sabika13
How do I solve this trig identity: cos2x/(1+sinx^2)=tan(pie/4 -x)
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matricked
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i think ur question is cos2x/(1+sin2x)
now cos2x =cos^2x -sin^2x=(cosx+sinx)(cosx-sinx)
and 1+sin2x = cos^2x + sin^2x +2sinxcosx =(cosx+sinx)^2
thus cos2x/(1+sinx)^2= (cosx+sinx)(cosx-sinx)/(cosx+sinx)^2
=(cosx- sinx)/((cosx+sinx)
divide both num and denom by cosx
and then we have
=(1-sinx/cosx)/(1+sinx/cosx)
=(1-tanx)/(1+tanx) = (tan(pi/4) -tanx)/(1+tan(pi/4)tanx)=tan(pie/4 -x)
sabika13
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for the second last step.. where did the square roots go? isnt it suppose to be: (1-sinx/cosx)(1-sinx/cos)/(1+sinx/cos)(1+sinx/cosx)
matricked
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actually the factors (cosx+sinx) are there both in num and denom hence they cancel each other
sabika13
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but its (cosx-sinx) in the num :s
matricked
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(cosx+sinx)(cosx-sinx) / (cosx+sinx)^2 =(cosx-sinx) / (cosx+sinx)
matricked
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i hope its clear now
matricked
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why 13 u hv included in ur id ? if u don't mind
sabika13
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if its (cosx+sinx)^2, it has to equal (cosx+sinx)(cosx+sinx)
sabika13
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idk random.
matricked
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(cosx+sinx)^2= (cosx+sinx)(cosx+sinx) (correct )
that is why one of the (cosx+sinx)
matricked
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cancel each other
sabika13
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how does cos^2x -sin^2x=(cosx+sinx)(cosx-sinx)
matricked
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USE ..A*A -B*B =(A)^2 -(B)^2 = (A+B)(A-B)...
sabika13
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okay, then why isnt :(cosx+sinx)^2 --> (cosx-sinx)(cosx+sinx)
matricked
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bcoz a^2 =a*a
matricked
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so (a+b)^2 =(a+b)(a+b)
matricked
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but if u multiply (a+b)(a-b) u get (a^2 -b^2)
sabika13
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so (a^2-b^2) does not equal (a-b)^2 ?
matricked
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sahi hai..
sabika13
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oh i get it..
sabika13
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=(1-tanx)/(1+tanx) = (tan(pi/4) -tanx)/(1+tan(pi/4)tanx)=tan(pie/4 -x) can u explain that? .. where did pi/4 come from ?
Dido525
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Tan (pi/4) = 1.
sabika13
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is that a trig identity?
Dido525
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No. If you do that on a calculator you get that.
Dido525
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Tan(pi/4) is always 1. Look on a unit circle.
sabika13
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Okay, thanks!!
Dido525
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Welcome :) .