First slide

Trigonometric Identities

Question

If $α, β, γ,$ and $δ$ are in arithmetic progression. Then $\sin (α + β + γ + δ) \neq$

Easy

A

$\sin (3 β + δ)$
B

$\sin (α + 3 γ)$
C

$\sin (2 α + 2 δ)$
D

$\sin (3 β + γ)$

Solution

$\begin{array}{l} L e t β = α + d, γ = α + 2 d, δ = α + 3 d \\ \Rightarrow α + β + γ + δ = 4 α + 6 d \to (1) \\ Now 3 β + δ = 3 (α + d) + α + 3 d = 4 α + 6 d \to (2) \\ By (1) and (2), we have α + β + γ + δ = 3 β + δ \\ Also α+3γ=α+3 (α + 2 d) = 4 α + 6 d \to (3) \\ B y (1) a n d (3) w e h a v e α + β + γ + δ = α +3 γ \\ α + β + γ + δ = 2 α + 2 δ \end{array}$

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