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 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: 17,201 Thanks: 1291 $\displaystyle 4^{\large\sin^2\!x}\! + 4^{\large\cos^2\!x}\! = \left(2^{\large\sin^2\!x}\! - 2^{\large\cos^2\!x}\right)^2\! + 4$ Thanks from topsquark, Country Boy and gab17
 November 17th, 2016, 09:21 AM #3 Math Team   Joined: Jan 2015 From: Alabama Posts: 2,489 Thanks: 630 $\displaystyle 4^{\sin^2(x)}+ 4^{\cos^2(x)}= 4^{\sin^2(x)}+ 4^{1- \sin^2(x)}= 4^{\sin^2(x)}+ 4/4^{\sin^2(x)}= 4$. 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 $\displaystyle \sin(x)= 1/2$. Thanks from topsquark Last edited by skipjack; November 17th, 2016 at 11:57 AM.
November 17th, 2016, 11:49 AM   #4
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Math Focus: Wibbly wobbly timey-wimey stuff.
Quote:
 Originally Posted by gab17 $\displaystyle 4^{\large\sin^2\!x}\! + 4^{\large\cos^2\!x}\! = 4$
You keep posting vaguely similar questions but there seems to be no rhyme or reason to them. What are you trying to understand? Or are you just curious?

-Dan

 November 17th, 2016, 12:01 PM #5 Global Moderator   Joined: Dec 2006 Posts: 17,201 Thanks: 1291 Country Boy slipped up.
 November 17th, 2016, 01:36 PM #6 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,489 Thanks: 888 Math Focus: Elementary mathematics and beyond $$4^{\sin^2x}+4^{\cos^2x}=4$$ $$4^{(1-\cos2x)/2}+4^{(1+\cos2x)/2}=4$$ $$p=\cos2x$$ $$2^{-p}+2^p=2$$ $$1+2^{2p}=2\cdot2^p$$ $$(2^p-1)^2=0$$ Thanks from topsquark
November 17th, 2016, 04:01 PM   #7
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Quote:
 Originally Posted by Country Boy $\displaystyle 4^{\sin^2(x)}+ 4^{\cos^2(x)}= 4^{\sin^2(x)}+ 4^{1- \sin^2(x)}= 4^{\sin^2(x)}+ 4/4^{\sin^2(x)}= 4$. 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 $\displaystyle \sin(x)= 1/2$.

$\displaystyle \color{red}{y=4^{\sin^2x}}$

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: 17,201 Thanks: 1291 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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