sin(x+y)=sin(x)cos(y)-cos(x)sin(y)
also, remember pythagorean rule, 
given that sin(Θ)=4/5 and cos(x)=-5/13
find sin(x) and cos(Θ)
sin(x)
cos(x)=-5/13
using pythagorean identity
(sin(x))^2+(-5/13)^2=1
sin(x)=+/- 12/13
in the 2nd quadrant, sin is positve so sin(x)=12/13
cos(Θ)
sin(Θ)=4/5
using pythagrean identity
(4/5)^2+(cos(Θ))^2=1
cos(Θ)=+/-3/5
in 1st quadrant, cos is positive
cos(Θ)=3/5
so sin(Θ+x)=sin(Θ)cos(x)+cos(Θ)sin(x)
sin(Θ+x)=(4/5)(-5/13)+(3/5)(12/13)
sin(Θ+x)=16/65
answer is 1st option
Answer:9/4
Step-by-step explanation:
99/8*2/11=9/4*1/1=9/4
You can cancel out 99 and 11 to 9 and 1 as they are multiples of eac hother and you can do the same for 8 and 2 to 4 and 1
Answer:
-3
Step-by-step explanation:
to eliminate 6x we need the same no. 6x but that should be negative. so
2X × -3 = -6X
and
6X+(-6X)=0
Answer:
5:2
Step-by-step explanation: