November 17th, 2016, 04:00 AM  #1 
Newbie Joined: Nov 2016 From: Hungary Posts: 3 Thanks: 0  Trigonometric 2.0
$\displaystyle 4^{\large\sin^2\!x}\! + 4^{\large\cos^2\!x}\! = 4$
Last edited by skipjack; November 17th, 2016 at 04:57 AM. Reason: to make it make sense 
November 17th, 2016, 04:55 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 16,194 Thanks: 1147 
$\displaystyle 4^{\large\sin^2\!x}\! + 4^{\large\cos^2\!x}\! = \left(2^{\large\sin^2\!x}\!  2^{\large\cos^2\!x}\right)^2\! + 4$

November 17th, 2016, 09:21 AM  #3 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,211 Thanks: 555 
$\displaystyle 4^{\sin^2(x)}+ 4^{\cos^2(x)}= 4^{\sin^2(x)}+ 4^{1 \sin^2(x)}= 4^{\sin^2(x)}+ 4/4^{\sin^2(x)}= 4$. Letting $\displaystyle y= 4^{\sin(x)}$, this is $\displaystyle y+ 4/y= 4$. Multiplying on both sides by y, $\displaystyle y^2+ 4= 4y$ or $\displaystyle y^2 4y+ 4= (y 2)^2= 0$. $\displaystyle y= 4^{\sin(x)}= 2$ so $\displaystyle \sin(x)= 1/2$.
Last edited by skipjack; November 17th, 2016 at 11:57 AM. 
November 17th, 2016, 11:49 AM  #4 
Math Team Joined: May 2013 From: The Astral plane Posts: 1,437 Thanks: 528 Math Focus: Wibbly wobbly timeywimey stuff.  
November 17th, 2016, 12:01 PM  #5 
Global Moderator Joined: Dec 2006 Posts: 16,194 Thanks: 1147 
Country Boy slipped up.

November 17th, 2016, 01:36 PM  #6 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,363 Thanks: 826 Math Focus: Elementary mathematics and beyond 
$$4^{\sin^2x}+4^{\cos^2x}=4$$ $$4^{(1\cos2x)/2}+4^{(1+\cos2x)/2}=4$$ $$p=\cos2x$$ $$2^{p}+2^p=2$$ $$1+2^{2p}=2\cdot2^p$$ $$(2^p1)^2=0$$ 
November 17th, 2016, 04:01 PM  #7  
Senior Member Joined: Apr 2014 From: Europa Posts: 567 Thanks: 174  Quote:
$\displaystyle \color{red}{y=4^{\sin^2x}}$ Last edited by skipjack; November 23rd, 2016 at 08:22 PM.  
November 23rd, 2016, 06:49 PM  #8 
Newbie Joined: Nov 2016 From: Canada Posts: 13 Thanks: 3 Math Focus: All 
Answer is n$\pi$ + (1)$^{\large\text{n}}$ * $\pi$ / 4, for all n Last edited by skipjack; November 23rd, 2016 at 08:29 PM. 
November 23rd, 2016, 08:31 PM  #9 
Global Moderator Joined: Dec 2006 Posts: 16,194 Thanks: 1147 
No, that formula gives only some of the answers.

November 24th, 2016, 07:42 AM  #10 
Newbie Joined: Nov 2016 From: Canada Posts: 13 Thanks: 3 Math Focus: All 
Answer is n*PI + (1)power of n*(PI/4), for all n belongs to Z and n*PI + (1)power of n*(PI/4), for all n belongs to Z Last edited by Impel Tutors; November 24th, 2016 at 07:47 AM. 

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