# Trigonometric Double Angle proof

Below is the trigonometric double angle problem. Please prove this.

If $$\tan θ=\sec 2α$$, prove that $$\sin 2θ=\dfrac{1-\tan 4α}{1+\tan 4α}$$

What I have tried:

R.H.S = $$\dfrac{1-\tan 4α}{1+\tan 4α}$$

= $$\dfrac{cos4α-\sin4α}{cos4α+sin4α}$$

= $$\dfrac{cos^2(2α) - sin^2(2α) - 2sin2αcos2α}{cos^2(2α) - sin^2(2α) + 2sin2αcos2α}$$

.....

......

Just the hit will do.

Sorry everybody, it was a false proof, just got the wrong questions. It was supposed to be

$$\sin 2θ=\dfrac{1-\tan^4α}{1+\tan^4α}$$

• ..and you have done what so far? Dec 15, 2013 at 17:20
• I tried to prove from the RHS, converting tan4α to sin4α/cos4α and then solving and applying double angle formula on cos4α with cos2.2α and then apply 1/cos2α=tanθ.... Dec 15, 2013 at 17:23
• Then you should add that to your question, perhaps under "I"ve tried the following:" or like that, otherwise most members will think you're tying others to do your homework for you and, besides downvoting you, perhaps you'll get your question closed. Dec 15, 2013 at 17:24
• @DonAntonio can you just provide the hint so that I can prove it myself Dec 15, 2013 at 17:34
• Haven't tried this myself but it might be worth using the half-angle identity: $$\sin(2\theta)=\frac{2\tan(\theta)}{1+\tan^2(\theta)}$$ Dec 15, 2013 at 17:47

Choose $\theta=\arctan (-2)$ and $\alpha=\dfrac{\pi}{3}$. Then $\tan \theta=\sec 2\alpha$.
But $\sin 2\theta=2\sin\theta\cos\theta=-2\cdot\sqrt{\dfrac{\tan^2\theta}{1+\tan^2\theta}}\cdot\sqrt{\dfrac1{1+\tan^2\theta}}=-2\cdot\sqrt{\dfrac45}\cdot\sqrt{\dfrac15} =-0.8$
and $\dfrac{1-\tan 4\alpha}{1+\tan 4\alpha}=\dfrac{1-\sqrt3}{1+\sqrt3}=2-\sqrt3\not=-0.8.$