1. ## Double angle equations

Hello all! I just registered and have a test over double angle and trigonometric addition formulas on Tuesday, but there are a few problems I can't figure out.

1. $\displaystyle sin4x=-2sin2x$

2. $\displaystyle (sin2x+cos2x)^2=0$

3. $\displaystyle 2sin2x=-tan2x$

Thanks in advance for any and all help!!!

2. Originally Posted by pilotguy
Hello all! I just registered and have a test over double angle and trigonometric addition formulas on Tuesday, but there are a few problems I can't figure out.
1. $\displaystyle \sin(4x)=-2\sin(2x)$

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

$\displaystyle 2\sin(2x)[\cos(2x) + 1] = 0$

set each factor equal to 0 and solve.

2. $\displaystyle [\sin(2x)+\cos(2x)]^2=0$

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

$\displaystyle \sin(2x) = -\cos(2x)$

for what angles do cosine and sine have opposite values?

3. $\displaystyle 2\sin(2x)=-\tan(2x)$

$\displaystyle 2\sin(2x) + \tan(2x) = 0$

$\displaystyle \sin(2x)[2 + \sec(2x)] = 0$

set each factor equal to 0 and solve.

3. Originally Posted by skeeter
1. $\displaystyle \sin(4x)=-2\sin(2x)$

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

$\displaystyle 2\sin(2x)[\cos(2x) + 1] = 0$

set each factor equal to 0 and solve.

2. $\displaystyle [\sin(2x)+\cos(2x)]^2=0$

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

$\displaystyle \sin(2x) = -\cos(2x)$

for what angles do cosine and sine have opposite values?

3. $\displaystyle 2\sin(2x)=-\tan(2x)$

$\displaystyle 2\sin(2x) + \tan(2x) = 0$

$\displaystyle \sin(2x)[2 + \sec(2x)] = 0$

set each factor equal to 0 and solve.
Thanks! But I'm still a little confused on 2. I don't really know off the top of my head that sine and cosine have opposite values at $\displaystyle 3\pi/4$ and $\displaystyle 7\pi/4$

Is there another way to solve it?

4. Originally Posted by pilotguy
Thanks! But I'm still a little confused on 2. I don't really know off the top of my head that sine and cosine have opposite values at $\displaystyle 3\pi/4$ and $\displaystyle 7\pi/4$

Is there another way to solve it?
another way to do it ...

$\displaystyle \sin(u) = -\cos(u)$

$\displaystyle \frac{\sin(u)}{\cos(u)} = -1$

$\displaystyle \tan(u) = -1$

or ...

$\displaystyle \sin(u) = -\cos(u)$

$\displaystyle \sin^2(u) = \cos^2(u)$

$\displaystyle \sin^2(u) = 1 - \sin^2(u)$

$\displaystyle 2\sin^2(u) - 1 = 0$

$\displaystyle \sin^2(u) = \frac{1}{2}$

$\displaystyle \sin(u) = \pm \frac{1}{\sqrt{2}}$

and you still need to know where cosine and sine have opposite values ... so, yes, you really need to know that info off the top of your head.

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