# Thread: Trig problem

1. ## Trig problem

how do you solve this? My friend and I stare at it for a while and still couldn't figure it out. We didn't do that well in our Pre-Calc class, so please help,

solve for @

sin@ + cos@ = 1

thanks

2. $cos(\frac{\pi}{2})=0, \;\ sin(\frac{\pi}{2})=1$

So, pi/2 is a solution...but not all of them.

cos and sin have a period of 2Pi.

So, ${\theta}=2{\pi}c+\frac{\pi}{2}, \;\ or \;\ {\theta}=2{\pi}c$

Where c=0,1,2,...........

3. What did you try?

For starters, a little thought...since both sine and cosine are, at most, unity (1), we would have trouble finding a solution if one of them managed to be negative. (Why is that?) This means we need to look ONLY where both not negative.

After saying that, you can substitute from the Pythagorean Identity, $\sin^{2}(x) + \cos^{2}(x) = 1 \implies \sin(x) = \sqrt{1-\cos^{2}(x)}$. Remember, $\sin(x) \ge 0$.

Substitute into the original equation, rearrange a little, square things, and you should be able to find a quadratic equation that will result in cos(x) = something.

OR - a way that does not require so much algebra but a little more knowledge of trigonometry.

$\sin(x) + \cos(x) = \sqrt{2}\left(\sin\left(x+\frac{\pi}{4}\right)\rig ht)$

The whole thing is just quite a bit easier in this form.

4. thanks, but sorry, I looked at the problem wrongly. It is actually cos@ + sin@ = 0. But we got it after realizing that we read it wrong.
thanks anyway. stupid me

5. Oh, well,... in THAT case...It's the same. Addition is Commutative. a + b = b + a??

6. Originally Posted by Billwaa
thanks, but sorry, I looked at the problem wrongly. It is actually cos@ + sin@ = 0. But we got it after realizing that we read it wrong.
thanks anyway. stupid me
Square the equation :

0=(cos(x)+sin(x))²=cos²(x)+sin²(x)+2sin(x)cos(x)
But cos²(x)+sin²(x)=1.
And 2sin(x)cos(x)=sin(2x)

So now we have to solve for sin(2x)=-1.