Proof of \(\displaystyle{\cos{\alpha}}+{\sin{\beta}}+{\cos{\gamma}}={4}\cdot{\sin{{\left({\frac{\alpha}{{2}}}+{45}°\right)}}}\cdot{\sin{{\frac{\beta}{{2}}}}}\cdot{\sin{{\left({\frac{\gamma}{{2}}}+{45}°\right)}}}\) I tried to simplify right

David Rodgers

David Rodgers

Answered question

2022-03-31

Proof of cosα+sinβ+cosγ=4sin(α2+45°)sinβ2sin(γ2+45°)
I tried to simplify right side:
4(12(cos(α2+γ2+π2)cos(α2γ2))sin(β2)=
=2sin(β)cos(α2+γ2+π2)+2cos(α2γ2)sin(β2)=
=sin(β2+α2+γ2+π2)sin(β2α2γ2π2)+sin(β2+α2γ2)+sin(β2α2+γ2)

Answer & Explanation

German Ferguson

German Ferguson

Beginner2022-04-01Added 18 answers

use that
sin(x2+π4)=12((sin(x2)+cos(x2))

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