Question
Solve identity,
(1-sin2x)/cos2x = cos2x/(1+sin2x)
I tried starting from the right side,
RS:
=(cos²x-sin²x)/(1+2sinxcosx)
=(cos²x-(1-cos²x))/(1+2sinxcosx)
and the right side just goes in circle. May I get a hint to start off?
(1-sin2x)/cos2x = cos2x/(1+sin2x)
I tried starting from the right side,
RS:
=(cos²x-sin²x)/(1+2sinxcosx)
=(cos²x-(1-cos²x))/(1+2sinxcosx)
and the right side just goes in circle. May I get a hint to start off?
Answers
First of all since the angle is 2x throughout, let's just use y for 2x
Secondly, you probably want to prove it as an identity, rather than solve it
RS
= cosy/(1+siny)<b> [(1-siny)/(1-siny)]</b>
= cosy(1-siny)/(1- sin^2 y)
= cosy(1-siny)/cos^2y
= (1-siny)/cosy)
= (1- sin 2x)/cos 2x
= LS
Secondly, you probably want to prove it as an identity, rather than solve it
RS
= cosy/(1+siny)<b> [(1-siny)/(1-siny)]</b>
= cosy(1-siny)/(1- sin^2 y)
= cosy(1-siny)/cos^2y
= (1-siny)/cosy)
= (1- sin 2x)/cos 2x
= LS
Cross multiply.
cos^2(2x) = 1- sin^2(2x) = cos^2 2x
q.e.d.
x can be anything.
That is why it is called an identity
cos^2(2x) = 1- sin^2(2x) = cos^2 2x
q.e.d.
x can be anything.
That is why it is called an identity
Got it, thanks!
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