(A) cos β (B) cos 2β (C) sin α (D) sin 2α
(B) cos 2β
As indicated by the inquiry,
![\[cos\left( \alpha +\beta \right)\text{ }=\text{ }0\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-e61ce9c25f8a2f55d3e076c9031cdfc3_l3.png "Rendered by QuickLaTeX.com")
Since, cos 90° = 0
We can compose,
![\[cos\left( \alpha +\beta \right)=\text{ }cos\text{ }90{}^\circ \]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ce35eb4201c402015a628e2a300de68e_l3.png "Rendered by QuickLaTeX.com")
By contrasting cosine condition on L.H.S and R.H.S,
We get,
![\[\left( \alpha +\beta \right)=\text{ }90{}^\circ \]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-2e1824519293998d8c17762753960dcf_l3.png "Rendered by QuickLaTeX.com")
α = 90°-β
Presently we need to lessen sin (α – β ),
In this way, we take,
![\[sin\left( \alpha -\beta \right)\text{ }=\text{ }sin\left( 90{}^\circ -\beta -\beta \right)\text{ }=\text{ }sin\left( 90{}^\circ -2\beta \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c4f07b7a75e4a6b8008fae1fbb0d1276_l3.png "Rendered by QuickLaTeX.com")
![\[sin\left( 90{}^\circ -\theta \right)\text{ }=\text{ }cos\text{ }\theta \]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-43f8ad94e52addfc58ca19a576e5dd93_l3.png "Rendered by QuickLaTeX.com")
In this way,
![\[sin\left( 90{}^\circ -2\beta \right)\text{ }=\text{ }cos\text{ }2\beta \]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-2acca3e297bb01bdc009b4d26beac50e_l3.png "Rendered by QuickLaTeX.com")
Hence, sin(α-β) = cos 2β