A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing
 3 years ago
(sinasin2a+sin3asin6a+sin4asin13a)/(sinacos2a+sin3asin6a+sin4acos13a)= tan9a
 3 years ago
(sinasin2a+sin3asin6a+sin4asin13a)/(sinacos2a+sin3asin6a+sin4acos13a)= tan9a

This Question is Closed

kushashwa23
 3 years ago
Best ResponseYou've already chosen the best response.2post it through equation forum please

lalaly
 3 years ago
Best ResponseYou've already chosen the best response.3\[sinAsin2A = \frac{1}{2}(\cos A  \cos 3A)\] \[\sin3Asin6A = \frac{1}{2}(\cos3A  \cos9A)\] \[\sin4Asin13A = \frac{1}{2}(\cos9A  \cos17A)\] \[sinAcos2A = \frac{1}{2}(sinA+\sin3A) \]\[\sin3Acos6A = \frac{1}{2}(\sin3A+\sin9A)\] \[\sin4Acos13A = \frac{1}{2}(\sin9A+\sin17A)\] \[(sinAsin2A + \sin3Asin6A + \sin4Asin13A) / (sinAcos2A + \sin3Acos6A + \sin4Acos13A) * 2 / 2\] \[= (\cos A  \cos 3A + \cos3A  \cos9A + \cos9A  \cos17A) / (sinA+\sin3A \sin3A+\sin9A \sin9A+\sin17A)\] \[= (\cos A  \cos 17A) / (\sin 17A\sin A)\] \[= 2 \sin 9A \sin 8A / 2 \cos 9A \sin 8A\] (cancel 2 and sin 8A) \[= \sin 9A / \cos 9A\] \[= \tan 9A\]
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.