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Verify the identity sin(x+π/2)=cosx
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\[\sin(x+\pi/2) = cosx\] use the addition formula for sine \[\sin(x+\pi/2)\] becomes \[sinxcospi/2 + cosxsinpi/2\] But, sinpi/2 = 1 and cospi/2 = 0 \[sinx(0) + cosx(1) = cosx\] cosx = cosx L.S = R.S
we know that \[π=180°\]so therefore\[\frac{π}{2}=90\]now we can rw-write this as\[\sin(x+90)\]\[\sin(A+B)=\sin A \cos B + \cos A \sin B\]so,\[\sin(x+90)=\sin(x) \cos (90) + \cos (x) \sin (90)\]we know that\[Sin(90)=1~~~~~~and~~~~~~Cos(90)=0\]so you can simplify this to,\[Sin(x)\times 0+Cos(x)\times 1\]\[0+Cos(x)\]\[Cos(x)\]See how I reduced the left side to Cos (x)?
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