At Westonci.ca, we make it easy to get the answers you need from a community of informed and experienced contributors. Discover in-depth solutions to your questions from a wide range of experts on our user-friendly Q&A platform. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.

help again...its for math offering 15 points

Help Againits For Math Offering 15 Points class=

Sagot :

Answer:

[tex]\tan \dfrac{\theta}2 = \dfrac{1}2[/tex]

Step by step Explanation:

[tex]\sin \theta = \dfrac{\text{Perpendicular} }{\text{Hypotenuse}} = \dfrac{12}{15}\\\\\\\cos \theta = \dfrac{\text{Base}}{\text{Hypotenuse}}= \dfrac{9}{15}\\\\\text{Now,}\\\\\tan \dfrac{\theta}2 = \dfrac{\sin \tfrac{\theta}2}{\cos \tfrac{\theta}2}\\\\\\~~~~~~~~=\dfrac{2\cos \tfrac{\theta}2 \sin \tfrac{\theta}2 }{2\cos^2 \tfrac{\theta}2}~~~~~~;\left[\text{Multiply by}~ 2\cos\tfrac{\theta}2 \right][/tex]

         [tex]=\dfrac{\sin \theta}{1+ \cos \theta}~~~~~~~~~~~~;[2 \sin x \cos x = \sin 2x ~ \text{and}~ 2\cos^2 x =1+\cos 2x]\\\\\\=\dfrac{\tfrac{12}{15}}{1+ \tfrac{9}{15}}\\\\\\=\dfrac{\tfrac{12}{15}}{\tfrac{24}{15}}\\\\\\=\dfrac{12}{15}\times \dfrac{15}{24}\\\\\\=\dfrac{12}{24}\\\\\\=\dfrac{1}2[/tex]