We want the range of an expression involving siny and cosy. A standard trick is to set t=sin2y, so t∈[0,1], and rewrite everything in terms of t. The result becomes a simple quadratic in t, whose minimum and maximum on [0,1] we can find easily. Let t=sin2y⇒t∈[0,1] Then cos2y=1−t⇒cos4y=(1−t)2 So B=sin2y+cos4y=t+(1−t)2. Expand: B=t+1−2t+t2=t2−t+1 Now find the minimum of B(t)=t2−t+1 on [0,1]. Vertex of the parabola: t=
1
2
,B(
1
2
)=(
1
2
)2−
1
2
+1=
1
4
−
1
2
+1=
3
4
. At the endpoints: B(0)=02−0+1=1,B(1)=12−1+1=1. So the minimum is