Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Mathematics
If sin θ+ cos θ=(1/2), then 16( sin (2 θ)+ cos (4 θ)+ sin (6 θ)) is equal to:
Q. If
sin
θ
+
cos
θ
=
2
1
, then
16
(
sin
(
2
θ
)
+
cos
(
4
θ
)
+
sin
(
6
θ
))
is equal to:
628
140
JEE Main
JEE Main 2021
Trigonometric Functions
Report Error
A
23
B
-27
C
-23
D
27
Solution:
sin
θ
+
cos
θ
=
2
1
sin
2
θ
+
cos
2
θ
+
2
sin
θ
cos
θ
=
4
1
sin
2
θ
=
−
4
3
Now:
cos
4
θ
=
1
−
2
sin
2
2
θ
=
1
−
2
(
−
4
3
)
2
=
1
−
2
×
16
9
=
−
8
1
sin
6
θ
=
3
sin
2
θ
−
4
sin
3
2
θ
=
(
3
−
4
sin
2
2
θ
)
⋅
sin
2
θ
=
[
3
−
4
(
16
9
)
]
⋅
(
−
4
3
)
⇒
[
4
3
]
×
(
−
4
3
)
=
−
16
9
16
[
sin
2
θ
+
cos
4
θ
+
sin
6
θ
]
16
(
−
4
3
−
8
1
−
16
9
)
=
−
23