Can you take a more clearer picture thx
It would be obtuse.
Hope this helps :)
Answer:
I think it is the first one. If I am wrong, sue me in the comments. If I am right, please
<h2><u>
MARK ME BRAINLIEST.</u></h2>
Hello,
Here is the demonstration in the book Person Guide to Mathematic by Khattar Dinesh.
Let's assume
P=cos(a)*cos(2a)*cos(3a)*....*cos(998a)*cos(999a)
Q=sin(a)*sin(2a)*sin(3a)*....*sin(998a)*sin(999a)
As sin x *cos x=sin (2x) /2
P*Q=1/2*sin(2a)*1/2sin(4a)*1/2*sin(6a)*....
*1/2* sin(2*998a)*1/2*sin(2*999a) (there are 999 factors)
= 1/(2^999) * sin(2a)*sin(4a)*...
*sin(998a)*sin(1000a)*sin(1002a)*....*sin(1996a)*sin(1998a)
as sin(x)=-sin(2pi-x) and 2pi=1999a
sin(1000a)=-sin(2pi-1000a)=-sin(1999a-1000a)=-sin(999a)
sin(1002a)=-sin(2pi-1002a)=-sin(1999a-1002a)=-sin(997a)
...
sin(1996a)=-sin(2pi-1996a)=-sin(1999a-1996a)=-sin(3a)
sin(1998a)=-sin(2pi-1998a)=-sin(1999a-1998a)=-sin(a)
So sin(2a)*sin(4a)*...
*sin(998a)*sin(1000a)*sin(1002a)*....*sin(1996a)*sin(1998a)
= sin(a)*sin(2a)*sin(3a)*....*sin(998)*sin(999) since there are 500 sign "-".
Thus
P*Q=1/2^999*Q or Q!=0 then
P=1/(2^999)
Answer:
10 and 3
Step-by-step explanation:
let the first number be x and the second be y
x=y+7------i
their sum is 13
x+y=13-----ii
substituting for x in ii
y+7+y=13
collecting like terms
2y=13-7
2y=6
y=3
recall ii and substitute for y
x+y=13
x+3=13
collecting like terms
x=10
