Question

LHS=cos
2
x+cos
2
(x+
3
π
​
)+cos
2
(x−
3
π
​
)
Weknowthat,cos2x=2cos
2
x−1
∴cos
2
x=
2
cos2x+1
​

∴cos
2
(x+
3
π
​
)=
2
cos2(x+
3
π
​
)+1
​

∴cos
2
(x−
3
π
​
)=
2
cos2(x−
3
π
​
)+1
​

∴LHS=1+
2
cos2x
​
+
2
1+cos(2x+
3
2π
​
)
​
+
2
1+cos(2x−
3
2π
​
)
​

=
2
1
​
[1+1+1+cos2x+cos(2x+
3
2π
​
)+cos(2x−
3
2π
​
)]
=
2
1
​
[3+cos2x+2cos
⎝
⎜
⎜
⎜
⎛
​

2
2x+2x+
3
2π
​
−
3
2π
​

​

⎠
⎟
⎟
⎟
⎞
​
].
⎝
⎜
⎜
⎜
⎛
​
cos
⎝
⎜
⎜
⎜
⎛
​

2
2x−2x+
3
2π
​
+
3
2π
​

​

⎠
⎟
⎟
⎟
⎞
​

⎠
⎟
⎟
⎟
⎞
​

=
2
1
​
[3+cos2x+2cos2x.cos
3
2π
​
]
=
2
1
​
[3+cos2x+2cos2x(−cos
3
π
​
)]
=
2
1
​
[3+cos2x−cosx]
=
2
3
​

=RHS
Answered By​

Answer 1

Answer:

??????????????????????????????????????????

Answer 2

Answer:

I am not understand sorry