Given: − π 4 < β < 0 < α < π 4 \text{Given: } -\frac{\pi}{4} < \beta < 0 < \alpha < \frac{\pi}{4} Given: − 4 π < β < 0 < α < 4 π
sin ( α + β ) = 1 3 , cos ( α − β ) = 2 3 \sin(\alpha+\beta) = \frac{1}{3}, \qquad \cos(\alpha-\beta) = \frac{2}{3} sin ( α + β ) = 3 1 , cos ( α − β ) = 3 2
To find: ⌊ ( sin α cos β + cos β sin α + cos α sin β + sin β cos α ) 2 ⌋ \text{To find: } \left\lfloor \left(\frac{\sin\alpha}{\cos\beta} + \frac{\cos\beta}{\sin\alpha} + \frac{\cos\alpha}{\sin\beta} + \frac{\sin\beta}{\cos\alpha}\right)^{2} \right\rfloor To find: ⌊ ( c o s β s i n α + s i n α c o s β + s i n β c o s α + c o s α s i n β ) 2 ⌋
( sin α cos β + cos α sin β ) + ( cos β sin α + sin β cos α ) \left(\frac{\sin\alpha}{\cos\beta}+\frac{\cos\alpha}{\sin\beta}\right) + \left(\frac{\cos\beta}{\sin\alpha}+\frac{\sin\beta}{\cos\alpha}\right) ( c o s β s i n α + s i n β c o s α ) + ( s i n α c o s β + c o s α s i n β )
sin α cos β + cos α sin β = sin α sin β + cos α cos β sin β cos β \frac{\sin\alpha}{\cos\beta}+\frac{\cos\alpha}{\sin\beta} = \frac{\sin\alpha\sin\beta+\cos\alpha\cos\beta}{\sin\beta\cos\beta} c o s β s i n α + s i n β c o s α = s i n β c o s β s i n α s i n β + c o s α c o s β
= cos ( α − β ) sin β cos β = \frac{\cos(\alpha-\beta)}{\sin\beta\cos\beta} = s i n β c o s β c o s ( α − β )
cos β sin α + sin β cos α = cos β cos α + sin β sin α sin α cos α \frac{\cos\beta}{\sin\alpha}+\frac{\sin\beta}{\cos\alpha} = \frac{\cos\beta\cos\alpha+\sin\beta\sin\alpha}{\sin\alpha\cos\alpha} s i n α c o s β + c o s α s i n β = s i n α c o s α c o s β c o s α + s i n β s i n α
= cos ( α − β ) sin α cos α = \frac{\cos(\alpha-\beta)}{\sin\alpha\cos\alpha} = s i n α c o s α c o s ( α − β )
E = cos ( α − β ) ( 1 sin β cos β + 1 sin α cos α ) E = \cos(\alpha-\beta)\left(\frac{1}{\sin\beta\cos\beta}+\frac{1}{\sin\alpha\cos\alpha}\right) E = cos ( α − β ) ( s i n β c o s β 1 + s i n α c o s α 1 )
sin β cos β = 1 2 sin 2 β , sin α cos α = 1 2 sin 2 α \sin\beta\cos\beta = \frac{1}{2}\sin 2\beta, \qquad \sin\alpha\cos\alpha = \frac{1}{2}\sin 2\alpha sin β cos β = 2 1 sin 2 β , sin α cos α = 2 1 sin 2 α
E = 2 cos ( α − β ) ( 1 sin 2 α + 1 sin 2 β ) E = 2\cos(\alpha-\beta)\left(\frac{1}{\sin 2\alpha}+\frac{1}{\sin 2\beta}\right) E = 2 cos ( α − β ) ( s i n 2 α 1 + s i n 2 β 1 )
