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November 17th, 2016, 04:00 AM   #1
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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
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November 17th, 2016, 04:55 AM   #2
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$\displaystyle 4^{\large\sin^2\!x}\! + 4^{\large\cos^2\!x}\! = \left(2^{\large\sin^2\!x}\! - 2^{\large\cos^2\!x}\right)^2\! + 4$
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November 17th, 2016, 09:21 AM   #3
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$\displaystyle 4^{\sin^2(x)}+ 4^{\cos^2(x)}= 4^{\sin^2(x)}+ 4^{1- \sin^2(x)}= 4^{\sin^2(x)}+ 4/4^{\sin^2(x)}= 4$. 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 $\displaystyle \sin(x)= 1/2$.
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Last edited by skipjack; November 17th, 2016 at 11:57 AM.
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November 17th, 2016, 11:49 AM   #4
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Quote:
Originally Posted by gab17 View Post
$\displaystyle 4^{\large\sin^2\!x}\! + 4^{\large\cos^2\!x}\! = 4$
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
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November 17th, 2016, 12:01 PM   #5
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Country Boy slipped up.
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November 17th, 2016, 01:36 PM   #6
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$$4^{\sin^2x}+4^{\cos^2x}=4$$

$$4^{(1-\cos2x)/2}+4^{(1+\cos2x)/2}=4$$

$$p=\cos2x$$

$$2^{-p}+2^p=2$$

$$1+2^{2p}=2\cdot2^p$$

$$(2^p-1)^2=0$$
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November 17th, 2016, 04:01 PM   #7
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Quote:
Originally Posted by Country Boy View Post
$\displaystyle 4^{\sin^2(x)}+ 4^{\cos^2(x)}= 4^{\sin^2(x)}+ 4^{1- \sin^2(x)}= 4^{\sin^2(x)}+ 4/4^{\sin^2(x)}= 4$. 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.
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November 23rd, 2016, 06:49 PM   #8
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Answer is

n$\pi$ + (-1)$^{\large\text{n}}$ * $\pi$ / 4, for all n

Last edited by skipjack; November 23rd, 2016 at 08:29 PM.
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November 23rd, 2016, 08:31 PM   #9
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No, that formula gives only some of the answers.
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November 24th, 2016, 07:42 AM   #10
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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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