wiedząc że tg x=⅓ oblicz wartość wyrażenia 5(2sin² x-1)
tg x = ⁴a₃b = ⅓
a=1
b=3
z twierdzenia Pitagorasa
a²+b²=c²
1²+3²=c²
1+9=c²
c²=10
√c=√10
sin x =a/c=1/√10*√10/√10=√10/10
5(2*sin² x -1)= 5 (2*(√10/10)²-1)=
=5(2*10/100-1)= 5(2/10-1)=
=5*(-0,8)= -4
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tg x = ⁴a₃b = ⅓
a=1
b=3
z twierdzenia Pitagorasa
a²+b²=c²
1²+3²=c²
1+9=c²
c²=10
√c=√10
sin x =a/c=1/√10*√10/√10=√10/10
5(2*sin² x -1)= 5 (2*(√10/10)²-1)=
=5(2*10/100-1)= 5(2/10-1)=
=5*(-0,8)= -4