I'm not quite sure how to do this?

[brittle junco]

I've tried converting θ= pi/2 - (ϕ+ψ), but then after expanding everything i still couldnt get an answer. Can anyone help me with this>

[grim surge]

1. Wait patiently for a helper to come along.
2. Once someone helps you, say thank you and close the thread with: ```diff
   +close

[radiant yew]

maybe try and let idk $\alpha = \theta + \phi$ or someth like that and use trig identities to solve, also maybe try converting part of the left hand side of the equation into some sort of factorised form because it looks a lot like $(a+b)^2 = a^2 + 2ab + b^2$, there's just a factor of an extra sine. I'm just mentioning approaches

[charred mortar]

madlad-mathemagician

[radiant yew]

i'm outside rn so i cant rlly do it in my head

or maybe i can

lemme see

oh wait there are hints

ok i think i got the approach down. Basically, you see that $sin(\alpha + \beta)$ term, since it has cosines in it but u dont acc see any cosines in the identity, u wanna express the $\beta$ in the cosine term as $\frac{\pi}{2} - \text{ some angle }$

[charred mortar]

madlad-mathemagician

[radiant yew]

and that angle is presumably $\alpha$ itself

[charred mortar]

madlad-mathemagician

[radiant yew]

so you get $\beta = \frac{\pi}{2} - \alpha$ and since you have $sin(\alpha + \beta) = sin(\frac{pi}{2}) = 1$ I'm pretty u gotta do something with this

[charred mortar]

madlad-mathemagician

[radiant yew]

and maybe use $(a+b+c)^2 = a^2 + b^2 + c^2 + 2bc - 2ca-2ab$

[charred mortar]

madlad-mathemagician

[radiant yew]

yh if u figure something out it should be pretty straightforward, just some trig. But im sure this is the approach here