Find the least value for sin⁡x−cos2x−1At first I found the first derivative to be y′=cos⁡x+2sin⁡xcos⁡xCritical point 0,−π6 (principal)y−sin⁡x+2(cos⁡2x)Then substitution of x by critical points I found minima. But my answer is incorrect.Correct minimum value is −94

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Aiden Cooper · Accepted Answer

f(x)=sin2x+sin⁡{x}−2=(sin⁡{x}+12)2−2.25≥−2.25  The equality occurs for sin⁡x=−12, which says that -2.25 is a minimal value.

Micah May · Accepted Answer

Yes your way is correct indeed for x=−π6 we havef(x)=−12−34−1=−94As an alternative, recall that cos2x=1−sin2x and therefore we havef(x)=sin⁡x−cos2x−1=sin2x+sin⁡x−2then set t=sin⁡x and consider the parabolag(t)=t2+t−2⇒g′(t)=2t+1=0⇒t=−12

Find the least value for \sin x - \cos^2 x -1 At first I found the first derivative to be

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