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It appears that there is no algebraic solution to this so I am solving it numerically.
From cos 2 θ = 1 − 2 sin 2 θ :
1 − 2 sin 2 1 5 ∘ ⟹ sin 1 5 ∘ = cos 3 0 ∘ = 2 1 − cos 3 0 ∘ = 4 2 − 3 = 2 1 2 4 − 2 3 = 2 1 2 ( 3 − 1 ) 2 = 2 2 3 − 1 ≈ 0 . 2 5 8 8 1 9 0 4 5
From tan 3 θ = 1 − 3 tan 2 θ 3 tan θ − tan 3 θ :
1 − 3 tan 2 1 0 ∘ 3 tan 1 0 ∘ − tan 3 1 0 ∘ 3 tan 3 1 0 ∘ − 3 tan 2 1 0 ∘ − 3 3 tan 1 0 ∘ + 1 = 3 1 = 0
Using Newton Raphson method , we get tan 1 0 ∘ ≈ 0 . 1 7 6 3 2 6 9 8 1 .
Therefore, 5 sin 1 5 ∘ + 5 tan 1 0 ∘ ≈ 5 ( 0 . 2 5 8 8 1 9 0 4 5 + 0 . 1 7 6 3 2 6 9 8 1 ) ≈ 2 . 1 7 6 .