# Thread: Confusion over trig equations and sine/cosine

1. ## Confusion over trig equations and sine/cosine

I'm trying to make sense of a video assigned by my professor on sine and cosine equations . The problem is as follows

sin 2x + cos x = 0

In the video he says sin 2x can be changed to 2 sin x + cos x = 0. And then he says 2 sin x can be substituted with

2 sin x cos x + cos x = 0, which can then be factored by grouping.

That's where I'm confused. Why does 2 sin x magically equal 2 sin x cos x all of a sudden? What am I missing here? I can solve the equation the rest of the way following his steps, but I don't understand why that substitution can be made. Any helpful replies would be most welcome.

2. ## Re: Confusion over trig equations and sine/cosine

Originally Posted by letoatreides3508
I'm trying to make sense of a video assigned by my professor on sine and cosine equations . The problem is as follows

sin 2x + cos x = 0

In the video he says sin 2x can be changed to 2 sin x + cos x = 0. And then he says 2 sin x can be substituted with

2 sin x cos x + cos x = 0, which can then be factored by grouping.

That's where I'm confused. Why does 2 sin x magically equal 2 sin x cos x all of a sudden? What am I missing here? I can solve the equation the rest of the way following his steps, but I don't understand why that substitution can be made. Any helpful replies would be most welcome.
either the video or your interpretation of it is incorrect. It should go like this

$\sin(2x)+\cos(x)=0$

$2\sin(x)\cos(x)+\cos(x)=0$

$\cos(x)\left(2\sin(x)+1\right)=0$

Now one of those factors is equal to zero. So either

$\large \cos(x)=0 \Rightarrow x=\pi(k+\frac 1 2) ~~k\in \mathbb{Z}$ or

$\large 2\sin(x)+1=0 \Rightarrow (x=-\frac \pi 4 + 2 k \pi) \vee (x = -\frac {3 \pi} 4 + 2 k \pi)~~k\in \mathbb{Z}$

3. ## Re: Confusion over trig equations and sine/cosine

Originally Posted by letoatreides3508
I'm trying to make sense of a video assigned by my professor on sine and cosine equations . The problem is as follows

sin 2x + cos x = 0

In the video he says sin 2x can be changed to 2 sin x + cos x = 0. And then he says 2 sin x can be substituted with

2 sin x cos x + cos x = 0, which can then be factored by grouping.

That's where I'm confused. Why does 2 sin x magically equal 2 sin x cos x all of a sudden? What am I missing here? I can solve the equation the rest of the way following his steps, but I don't understand why that substitution can be made. Any helpful replies would be most welcome.
No, \displaystyle \begin{align*} \sin{(2x)} \equiv 2\sin{(x)}\cos{(x)} \end{align*}, NOT \displaystyle \begin{align*} 2\sin{(x)} \end{align*}.

4. ## Re: Confusion over trig equations and sine/cosine

Ok so why does sin(2x) break down into 2sin(x)cos(x).. Is this like a change of base kind of thing?

5. ## Re: Confusion over trig equations and sine/cosine

It's called a double angle identity for sine. Its derivation is a result of the compound angle identities, the proof of which can be found here.

\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*}

and in the case where \displaystyle \begin{align*} \alpha = \beta = x \end{align*} that gives

\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) } \\ \sin{ \left( x + x \right) } &\equiv \sin{ (x) } \cos{ (x )} + \cos{(x)} \sin{(x)} \\ \sin{(2x)} &\equiv 2\sin{(x)}\cos{(x)} \end{align*}

6. ## Re: Confusion over trig equations and sine/cosine

Thank you so much... I had another queistion if you could indulge me.

I'm solving equations using the formula pi/2 + 2k( pi), k int. When using this formula, I treat it like any other function, where k is the y value, right?