# Thread: Show me where I went wrong, trig equation

1. ## Show me where I went wrong, trig equation

$cos\theta = cos(\theta+60)$

$cos\theta = cos\theta cos60 - sin\theta sin60$

$cos\theta = \frac{1}{2}cos\theta-\frac{\sqrt{3}}{2} sin\theta$

than multiply eachterm by '2', so I get;

$2cos\theta = cos\theta - \sqrt{3}sin\theta$

$2cos\theta - cos\theta = - \sqrt{3}sin\theta$

$cos\theta = - \sqrt{3}sin\theta$

$\frac{cos\theta}{sin\theta} = -\sqrt{3}$

$tan\theta = -\sqrt{3}$

my book says the answer is $tan\theta = -\frac{1}{\sqrt{3}}$

Can someone please show me what I have done wrong, cause I have been through my working quite a few times still get the same answer!

Thank you

2. Originally Posted by Tweety
$cos\theta = cos(\theta+60)$

$cos\theta = cos\theta cos60 - sin\theta sin60$

$cos\theta = \frac{1}{2}cos\theta-\frac{\sqrt{3}}{2} sin\theta$

than multiply eachterm by '2', so I get;

$2cos\theta = cos\theta - \sqrt{3}sin\theta$

$2cos\theta - cos\theta = - \sqrt{3}sin\theta$

$cos\theta = - \sqrt{3}sin\theta$

$\frac{cos\theta}{sin\theta} = -\sqrt{3}$

$tan\theta = -\sqrt{3}$

my book says the answer is $tan\theta = -\frac{1}{\sqrt{3}}$

Can someone please show me what I have done wrong, cause I have been through my working quite a few times still get the same answer!

Thank you
Only your last step is wrong

$\frac{cos(x)}{sin(x)} \neq tan(x)$

$\frac{cos(x)}{sin(x)} = cot(x) = \frac{1}{tan(x)}$

If you take the reciprocal of the answer you have it will be right

3. yes the last step cos / sin isnot tan ,, sin / cos is tan so it will be ( cos / sin ) = - square root ( 3 )
=> 1/tan = -square root (3) => tan = 1/-squr root 3
got it ?

4. AAH thanks, can't actually believe that I did not spot that myself!

I know that cos/sin does not equal tan, but was not looking at it properly.

5. it happens tweety