Answer Type
COS I COS Y
Answers
Answer:I: cosx+cosy=
3
1
........1
sinx+siny=
4
1
........(2)
Squaring and adding (1) and (2), we get
cos
2
x+cos
2
y+2cosxcosy+sin
2
x+sin
2
y+2sinxsiny=
9
1
+
16
1
......(3)
2+2cos(x−y)=
144
25
2+4cos
2
2
x−y
−2=
144
25
cos
2
2
x−y
=
144×4
25
Squaring and subtracting (2) from (1), we get
cos
2
x+cos
2
y+2cosxcosy−sin
2
x−sin
2
y−2sinxsiny=
9
1
−
16
1
cos2x+cos2y+2cos(x+y)=
144
7
2cos(x+y)cos(x−y)+2cos(x+y)=
144
7
2cos(x+y)(cos(x−y)+1)=
144
7
2cos(x+y)(2cos
2
(x−y)/2−1+1)=
144
7
Substituting value of (3)
2cos(x+y)⋅2⋅
144×4
25
=
144
7
cos(x+y)=
25
7
II: sinx+siny=
4
1
.........(1)
sinx−siny=
5
1
..........(2)
Dividing (2) by (1), we get
sinx−siny
sinx+siny
=
4
5
⇒
2cos(
2
x+y
)sin(
2
x−y
)
2sin(
2
x+y
)cos(
2
x−y
)
=
4
5
⇒
cot(
2
x+y
)
cot(
2
x−y
)
=
4
5
⇒4cot(
2
x−y
)=5cot(
2
x+y
)
Step-by-step explanation:
Answer:
here's ur answer dude
Step-by-step explanation:
cosx+cosy+cosα=0
cosx+cosy=−cosα
[cosC+cosD=2cos(
2
C+D
)cos(
2
C−D
)]
2cos(
2
x+y
)cos(
2
x−y
)=−cosα...(i)
sinx+siny+sinα=0
sinx+siny=−sinα
[sinC+sinD=2sin(
2
C+D
)cos(
2
C−D
)]