How do you prove #tan^2(1/2theta)=(tantheta-sintheta)/(tantheta+sintheta)#?

Question

Madelyn Adams · Accepted Answer

see below

#tan^2(1/2 theta)=(tan theta-sin theta)/(tan theta + sin theta)# Right Side #=(tan theta-sin theta)/(tan theta + sin theta)# #=(sin theta / cos theta -sin theta)/(sin theta/cos theta + sin theta)# #=((sin theta-sin theta cos theta) / cos theta) / ((sin theta+sin theta cos theta)/cos theta)# #=(sin theta-sin theta cos theta) / cos theta * cos theta/(sin theta+sin theta cos theta)# #=(sin theta-sin theta cos theta)/(sin theta+sin theta cos theta)# #=(sin theta(1-cos theta))/(sin theta(1+cos theta))# #=(1-cos theta)/(1+cos theta)# #=(tan (1/2 theta))^2# #=tan^2(1/2 theta)# #=# Left Side

Emery Adkins · Answer

undefined To prove the identity $ \tan^2\left(\frac{1}{2}\theta\right) = \frac{\tan\theta - \sin\theta}{\tan\theta + \sin\theta} $, we'll use trigonometric identities and properties. The key identities we'll use involve the half-angle and double-angle formulas for tangent and sine, as well as the fundamental identity $ \sin^2\theta + \cos^2\theta = 1 $.

**Step 1: Start with the right side of the identity.**

Given:
$$ \frac{\tan\theta - \sin\theta}{\tan\theta + \sin\theta} $$

**Step 2: Use the definition of $ \tan\theta = \frac{\sin\theta}{\cos\theta} $.**

This transforms the expression into:
$$ \frac{\frac{\sin\theta}{\cos\theta} - \sin\theta}{\frac{\sin\theta}{\cos\theta} + \sin\theta} $$

To simplify, get a common denominator in both the numerator and the denominator:
$$ \frac{\frac{\sin\theta - \sin\theta\cos\theta}{\cos\theta}}{\frac{\sin\theta + \sin\theta\cos\theta}{\cos\theta}} $$

Simplifying the fractions by multiplying both numerator and denominator by $\cos\theta$, we get:
$$ \frac{\sin\theta(1 - \cos\theta)}{\sin\theta(1 + \cos\theta)} $$

Since $ \sin\theta $ is in both the numerator and the denominator, it can be canceled out (assuming $ \theta $ is not at an angle where $ \sin\theta = 0 $), leading to:
$$ \frac{1 - \cos\theta}{1 + \cos\theta} $$

**Step 3: Apply the half-angle identity for tangent.**

The half-angle identity for $ \tan $ is given by:
$$ \tan^2\left(\frac{\theta}{2}\right) = \frac{1 - \cos\theta}{1 + \cos\theta} $$

This matches our derived expression exactly.

Thus, we have shown that:
$$ \tan^2\left(\frac{1}{2}\theta\right) = \frac{\tan\theta - \sin\theta}{\tan\theta + \sin\theta} $$

This concludes the proof. Note that this identity and the steps taken assume $ \theta $ is within a range that prevents division by zero and undefined expressions.