Proving $\tan (\frac{\pi }{4} - x) = \frac{{1 - \sin 2x}}{{\cos 2x}}$

How do I prove the identity:

$$\tan \left(\frac{\pi }{4} - x\right) = \frac{{1 - \sin 2x}}{{\cos 2x}}$$

Any common strategies on solving other identities would also be appreciated.

I chose to expand the left hand side of the equation and got stuck here: $$\frac{\cos x-\sin x}{\cos x+\sin x}$$

• Try multiply both the numerator and denominator by the numerator. – user60610 Mar 31 '13 at 12:55
• $\tan (\frac{\pi }{4} - x)=\frac{\sin (\frac{\pi }{4} - x)}{\cos (\frac{\pi }{4} - x)}$ and expand using the laws of $\sin(A-B)$ and $\cos(A-B)$. – Mhenni Benghorbal Mar 31 '13 at 13:06
• Your RHS has a typo. It should be $\frac{(1-\sin2x)}{\cos2x}$ – Manoj Pandey Mar 31 '13 at 13:16
• @Assad Could you verify that indeed it should be $\sin 2x$ instead of $2 \sin x$ in the identity? – TMM Mar 31 '13 at 13:18
• It is sin2x, i've just checked, sorry for the typo! – seeker Mar 31 '13 at 13:19

Use the identity $$\tan(x-y)=\frac{\tan x-\tan y}{1+\tan x\tan y}$$. $$\text{Multiply the expression} \ \frac{\cos x-\sin x}{\cos x+\sin x}\ \text{by} \ \frac{\cos x-\sin x}{\cos x-\sin x}.$$ You'll get an identity in the denominator immediately and finish the numerator.

• My answer was going in the wrong direction ;) I deleted it. +1 – Clive Newstead Mar 31 '13 at 12:58
• This seems to be the right answer, how did you recognise that you needed to do that? – seeker Mar 31 '13 at 13:00
• you have to use the sum and difference formulas. Sine, cosine, and tangent functions have sum and difference identites. Generally they are good to memorize, but most precalculus, trig, and calculus texts have a chart. As soon as you see a sum in the argument, you should be aware there is an identity related to that function. After that with your expression, for the novice, it is not always obvious to see where to go from there. Multiplying by "1" is a trick used to manipulate your expression to get something else. It becomes more obvious after experience. – Eleven-Eleven Mar 31 '13 at 13:04
• Thank you, that's made it a little easier to understand! – seeker Mar 31 '13 at 13:05
• @Mhenni has another good way to start if you don't know the tangent identity we used. – Eleven-Eleven Mar 31 '13 at 13:07

As @ChristopherErnst suggests in a comment, some things become "more obvious after experience". Here are two alternative approaches to your problem that bear this out.

If you find yourself working with double- and half-angle arguments often, you might get the immediate sense that the right-hand side would be better if the roles of sine and cosine were reversed, since

$$\frac{1-\cos 2\theta}{\sin 2\theta} = \frac{2\sin^2\theta}{2\sin\theta\cos\theta} = \frac{\sin\theta}{\cos\theta} = \tan\theta \qquad\qquad(*)$$

As it turns out, reversing those roles is as easy as replacing the arguments with their complements:

$$\cos x = \sin\left(\frac{\pi}{2}-x\right) \qquad \sin x = \cos\left(\frac{\pi}{2}-x\right)$$

So, writing $2x= \frac{\pi}{2}-2\theta$ (equivalently, $2\theta = \frac{\pi}{2}-2x$) turns your identity into $(*)$.

Another approach immediately recognizes "$1-\sin 2x$" as a perfect square:

$$1-\sin 2x = (\cos^2 x + \sin^2 x) - 2 \cos x \sin x = \left( \cos x - \sin x \right)^2$$

(This isn't something I've seen exploited all that often, but it has come up with unusual frequency in some trig manipulations in my current research, so it kinda jumps out at me.) It's a convenient counterpart to the difference-of-squares version of the cosine double-angle identity:

$$\cos 2x = \cos^2 x - \sin^2 x = \left( \cos x - \sin x \right)\left( \cos x + \sin x \right)$$

Thus, the right-hand side of your identity reduces nicely ...

$$\frac{\left( \cos x - \sin x \right)^2}{\left( \cos x - \sin x \right)\left( \cos x + \sin x \right)} = \frac{\cos x - \sin x}{\cos x + \sin x}$$

... to the expression you have already shown to be equal to $\tan\left(\frac{\pi}{4}-x\right)$.