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Verify the identity. cos (x + pi/2) = -sin x
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@gorv
pi/2=90 degree
now if we add something it will go in second quaderent
in second quad cos give negative value
cos(pi/2+x)=-sinx
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\[\cos \left( A+B \right)=\cos A \cos B-\sin A \sin B\] \[\cos \frac{ \pi }{ 2 }=0,\sin \frac{ \pi }{ 2 }=1\]
or u can use formula cos(a+b)
so I'm having trouble figuring out where to plug things into but i'll try cos A + B = cos pi/2 cos 0 - sin pi/2 sin 1 I don't think any of that was right @gorv
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