My Math Forum Help me with a very difficult question
 User Name Remember Me? Password

 Trigonometry Trigonometry Math Forum

 March 28th, 2017, 09:53 AM #1 Newbie   Joined: Mar 2017 From: Azerbaijan Posts: 1 Thanks: 0 Help me with a very difficult question (tan15°)×(tan25°)×(tan35°)×(tan85°) I know the answer, but I do not know the solution answer: 1 help me please. Last edited by skipjack; April 24th, 2017 at 08:56 PM.
 April 24th, 2017, 04:26 PM #2 Banned Camp   Joined: Apr 2017 From: durban Posts: 22 Thanks: 0 Math Focus: Algebra * Joined: Mar 2017 From: Azerbaijan Posts: 1 Thanks: 0 Help me with a very difficult question (tan15°)×(tan25°)×(tan35°)×(tan85°) I know the answer but I do not know the solution answer: 1 help me please. It is very possible to get the value of 15 From special angle 45 - 30 and when solve will produce 2 - root 3 but this process will nt work for 25, 35, and 85 Last edited by skipjack; April 24th, 2017 at 08:57 PM.
 April 24th, 2017, 04:32 PM #3 Banned Camp   Joined: Apr 2017 From: durban Posts: 22 Thanks: 0 Math Focus: Algebra Challenging Trigonometry Problems!? 1. If tan 15 tan 25 tan 35 = tan t, and t is between 0 and 180, find t. (All angle measures are in degrees.) 2. Find the number of solutions, in degrees, to the equation sin^10 x + cos^10 x = (29/16)cos^4 2x, where x is between 0 and 2007 degrees, inclusive. 3. Find both the maximum and minimum values of a) f(x) = sin x + cos x + tan x + csc x + sec x + cot x b) |f(x)| = |sin x + cos x + tan x + csc x + sec x + cot x|
 April 24th, 2017, 10:04 PM #4 Global Moderator   Joined: Dec 2006 Posts: 18,962 Thanks: 1606 \displaystyle \begin{align*}\tan15^\circ\tan25^\circ\tan35^\circ \tan85^\circ &= \frac{8\sin15^\circ\sin25^\circ\sin35^\circ\sin85^ \circ}{8\cos15^\circ\cos25^\circ\cos35^\circ\cos85 ^\circ} \\ &= \frac{(\cos10^\circ - \cos40^\circ)(2\cos50^\circ + 1)}{(\cos40^\circ + \cos10^\circ)(2\cos50^\circ - 1)}\\ &= \frac{2\cos10^\circ\cos50^\circ + \cos10^\circ - 2\cos40^\circ\cos50^\circ - \cos40^\circ}{2\cos40^\circ\cos50^\circ - \cos40^\circ + 2\cos10^\circ\cos50^\circ - \cos10^\circ} \\ &= \frac{2\cos10^\circ\cos50^\circ + \cos10^\circ - \cos10^\circ - \cos40^\circ}{\cos10^\circ - \cos40^\circ + 2\cos10^\circ\cos50^\circ - \cos10^\circ} \\ &= 1\end{align*} Thanks from Maschke
 April 25th, 2017, 05:32 AM #5 Banned Camp   Joined: Apr 2017 From: durban Posts: 22 Thanks: 0 Math Focus: Algebra That right skipjack. Dnt look at it from that corner

 Tags difficult, question

Search tags for this page

### sin^10x cos^10x=29/16 cos^4 2x

Click on a term to search for related topics.
 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post huy Math 25 November 27th, 2017 05:59 PM student777 Math 1 January 11th, 2017 11:52 AM Iamthatdude Math Events 1 August 28th, 2013 12:13 PM ahmed-ar Calculus 3 December 9th, 2011 12:44 PM DragonScion Advanced Statistics 0 October 29th, 2011 02:54 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top

Copyright © 2018 My Math Forum. All rights reserved.