1this is easy dude...
first because of mistype i was not able to solve :P
sin4x+4cos2x
= sin4x+4(1-sin2x)
= (sin2x-2)2
square root will give sin2x-2
same way other...
so final answer is sin2x-cos2x = -cos2x
i think u mistype again.!
√sin4x+4cos2x-√4sin2x+cos4x = cos2x
*mistake in typing.. corrected!
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12 Answers
1
62no buddy.. u have a small mistake.. u correct it :)
almost correct.. but not yet!
1finally its
-cos2x=cos2x
=> 2cos2x=0
=> cos2x=0
=> 2x= (2n±1)π/2
=> x=(2n±1)π/4.
(correct me,if wrong.)
62no sky :)
u missed it.. actually there is a small error in kumar's solutin... u have to find it :)
U mistook what i meant :)
There is a mistake in kumar's solution.. a very slight one!
62the mistake is essentially in taking a square root.. in the solution of kumar...
It will be great if kumar or some other person could rectify it :)
11= (sin2x-2)2
square root will give sin2x-2
Here is the mistake...
square root will give 2- sin2x and not
sin2x - 2
Isnt this the mistake...
So the relation that we have to prove is proved
11Solution
√Sin4x+4cos2x - √cos4x+4sin2x
√Sin4x + 4 - 4sin2x - √cos4x + 4 - 4 cos2x
√(2-sin2)2 - √(2-cos2)2
-Sin2x + 2 + cos2x - 2
-Sin2x + cos2x
= Cos2A