If α,β,γ,and δare in arithmetic progression. Then $$sin$$α$$+$$β$$+$$γ$$+$$δ≠

Question

If α,β,γ,and δare in arithmetic progression.  Then $$sin$$α$$+$$β$$+$$γ$$+$$δ≠

Infinity Learn · Accepted Answer

Let   β$$=$$α$$+d$$,  γ$$=$$α$$+2d$$,  δ$$=$$α$$+3d$$          ⇒ α$$+$$β$$+$$γ$$+$$δ $$=$$ $$4$$α$$+6d$$ →$$1Now$$          $$3$$β$$+$$δ$$=3$$α$$+d+$$α$$+3d=4$$α$$+6d$$→$$2By$$ $$1$$  and $$2$$ , we have   α$$+$$β$$+$$γ$$+$$δ$$=$$          $$3$$β$$+$$δ       Also α$$+3$$γ$$=$$α$$+3$$α$$+2d$$ $$=$$ $$4$$α$$+6d$$→$$3$$       By $$1$$ and $$3$$ we have α$$+$$β$$+$$γ$$+$$δ$$=$$α$$+3$$γα$$+$$β$$+$$γ$$+$$δ$$=2$$α$$+2$$δ

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

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If α,β,γ,and δare in arithmetic progression. Then sinα+β+γ+δ≠

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