How do you simplify #sin(theta+pi)/cos(theta-pi)#?

Question

Elijah Alexander · Accepted Answer

This simplifies to #tantheta#.

Use the formulas #sin(A + B) = sinAcosB + cosAsinB# and #cos(A - B) = cosAcosB + sinAsinB#. #=>(sinthetacospi + costhetasinpi)/(costhetacospi + sinthetasinpi)# #=>(sintheta(-1) + costheta(0))/(costheta(-1) + sin theta(0))# #=> (-sintheta)/(-costheta)# Use the identity that #tanbeta = sinbeta/cosbeta#. #=> tantheta# I hope this is useful!

Dylan Arnold · Answer

undefined To simplify $ \frac{\sin(\theta + \pi)}{\cos(\theta - \pi)} $, use trigonometric identities. One of the identities states that $ \sin(\alpha + \beta) = \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta) $ and $ \cos(\alpha - \beta) = \cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta) $.

Apply these identities:

$ \sin(\theta + \pi) = \sin(\theta)\cos(\pi) + \cos(\theta)\sin(\pi) $

$ \cos(\theta - \pi) = \cos(\theta)\cos(\pi) + \sin(\theta)\sin(\pi) $

Since $ \cos(\pi) = -1 $ and $ \sin(\pi) = 0 $, substitute these values:

$ \sin(\theta + \pi) = \sin(\theta)(-1) + \cos(\theta)(0) = -\sin(\theta) $

$ \cos(\theta - \pi) = \cos(\theta)(-1) + \sin(\theta)(0) = -\cos(\theta) $

Now, substitute these results into the original expression:

$ \frac{\sin(\theta + \pi)}{\cos(\theta - \pi)} = \frac{-\sin(\theta)}{-\cos(\theta)} = \frac{\sin(\theta)}{\cos(\theta)} $

Thus, $ \frac{\sin(\theta + \pi)}{\cos(\theta - \pi)} $ simplifies to $ \frac{\sin(\theta)}{\cos(\theta)} $.