+0  
 
0
1
355
2
avatar+91 

How does one verify \(\frac{2tanx}{tan2x}=2-{sec}^{2}x\)

 Nov 12, 2017
 #1
avatar
+2

This is long and detailed answer courtesy of "Mathematica 11 Home Edition"

 

Verify the following identity:
2 (tan(x))/(tan(2 x)) = 2 - sec(x)^2

Multiply both sides by tan(2 x):
2 tan(x) = ^?tan(2 x) (2 - sec(x)^2)

 

Write secant as 1/cosine and tangent as sine/cosine:
2 (sin(x))/(cos(x)) = ^?(2 - 1/(cos(x)) ^2 ) (sin(2 x))/(cos(2 x))

(2 - (1/(cos(x)))^2) ((sin(2 x))/(cos(2 x))) = ((2 - cos(x)^(-2)) sin(2 x))/(cos(2 x)):
(2 sin(x))/(cos(x)) = ^?(sin(2 x) (2 - 1/cos(x)^2))/(cos(2 x))

 

Put 2 - 1/cos(x)^2 over the common denominator cos(x)^2: 2 - 1/cos(x)^2 = (2 cos(x)^2 - 1)/cos(x)^2:
(2 sin(x))/(cos(x)) = ^?((2 cos(x)^2 - 1)/cos(x)^2 sin(2 x))/(cos(2 x))

Cross multiply:
2 cos(x)^2 cos(2 x) sin(x) = ^?cos(x) sin(2 x) (2 cos(x)^2 - 1

 

Divide both sides by cos(x):
2 cos(x) cos(2 x) sin(x) = ^?sin(2 x) (2 cos(x)^2 - 1)

cos(2 x) = 1 - 2 sin(x)^2:


2 cos(x) 1 - 2 sin(x)^2 sin(x) = ^?sin(2 x) (2 cos(x)^2 - 1)

2 cos(x) (1 - 2 sin(x)^2) sin(x) = 2 cos(x) sin(x) - 4 cos(x) sin(x)^3:
2 cos(x) sin(x) - 4 cos(x) sin(x)^3 = ^?sin(2 x) (2 cos(x)^2 - 1)

cos(x)^2 = 1 - sin(x)^2:


2 cos(x) sin(x) - 4 cos(x) sin(x)^3 = ^?sin(2 x) (2 1 - sin(x)^2 - 1)

2 (1 - sin(x)^2) = 2 - 2 sin(x)^2:
2 cos(x) sin(x) - 4 cos(x) sin(x)^3 = ^?sin(2 x) (2 - 2 sin(x)^2 - 1)

-1 + 2 - 2 sin(x)^2 = 1 - 2 sin(x)^2:


2 cos(x) sin(x) - 4 cos(x) sin(x)^3 = ^?1 - 2 sin(x)^2 sin(2 x)

sin(2 x) = 2 sin(x) cos(x):


2 cos(x) sin(x) - 4 cos(x) sin(x)^3 = ^?2 cos(x) sin(x) (1 - 2 sin(x)^2)

(1 - 2 sin(x)^2)×2 cos(x) sin(x) = 2 cos(x) sin(x) - 4 cos(x) sin(x)^3:
2 cos(x) sin(x) - 4 cos(x) sin(x)^3 = ^?2 cos(x) sin(x) - 4 cos(x) sin(x)^3

The left hand side and right hand side are identical:
 

 Nov 12, 2017
 #2
avatar+91 
+1

Thank you!

The spell that would turn you into kelp has been reversed!

dom6547  Nov 12, 2017

21 Online Users

avatar
avatar
avatar
avatar