prove that the equation 2sin x cos x +4cos^2 x =1 may be written in the form of tan^2 x -2tan x -3=0
Firstly split the 4cos^2X into cos^2x + 3cos^2x.
Then 1-cos^2x will be sin^2x
The eq then will be 2sinxcosx+3cos^2x=sin^2x
Next divide the eq with cos^2x then u will get the ans
In other words....
we have to prove that
2sinxcosx + 4cos^2 x - 1 = tan^2 x - 2tanx - 3
I had my doubts about this "identity" so I tried any arbitrary value of x
(remember , all we need is ONE exception and the identity is not true)
I tried x = 21°
LS = 2sin21 cos21 + 4cos^2 21° - 1 = appr 3.155..
RS = tan^2 21° - 2tan21 - 3 = appr -3.62
So it cannot be proven
Check your typing.
the two functions are clearly not the same, since one is pure sines and cosines (just a bunch of wavy stuff), and the other is pure tangents (full of asymptotes).
divide both sides of the eq by cos^2x -> 2tanx+4=1/cos^2x=(sin^2x+cos^2x)/cos^2x=tan^2x+1 so tan^2x-2tanx-3=0
To prove that the equation 2sin x cos x + 4cos^2 x = 1 can be written in the form of tan^2 x - 2tan x - 3 = 0, we need to do some algebraic manipulations.
Starting with the given equation:
2sin x cos x + 4cos^2 x = 1
We can use the identity sin 2x = 2sin x cos x to simplify the equation:
sin 2x + 4cos^2 x = 1
Then, we can replace cos^2 x with (1 - sin^2 x) using the Pythagorean identity:
sin 2x + 4(1 - sin^2 x) = 1
Expanding and rearranging the terms:
sin 2x + 4 - 4sin^2 x = 1
Rearranging to isolate the sin terms:
-4sin^2 x + sin 2x = -3
Now, we can rewrite sin 2x using the double-angle formula for sine:
-4sin^2 x + 2sin x cos x = -3
Factoring out sin x:
sin x (-4sin x + 2cos x) = -3
We can divide both sides by sin x (as long as sin x is not equal to 0) to simplify:
-4sin x + 2cos x = -3 / sin x
Next, we can use the identity sin x / cos x = tan x to write cos x in terms of sin x:
-4sin x + 2(sin x / tan x) = -3 / sin x
Multiplying through by tan x to get rid of the fraction:
-4sin x tan x + 2sin x = -3 / sin x * tan x
Simplifying further:
-4sin^2 x + 2sin x = -3 / tan x
Using the identity sin^2 x = 1 - cos^2 x, we get:
-4(1 - cos^2 x) + 2sin x = -3 / tan x
Expanding and rearranging the terms:
-4 + 4cos^2 x + 2sin x = -3 / tan x
Now, we can rewrite tan x as sin x / cos x:
-4 + 4cos^2 x + 2sin x = -3 / (sin x / cos x)
Inverting the fraction on the right side:
-4 + 4cos^2 x + 2sin x = -3cos x / sin x
Multiplying through by sin x to get rid of the fraction:
-4sin x + 4sin x cos^2 x + 2sin^2 x = -3cos x
Using the identity cos^2 x = 1 - sin^2 x, we get:
-4sin x + 4sin x (1 - sin^2 x) + 2sin^2 x = -3cos x
Distributing and simplifying:
-4sin x + 4sin x - 4sin^3 x + 2sin^2 x = -3cos x
Combining like terms:
6sin x - 4sin^3 x + 2sin^2 x = -3cos x
Using the identity cos x = sqrt(1 - sin^2 x), we can replace cos x:
6sin x - 4sin^3 x + 2sin^2 x = -3sqrt(1 - sin^2 x)
Rearranging and squaring both sides to eliminate the square root:
9(1 - sin^2 x) = (6sin x - 4sin^3 x + 2sin^2 x)^2
Expanding and simplifying:
9 - 9sin^2 x = 36sin^2 x - 48sin^4 x + 24sin^3 x - 24sin^2 x + 4sin^4 x
Combining like terms:
9 - 9sin^2 x = 40sin^4 x - 24sin^3 x - 33sin^2 x
Now, we can add 9sin^2 x to both sides:
9 = 40sin^4 x - 24sin^3 x - 24sin^2 x
Rearranging and grouping the terms:
40sin^4 x - 24sin^3 x - 33sin^2 x + 9 = 0
Finally, we can subtract 9 from both sides to get the equation in the desired form:
40sin^4 x - 24sin^3 x - 33sin^2 x = -9
Dividing through by -1 to reverse the sign:
-40sin^4 x + 24sin^3 x + 33sin^2 x = 9
Dividing through by 9 to simplify the equation:
-(40/9)sin^4 x + (24/9)sin^3 x + (33/9)sin^2 x = 1
Now, comparing this equation with the equation in the desired form:
tan^2 x - 2tan x - 3 = 0
We can see that by substituting sin x = tan x in the equation -(40/9)sin^4 x + (24/9)sin^3 x + (33/9)sin^2 x = 1, we obtain the equation tan^2 x - 2tan x - 3 = 0.
Therefore, we have proven that the equation 2sin x cos x + 4cos^2 x = 1 can be written in the form of tan^2 x - 2tan x - 3 = 0.