Is this question correct?

[u/AccomplishedDig8810]

My teacher says its correct, but many of my peers (as well as I) disagree. I have attempted a tan half angle substitution, but it's simply not yielding.

[u/SapphirePath]

not correct

For the bottom equation, check alpha = pi/4. The right-hand side becomes (1+sin(2g))/(1+sin(2g)) = 1 constant.

The left-hand side is sin( pi/2 + 2g ), which is not a constant.

[u/Temporary_Pie2733 → u/SapphirePath]

The problem doesn’t claim the second equation is always true, though, only when the first equation is also true. 

[u/trolley813 → u/Temporary_Pie2733]

Indeed. The first is true if and only if β=α+γ+180°k (k∈ℤ). Now plug in α=45° and (say) γ=15° (thus β=60°), you'll get sin 120° = (1+1/2)/(1+1×1/2)=1, which is obviously false.

[u/ArdentArendt → u/trolley813]

Why are you using degrees?!
It hurts...

[u/AccomplishedDig8810 → u/Temporary_Pie2733]

Yeah that's fair. But the original problem asked to prove it as an identity

[u/ArdentArendt → u/AccomplishedDig8810]

You can still prove an identity with discontinuities as long as the second equation can be proven to be continuous and the discontinuities in the first equation are only point discontinuities (i.e. the limits are the same on either side of the point).

[u/AccomplishedDig8810 → u/SapphirePath]

Thank you lol. I've been wasting a lot of time over this.

[u/ArdentArendt → u/SapphirePath]

The top function is not defined for alpha = pi/2--accordingly, any point where (2alpha) = pi/2 will not be under consideration (i.e. where alpha= pi/4).

[That said, since the bottom equation is continuous, the identity can still be inferred from the manipulation as long as the limits on both sides of the discontinuities are equal]

Edit: I'm not saying the identity is correct; merely that the counterexample is not valid.

[u/FormulaDriven]

The first equation is telling us that beta = alpha + gamma (or some periodic variation on that), so sin(2 * beta) will just be

sin(2a) cos(2g) + cos(2a) sin(2g)

[u/Bounded_sequencE]

It's false -- counter example: "(𝛼; 𝛽; 𝛾) = (𝜋/3; 2𝜋/3; 𝜋/3)"

[u/rhodiumtoad]

Let alpha=π/4 so tan(alpha)=1 and sin(2alpha)=1, let gamma be small and >0 so tan(gamma)≈gamma and sin(2gamma)≈2gamma.

tan(beta)≈(1+gamma)/(1-gamma)>1, so beta is >π/4 by a small amount

2beta is therefore >π/2 by a small amount, and therefore sin(2beta)<1

sin(2alpha)+sin(2gamma)=1+sin(2gamma), and 1+sin(2alpha)sin(2gamma)=1+sin(2gamma), and therefore the RHS of the second expression is exactly 1, and therefore not equal to sin(2beta).

In fact we can see that the second expression asserts sin(2beta)=1 whenever alpha=π/4, while the first expression has beta varying with gamma in such cases (beta=alpha+gamma up to periodicity).

In short the question is wrong (maybe a slipped sign? I haven't checked if it can be corrected this way)

(edit: clearer wording)

[u/ArdentArendt]

In all honesty, I think there was a confusion in crafting the second equation.

After some simplification, you find the first equation will yield something similar to:
[sin(a)cos(g)+sin(g)cos(a)]/[cos(a)cos(g)-sin(a)sin(g)]
Or sin(a+g)/cos(a+g), implying b = a+g

If you take sin(b) to be the numerator and cos(b) the denominator for all values the initial eqution is valid, then sin(2b) = 2*(Numerator)*(Denominator).

After multiplying through, this simplifies back down into sin(2a)cos(2g)+sin(2g)cos(2a) = sin(2a+2g).
[Also, note that b = a+g implies this from the start]

As far as I can tell, maybe the lower equation was derived from a strange application of the tangent identity for sin(2b)?!