Sin^2(x) + tan^2(x) = √2

  • #1
47
6

Homework Statement


The problem given is sin2(x) + tan2(x) = √2

2. Homework Equations


The relevant equations would be any trigonometric identities

The Attempt at a Solution



sin2(x) + tan2(x) = √2

sin2(x) + (sin2(x)/cos2(x) ) = √2

[ cos2(x) sin2(x) + sin2(x) ]/ cos2(x) = √2

[ (1- sin2(x)) sin2(x) + sin2(x)] / [ 1 - sin2(x) ] = √2

[ (2 - sin2(x) ) sin2(x) ] / [ 1 - sin2(x) ] = √2

(2 - sin2(x) ) tan2(x) = √2

tan2(x) = ( √2 / [ 2 - sin2(x) ] )

I take this and substitute into the first equation:

sin2(x) + ( √2 / [ 2 - sin2(x) ] ) = √2

( 2 - sin2(x) ) sin2(x) / [ 2 - sin2(x) ] + √2 / [ 2 - sin2(x) ] = √2

( 2 - sin2(x) ) sin2(x) + √2 / [ 2 - sin2(x) ] = √2

( 2 - sin2(x) ) sin2(x) + √2 = ( √2 ) 2 - sin2(x)

( 2 - sin2(x) ) sin2(x) + √2 = 2√2 - √2 sin2(x)

( 2 - sin2(x) ) sin2(x) = √2 - √2 sin2(x)

( 2 - sin2(x) ) sin2(x) = √2 (1 - sin2(x) )

( ( 2 - sin2(x) ) sin2(x) / (1 - sin2(x) ) )= √2


Here is where I get stuck. I do not know what steps to take next. Please give me hints on this and do not hesitate to point out any mistakes in my work. They are very likely.
 

Answers and Replies

  • #2
14,386
11,697

Homework Statement


The problem given is sin2(x) + tan2(x) = √2

2. Homework Equations


The relevant equations would be any trigonometric identities

The Attempt at a Solution



sin2(x) + tan2(x) = √2

sin2(x) + (sin2(x)/cos2(x) ) = √2

[ cos2(x) sin2(x) + sin2(x) ]/ cos2(x) = √2

[ (1- sin2(x)) sin2(x) + sin2(x)] / [ 1 - sin2(x) ] = √2

[ (2 - sin2(x) ) sin2(x) ] / [ 1 - sin2(x) ] = √2
I can follow you until here. What's next is a backward substitution which I don't think will get you very far. At least the next steps are what could be done more easily, because you already have a quadratic equation in ##t := \sin^2(x)## which can be solved.
(2 - sin2(x) ) tan2(x) = √2

tan2(x) = ( √2 / [ 2 - sin2(x) ] )

I take this and substitute into the first equation:

sin2(x) + ( √2 / [ 2 - sin2(x) ] ) = √2

( 2 - sin2(x) ) sin2(x) / [ 2 - sin2(x) ] + √2 / [ 2 - sin2(x) ] = √2

( 2 - sin2(x) ) sin2(x) + √2 / [ 2 - sin2(x) ] = √2

( 2 - sin2(x) ) sin2(x) + √2 = ( √2 ) 2 - sin2(x)

( 2 - sin2(x) ) sin2(x) + √2 = 2√2 - √2 sin2(x)

( 2 - sin2(x) ) sin2(x) = √2 - √2 sin2(x)

( 2 - sin2(x) ) sin2(x) = √2 (1 - sin2(x) )

( ( 2 - sin2(x) ) sin2(x) / (1 - sin2(x) ) )= √2


Here is where I get stuck. I do not know what steps to take next. Please give me hints on this and do not hesitate to point out any mistakes in my work. They are very likely.
 
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Likes Charles Link, ForceBoy, SunThief and 1 other person
  • #3
47
6
Wow, I didn't see that! Thank you very much. This was really helpful.
 

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