α + β ∈ ( 0 , π 4 ) , α − β ∈ ( 0 , π 2 ) \alpha+\beta \in \left(0,\frac{\pi}{4}\right), \qquad \alpha-\beta \in \left(0,\frac{\pi}{2}\right) α + β ∈ ( 0 , 4 π ) , α − β ∈ ( 0 , 2 π )
cos ( α + β ) = 2 2 3 , sin ( α − β ) = 5 3 \cos(\alpha+\beta)=\frac{2\sqrt{2}}{3}, \qquad \sin(\alpha-\beta)=\frac{\sqrt{5}}{3} cos ( α + β ) = 3 2 2 , sin ( α − β ) = 3 5
2 α = ( α + β ) + ( α − β ) 2\alpha = (\alpha+\beta)+(\alpha-\beta) 2 α = ( α + β ) + ( α − β )
sin 2 α = sin ( α + β ) cos ( α − β ) + cos ( α + β ) sin ( α − β ) \sin 2\alpha = \sin(\alpha+\beta)\cos(\alpha-\beta)+\cos(\alpha+\beta)\sin(\alpha-\beta) sin 2 α = sin ( α + β ) cos ( α − β ) + cos ( α + β ) sin ( α − β )
sin 2 α = 1 3 ⋅ 2 3 + 2 2 3 ⋅ 5 3 = 2 + 2 10 9 \sin 2\alpha = \frac{1}{3}\cdot\frac{2}{3}+\frac{2\sqrt{2}}{3}\cdot\frac{\sqrt{5}}{3} = \frac{2+2\sqrt{10}}{9} sin 2 α = 3 1 ⋅ 3 2 + 3 2 2 ⋅ 3 5 = 9 2 + 2 10
2 β = ( α + β ) − ( α − β ) 2\beta = (\alpha+\beta)-(\alpha-\beta) 2 β = ( α + β ) − ( α − β )
sin 2 β = sin ( α + β ) cos ( α − β ) − cos ( α + β ) sin ( α − β ) \sin 2\beta = \sin(\alpha+\beta)\cos(\alpha-\beta)-\cos(\alpha+\beta)\sin(\alpha-\beta) sin 2 β = sin ( α + β ) cos ( α − β ) − cos ( α + β ) sin ( α − β )
= 2 − 2 10 9 = \frac{2-2\sqrt{10}}{9} = 9 2 − 2 10
1 sin 2 α + 1 sin 2 β = 9 2 + 2 10 + 9 2 − 2 10 \frac{1}{\sin 2\alpha}+\frac{1}{\sin 2\beta} = \frac{9}{2+2\sqrt{10}}+\frac{9}{2-2\sqrt{10}} s i n 2 α 1 + s i n 2 β 1 = 2 + 2 10 9 + 2 − 2 10 9
= 9 ⋅ ( 2 − 2 10 ) + ( 2 + 2 10 ) ( 2 + 2 10 ) ( 2 − 2 10 ) = 9\cdot\frac{(2-2\sqrt{10})+(2+2\sqrt{10})}{(2+2\sqrt{10})(2-2\sqrt{10})} = 9 ⋅ ( 2 + 2 10 ) ( 2 − 2 10 ) ( 2 − 2 10 ) + ( 2 + 2 10 )
= 9 ⋅ 4 4 − 40 = 9\cdot\frac{4}{4-40} = 9 ⋅ 4 − 40 4
= 9 ⋅ 4 − 36 = − 1 = 9\cdot\frac{4}{-36} = -1 = 9 ⋅ − 36 4 = − 1
E = 2 cos ( α − β ) ⋅ ( − 1 ) = 2 ⋅ 2 3 ⋅ ( − 1 ) = − 4 3 E = 2\cos(\alpha-\beta)\cdot(-1) = 2\cdot\frac{2}{3}\cdot(-1) = -\frac{4}{3} E = 2 cos ( α − β ) ⋅ ( − 1 ) = 2 ⋅ 3 2 ⋅ ( − 1 ) = − 3 4
E 2 = ( − 4 3 ) 2 = 16 9 E^{2} = \left(-\frac{4}{3}\right)^{2} = \frac{16}{9} E 2 = ( − 3 4 ) 2 = 9 16
16 9 = 1.777 … ⇒ ⌊ E 2 ⌋ = 1 \frac{16}{9} = 1.777\ldots \;\Rightarrow\; \left\lfloor E^{2} \right\rfloor = 1 9 16 = 1.777 … ⇒ ⌊ E 2 ⌋ = 1
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