Which sum or difference identity would you use to verify that cos (180° - ∅) = -cos ∅?

a.
sin (α -β) = sin α cos β – cos α sin β
b.
cos (α -β) = cos α cos β – sin α sin β
c.
cos (α -β) = cos α cos β + sin α sin β
d.
sin (α + β) = sin α cos β + cos α sin β

Since right side of above formula is [tex]180^{\circ}-\phi[/tex] which is same as [tex]\cos (\alpha-\beta)[/tex], we will use the identity [tex]\cos (\alpha-\beta)[/tex]

[tex]\cos (\alpha-\beta)=\cos\alpha\cdot\cos\beta+\sin\alpha\cdot\sin\beta[/tex]

[tex]\cos (180^{\circ}-\phi)=\cos 180^{\circ}\cdot\cos\phi+\sin 180^{\circ}\cdot\sin\phi[/tex]

We know [tex]\cos 180^{\circ}=-1[/tex] and [tex]\sin 180^{\circ}=0[/tex]

Substitute above, we get,

[tex]\cos (180^{\circ}-\phi)=-\cos \phi[/tex]

Thus, Option (b) is correct.

Which sum or difference identity would you use to verify that cos (180° - ∅) = -cos ∅? a. sin (α -β) = sin α cos β – cos α sin β b. cos (α -β) = cos α cos β – sin α sin β c. cos (α -β) = cos α cos β + sin α sin β d. sin (α + β) = sin α cos β + cos α sin β

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Which sum or difference identity would you use to verify that cos (180° - ∅) = -cos ∅?

a.
sin (α -β) = sin α cos β – cos α sin β
b.
cos (α -β) = cos α cos β – sin α sin β
c.
cos (α -β) = cos α cos β + sin α sin β
d.
sin (α + β) = sin α cos β + cos α sin β