# Thread: another easy trig problem...

1. ## another easy trig problem...

hello again!

ok, so here's a problem where it asks to change the equation so that "only one trigonometric ratio is present."

the equation is...

$\displaystyle sin\Theta=cos\Theta$

now, i'm only aware of one trig. identity that explains sin in terms of cos and it is called a half-angle identity.

and it looks like this...

$\displaystyle sin\dfrac{\Theta}{2}=\pm\sqrt\dfrac{1-cos\Theta}{2}$

Is this the identity i would use so that only one trig. ratio is present?

if so, could i just say...

$\displaystyle sin\Theta=\pm\sqrt\dfrac{1-cos2\Theta}{2}$

or maybe...

$\displaystyle sin\Theta=\pm2\sqrt\dfrac{1-cos\Theta}{2}$

2. Very good, but $\displaystyle \sin^{2}(x) + \cos^{2}(x) = 1$ should also be familiar and will lead to the same conclusion. Isn't is lovely that unique solutions don't care how you find them?

3. Originally Posted by jonnygill
hello again!

ok, so here's a problem where it asks to change the equation so that "only one trigonometric ratio is present."

the equation is...

$\displaystyle sin\Theta=cos\Theta$

now, i'm only aware of one trig. identity that explains sin in terms of cos and it is called a half-angle identity.

and it looks like this...

$\displaystyle sin\dfrac{\Theta}{2}=\pm\sqrt\dfrac{1-cos\Theta}{2}$

Is this the identity i would use so that only one trig. ratio is present?

if so, could i just say...

$\displaystyle sin\Theta=\pm\sqrt\dfrac{1-cos2\Theta}{2}$

or maybe...

$\displaystyle sin\Theta=\pm2\sqrt\dfrac{1-cos\Theta}{2}$
I think you are over-thinking this one. Take your original equation:
$\displaystyle \displaystyle sin( \theta ) = cos( \theta )$

What happens if you divide both sides by $\displaystyle \displaystyle cos( \theta )$?

-Dan