# another easy trig problem...

• Feb 17th 2011, 08:10 PM
jonnygill
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...

$\displaystyle sin\Theta=\pm2\sqrt\dfrac{1-cos\Theta}{2}$
• Feb 17th 2011, 08:53 PM
TKHunny
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?
• Feb 17th 2011, 08:53 PM
topsquark
Quote:

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