cos x - sin x cos x + sin x = sec 2x - tan 2x help, please
S sadler New member Joined Jan 24, 2010 Messages 1 Jan 24, 2010 #1 cos x - sin x cos x + sin x = sec 2x - tan 2x help, please
D daon Senior Member Joined Jan 27, 2006 Messages 1,284 Jan 24, 2010 #2 Start by multiplying by the conjugate of the denominator: \(\displaystyle \frac{\cos(x) - \sin(x)}{\cos(x)+\sin(x)} = \frac{\cos(x) - \sin(x)}{\cos(x)+\sin(x)} \cdot \frac{\cos(x) - \sin(x)}{\cos(x)-\sin(x)}\) In the numerator you will end up with: \(\displaystyle 1-2\sin(x)\cos(x) = 1-sin(2x)\) In the denominator you will end up with: \(\displaystyle \cos^2(x) - \sin^2(x) = \cos(2x)\) You should be able to end it
Start by multiplying by the conjugate of the denominator: \(\displaystyle \frac{\cos(x) - \sin(x)}{\cos(x)+\sin(x)} = \frac{\cos(x) - \sin(x)}{\cos(x)+\sin(x)} \cdot \frac{\cos(x) - \sin(x)}{\cos(x)-\sin(x)}\) In the numerator you will end up with: \(\displaystyle 1-2\sin(x)\cos(x) = 1-sin(2x)\) In the denominator you will end up with: \(\displaystyle \cos^2(x) - \sin^2(x) = \cos(2x)\) You should be able to end it
