrow\ \frac{2}{\sqrt{2}}cos\left(x-\frac{\pi}{4}\right)=\frac{1}{2}$

$\displaystyle \Rightarrow\ cos\left(x-\frac{\pi}{4}\right)=\frac{\sqrt{2}}{4}$

$\displaystyle x=cos^{-1}\frac{\sqrt{2}}{4}+\frac{\pi}{4}$

and this gives the 2nd quadrant solution.

Hint: You'll need to reverse the signs of cos(x) and sin(x)
• Jun 21st 2011, 06:07 AM
Soroban
Re: sin(x) + cos(x)= 1/2
Hello, neilofbodom!

Another approach . . .

Quote:

$\displaystyle \text{Solve: }\;\sin x + \cos x \:=\: \frac{1}{2}$

$\displaystyle \text{Divide by }\sqrt{2}: \;\;\frac{1}{\sqrt{2}}\sin x + \frac{1}{\sqrt{2}}\cos x \;=\;\frac{1}{2\sqrt{2}}$

$\displaystyle \text{Note that: }\:\sin\tfrac{\pi}{4} = \tfrac{1}{\sqrt{2}},\;\;\cos\tfrac{\pi}{4} = \tfrac{1}{\sqrt{2}}$

$\displaystyle \text{So we have: }\;\underbrace{\cos\tfrac{\pi}{4}\sin x + \sin\tfrac{\pi}{4}\cos x} \;=\;\frac{1}{2\sqrt{2}}$
$\displaystyle \text{Then: }\qquad\qquad\quad\sin\left(x + \tfrac{\pi}{4}\right) \;=\;\frac{1}{2\sqrt{2}}$

$\displaystyle \text{Hence: }\;x + \frac{\pi}{4} \;=\;\sin^{-1}\!\left(\frac{1}{2\sqrt{2}}\right)$

$\displaystyle \text{Therefore: }\;x \;=\;\sin^{-1}\!\left(\frac{1}{2\sqrt{2}}\right) - \frac{\pi}{4} \;\approx\;-0.424$

• Jun 21st 2011, 06:43 AM
Prove It
Re: sin(x) + cos(x)= 1/2
Quote:

Originally Posted by Soroban
Hello, neilofbodom!

Another approach . . .

$\displaystyle \text{Divide by }\sqrt{2}: \;\;\frac{1}{\sqrt{2}}\sin x + \frac{1}{\sqrt{2}}\cos x \;=\;\frac{1}{2\sqrt{2}}$

$\displaystyle \text{Note that: }\:\sin\tfrac{\pi}{4} = \tfrac{1}{\sqrt{2}},\;\;\cos\tfrac{\pi}{4} = \tfrac{1}{\sqrt{2}}$

$\displaystyle \text{So we have: }\;\underbrace{\cos\tfrac{\pi}{4}\sin x + \sin\tfrac{\pi}{4}\cos x} \;=\;\frac{1}{2\sqrt{2}}$
$\displaystyle \text{Then: }\qquad\qquad\quad\sin\left(x + \tfrac{\pi}{4}\right) \;=\;\frac{1}{2\sqrt{2}}$

$\displaystyle \text{Hence: }\;x + \frac{\pi}{4} \;=\;\sin^{-1}\!\left(\frac{1}{2\sqrt{2}}\right)$

$\displaystyle \text{Therefore: }\;x \;=\;\sin^{-1}\!\left(\frac{1}{2\sqrt{2}}\right) - \frac{\pi}{4} \;\approx\;-0.424$

Of course, to get all possible solutions...

\displaystyle \displaystyle \begin{align*} \sin{\left(x + \frac{\pi}{4}\right)} &= \frac{\sqrt{2}}{4} \\ x + \frac{\pi}{4} &= \left\{\arcsin{\left(\frac{\sqrt{2}}{4}\right)}, \pi - \arcsin{\left(\frac{\sqrt{2}}{4}\right)}\right\} + 2\pi n\textrm{ where }n\in\mathbf{Z} \\ x &= \left\{\arcsin{\left(\frac{\sqrt{2}}{4}\right)} - \frac{\pi}{4}, \frac{3\pi}{4} - \arcsin{\left(\frac{\sqrt{2}}{4}\right)}\right\} + 2\pi n \end{align*}
• Jun 21st 2011, 07:18 AM
Re: sin(x) + cos(x)= 1/2
Quote:

Originally Posted by neilofbodom
Id like to know the solution to

sin(x) + cos(x)= 1/2

rather than the answer, i'd like to know how to work it out since i got a wrong answer.

thanks,
neilofbodom

You could also use simple geometry as shown in the attachment,
using the fact that the sum of 2 sides of a triangle is greater than the 3rd side.
This rules out the 1st and 3rd quadrants.

Then you can find the 2nd quadrant solution using

$\displaystyle sinx+cosx=sinx+sin\left(\frac{\pi}{2}-x\right)=2cos\frac{\pi}{4}cos\left(x-\frac{\pi}{4}\right)$

to get

$\displaystyle x=arccos\left(\frac{\sqrt{2}}{4}\right)+\frac{\pi} {4}$

Then to get the 4th quadrant solution, rotate the triangle to the position
shown in the attachment to get

$\displaystyle x=\frac{9\pi}{4}-arccos\left(\frac{\sqrt{2}}{4}\right)$