Show that 1-[cos2A / cos^2A] = tan^2A
R Raul New member Joined Nov 27, 2006 Messages 1 Nov 27, 2006 #1 Show that 1-[cos2A / cos^2A] = tan^2A
M Mrspi Senior Member Joined Dec 17, 2005 Messages 2,128 Nov 27, 2006 #2 Re: confused Raul said: Show that 1-[cos2A / cos^2A] = tan^2A Click to expand... From the "double-angle" identities, we see that cos 2A = cos^2 A - sin^2 A Substitute into the left-hand side: 1 - [ (cos^2 A - sin^2 A) / cos^2 A] Write 1 as cos^2 A / cos^2 A: (cos^2 A / cos^2 A) - [ (cos^2 A - sin^2 A) / cos^2 A ] Now, the two fractions have the same denominator. Subtract the numerators, and place the result over the common denominator: cos^2 A - (cos^2 A - sin^2 A) ---------------------------------- xxxxxxxxcos^2 A Ok...you take it from here. I hope this helps you.
Re: confused Raul said: Show that 1-[cos2A / cos^2A] = tan^2A Click to expand... From the "double-angle" identities, we see that cos 2A = cos^2 A - sin^2 A Substitute into the left-hand side: 1 - [ (cos^2 A - sin^2 A) / cos^2 A] Write 1 as cos^2 A / cos^2 A: (cos^2 A / cos^2 A) - [ (cos^2 A - sin^2 A) / cos^2 A ] Now, the two fractions have the same denominator. Subtract the numerators, and place the result over the common denominator: cos^2 A - (cos^2 A - sin^2 A) ---------------------------------- xxxxxxxxcos^2 A Ok...you take it from here. I hope this helps you.
S soroban Elite Member Joined Jan 28, 2005 Messages 5,586 Nov 27, 2006 #3 Hello, Raul! Another approach . . . Show that: \(\displaystyle \L\,1\,-\,\frac{\cos(2A)}{\cos^2(A)}\: = \:\tan^2(A)\) Click to expand... Use the identity: \(\displaystyle \,\cos(2\theta) \:=\:2\cos^2(\theta)\,-\,1\) We have: \(\displaystyle \L\,1\,-\,\frac{2\cos^2(A)\,-\,1}{\cos^2(A)} \;=\; 1 \,-\,\left[\frac{2\cos^2(A)}{\cos^2(A)}\,-\,\frac{1}{\cos^2(A)}\right] \;=\;1\,-\,\left[2\,-\,\sec^2(A)\right]\) . . \(\displaystyle \L=\;1\,-\,2\,+\,\sec^2(A\)\;=\;\sec^2(A)\,-\,1\;=\;\tan^2(A)\)
Hello, Raul! Another approach . . . Show that: \(\displaystyle \L\,1\,-\,\frac{\cos(2A)}{\cos^2(A)}\: = \:\tan^2(A)\) Click to expand... Use the identity: \(\displaystyle \,\cos(2\theta) \:=\:2\cos^2(\theta)\,-\,1\) We have: \(\displaystyle \L\,1\,-\,\frac{2\cos^2(A)\,-\,1}{\cos^2(A)} \;=\; 1 \,-\,\left[\frac{2\cos^2(A)}{\cos^2(A)}\,-\,\frac{1}{\cos^2(A)}\right] \;=\;1\,-\,\left[2\,-\,\sec^2(A)\right]\) . . \(\displaystyle \L=\;1\,-\,2\,+\,\sec^2(A\)\;=\;\sec^2(A)\,-\,1\;=\;\tan^2(A)\)
