# trigonometric function

#### Isa16

Given the following information:
f(x)=-sinx
g(x)=cosx

find x(s) in order to have f=g

#### romsek

MHF Helper
This is pretty simple to solve. Have you tried anything?

#### Isa16

yes I did, but probably wrongly.
Could you tell me at least how to start?

#### romsek

MHF Helper
yes I did, but probably wrongly.
Could you tell me at least how to start?
$-\sin(x) = \cos(x)$

$\dfrac{\sin(x)}{\cos(x)} = -1$

$\tan(x) = -1$

solve that for $x$

#### Isa16

Thanks Romsek, you're right it is pretty simple.
But could you please tell me if I found the correct value of x?
x= 135º or 315º

#### romsek

MHF Helper
Thanks Romsek, you're right it is pretty simple.
But could you please tell me if I found the correct value of x?
x= 135º or 315º
both of those values satisfy the original equation but so do those plus multiples of $360^\circ$

#### Isa16

I realized this after a while.
it goes like x={135º +n360º or 315º +n360º, n ∈ Z}.

Thanks a lot!
~Have a good day!

#### Prove It

MHF Helper
$-\sin(x) = \cos(x)$

$\dfrac{\sin(x)}{\cos(x)} = -1$

$\tan(x) = -1$

solve that for $x$
While this method is correct, I tend to be wary of solving equations through division, as you never know if you are dividing by 0 and thus losing solutions. In this case I would probably lean more towards making use of a compound angle identity \displaystyle \begin{align*} \sin{ \left( \alpha + \beta \right) } \equiv \sin{ \left( \alpha \right) } \cos{ \left( \beta \right) } + \cos{ \left( \alpha \right) } \sin{ \left( \beta \right) } \end{align*}.

So with this equation we have

\displaystyle \begin{align*} -\sin{ \left( x \right) } &= \cos{ \left( x \right) } \\ 0 &= \sin{ \left( x \right) } + \cos{ \left( x \right) } \\ 0 \cdot k &= \sin{ \left( x \right) } \cdot k + \cos{ \left( x \right) } \cdot k \\ 0 &= \sin{ \left( x \right) } \cos{ \left( y \right) } + \cos{ \left( x \right) } \sin{ \left( y \right) } \end{align*}

and here \displaystyle \begin{align*} \cos{ \left( y \right) } = \sin{ \left( y \right) } = k \end{align*}.

As \displaystyle \begin{align*} \cos{ \left( y \right) } = \sin{ \left( y \right) } \end{align*} the only possible value of y in the first cycle is \displaystyle \begin{align*} \frac{\pi}{4} \end{align*}, where both the sine and cosine are \displaystyle \begin{align*} k = \frac{1}{\sqrt{2}} \end{align*}, thus

\displaystyle \begin{align*} 0 &= \sin{ \left( x \right) } \cos{ \left( \frac{\pi}{4} \right) } + \cos{ \left( x \right) } \sin{ \left( \frac{\pi}{4} \right) } \\ 0 &= \sin{ \left( x + \frac{\pi}{4} \right) } \\ n\,\pi &= x + \frac{\pi}{4} \textrm{ where } n \in \mathbf{Z} \\ x &= n\,\pi - \frac{\pi}{4} \\ x &= \frac{\left( 4\,n - 1 \right) \, \pi}{4} \end{align*}