Having trouble verifying this identity...
cscxtanx + secx= 2cosx
This is what I've been trying
(1/sinx)(sinx/cosx) + secx = 2cosx
(1/cosx)+(1/cosx) = 2 cosx
2 secx = 2cosx
that's what i ended up with, but i know it's now right. did i mess up with the algebra?
Looks good. Lets try it with x=45deg
sqrt2*1+sqrt2=
2sqrt2 which is not equal to 2cos45
Thanks!
To verify the given identity csc(x)tan(x) + sec(x) = 2cos(x), let's simplify the left-hand side (LHS) of the equation using trigonometric identities.
Start with the provided expression:
csc(x)tan(x) + sec(x)
Now, let's express each term in terms of sine and cosine functions:
csc(x)tan(x) = (1/sin(x))(sin(x)/cos(x)) = sin(x)/cos(x)
sec(x) = 1/cos(x)
Substitute these expressions back into the original equation:
sin(x)/cos(x) + 1/cos(x)
Now, we have a common denominator, which is cos(x). Let's combine the terms:
(sin(x) + 1)/cos(x)
Now, the expression on the LHS is (sin(x) + 1)/cos(x).
To simplify this expression further, we can use the Pythagorean identity:
sin^2(x) + cos^2(x) = 1
Rearrange this equation:
sin^2(x) = 1 - cos^2(x)
Divide both sides by cos^2(x):
sin^2(x)/cos^2(x) = (1 - cos^2(x))/cos^2(x)
Since sin(x)/cos(x) = tan(x), we can substitute this identity into the equation:
tan^2(x) = (1 - cos^2(x))/cos^2(x)
Now, let's simplify the right-hand side (RHS) expression:
2cos(x)
Now, we have an equation in terms of tan(x) and cos(x). To verify if the equation is true, we can simplify both sides of the equation and see if they are equal.
Square both sides of the equation:
tan^2(x) = 4cos^2(x)
Using the Pythagorean identity sin^2(x) + cos^2(x) = 1, substitute sin^2(x) = 1 - cos^2(x):
(1 - cos^2(x))/cos^2(x) = 4cos^2(x)
Multiply both sides by cos^2(x):
1 - cos^2(x) = 4cos^4(x)
Rearrange the equation:
4cos^4(x) + cos^2(x) - 1 = 0
This is now a quadratic equation in terms of cos(x). Simplify further:
(2cos^2(x) - 1)(2cos^2(x) + 1) = 0
Now, let's find when the equation is true by solving for cos(x):
2cos^2(x) - 1 = 0 → cos^2(x) = 1/2 → cos(x) = ±√(1/2) → cos(x) = ±(1/√2)
2cos^2(x) + 1 = 0 → cos^2(x) = -1/2 → (no real solutions)
Therefore, cos(x) = 1/√2 or cos(x) = -1/√2.
These are the solutions for cos(x). You can check these values by substituting them back into the original equation and verifying if LHS is equal to RHS.
So, to answer your question, your original statement that 2sec(x) = 2cos(x) is incorrect. There was an error while simplifying the expression.