Question

Prove the identity \dfrac{\sin (2α -β) }{\sinα } -2\cos(α -β )=-\sinβ\cscα

Collected in the board: Trigonometry

Steven Zheng posted 2 months ago


Answer

\dfrac{\sin (2α -β) }{\sinα } -2\cos(α -β )=-\sinβ\cscα

\dfrac{\sin (2α -β) }{\sinα } -2\cos(α -β )=- \dfrac{\sinβ}{\sinα}

\sin(2α-β)-2\sinα\cos(α-β)=-\sinβ
(1)

\sin(2α-β)-2\sinα\cos(α-β)

=\sinα\cos(α-β)+\cosα\sin(α-β)-2\sinα\cos(α-β)

=\cosα\sin(α-β)- \sinα\cos(α-β)

=\sin(α-β -α) =-\sinβ

Steven Zheng posted 2 months ago

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