a)
Wzory, z których korzystamy:
(1) sin(90∘−α)=cosα
(2) cos(90∘−α)=cosα
(3) sin2α+cos2α=1
Mamy więc:
(sin12∘+cos12∘)(sin12∘−cos12∘)+2sin278∘=
=(sin12∘)2−(cos12∘)2+2sin278∘=
=[sin(90∘−78∘)]2−[cos(90∘−78∘)]2+2sin278∘=(1)(2)
=(cos78∘)2−(sin78∘)2+2sin278∘=cos278∘−sin278∘+2sin278∘=
=cos278∘+sin278∘=(3)1
b)
Wzory, z których korzystamy:
(1) cos(90∘−α)=cosα
(2) sin(90∘−α)=cosα
(3) sin2α+cos2α=1
Mamy więc:
(cos17∘+cos73∘)2+(sin17∘−sin73∘)2=
=[cos(90∘−73∘)+cos73∘]2+[sin(90∘−73∘)−sin73∘]2=(1)(2)
=(sin73∘+cos73∘)2+(sin73∘−sin73∘)2=
=(sin73∘)2+2sin73∘⋅cos73∘+(cos73∘)2+
+(sin73∘)2−2sin73∘⋅cos73∘+(cos73∘)2=
=2sin273∘+2cos273∘=2(sin273∘+cos273∘)=(3)2⋅1=2
c)
Wzory, z których korzystamy:
(1) tg(90∘−α)=ctg α
(2) tg α⋅ctg α=1, α=k⋅90∘, k∈Z
Mamy więc:
tg 41∘⋅tg 42∘⋅…⋅tg 49∘=
=tg 41∘⋅tg 42∘⋅tg 43∘⋅tg 44∘⋅tg 45∘⋅tg 46∘⋅tg 47∘⋅tg 48∘⋅tg 49∘=
=tg (90∘−49∘)⋅tg (90∘−48∘)⋅tg (90∘−47∘)⋅tg (90∘−46∘)⋅tg 45∘⋅
⋅tg 46∘⋅tg 47∘⋅tg 48∘⋅tg 49∘=(1)
=ctg 49∘⋅ctg 48∘⋅ctg 47∘⋅ctg 46∘⋅tg 45∘⋅tg 46∘⋅tg 47∘⋅tg 48∘⋅tg 49∘=
=tg 49∘⋅ctg 49∘⋅tg 48∘⋅ctg 48∘⋅tg 47∘⋅ctg 47∘⋅tg 46∘⋅ctg 46∘⋅tg 45∘=(2)
=1⋅1⋅1⋅1⋅1=1
d)
Wzory, z których korzystamy:
(1) ctg(90∘−α)=tg α
(2) tg(90∘−α)=ctg α
(3) tg α⋅ctg α=1, α=k⋅90∘, k∈Z
Mamy więc:
(ctg 10∘+ctg 80∘)2−(tg 10∘−tg 80∘)2=
=[ctg (90∘−80∘)+ctg 80∘)2−[tg (90∘−80∘)−tg 80∘]2=(1)(2)
=(tg 80∘+ctg 80∘)2−(tg 80∘−tg 80∘)2=
=(tg 80∘)2+2tg 80∘⋅ctg 80∘+(ctg 80∘)2−
−[(tg 80∘)2−2tg 80∘⋅ctg 80∘+(ctg 80∘)2]=(3)
=tg2 80∘+2⋅1+ctg2 80∘−tg2 80∘+2⋅1−ctg2 80∘=2+2=4
e)
Wzory, z których korzystamy:
(1) ctg(90∘−α)=tg α
(2) tg α⋅ctg α=1, α=k⋅90∘, k∈Z
Mamy więc:
ctg 10∘⋅ctg 20∘⋅ctg 30∘⋅ctg 40∘⋅ctg 50∘⋅ctg 60∘⋅ctg 70∘⋅ctg 80∘=
=ctg (90∘−80∘)⋅ctg (90∘−70∘)⋅ctg (90∘−60∘)⋅ctg (90∘−50∘)⋅
⋅ctg 50∘⋅ctg 60∘⋅ctg 70∘⋅ctg 80∘=(1)
=tg 80∘⋅tg 70∘⋅tg 60∘⋅tg 50∘⋅ctg 50∘⋅ctg 60∘⋅ctg 70∘⋅ctg 80∘=
=tg 80∘⋅ctg 80∘⋅tg 70∘⋅ctg 70∘⋅tg 60∘⋅ctg 60∘⋅tg 50∘⋅ctg 50∘ =(2)
=1⋅1⋅1⋅1=1
