sin3θ⋅cos3θ+cos3θ.sin3θ=
Answers
Given equation
sin
3
θ
+
cos
3
θ
=
1
−
sin
2
θ
⇒
3
sin
θ
−
4
sin
3
θ
+
4
cos
3
θ
−
3
cos
θ
−
(
1
−
sin
2
θ
)
=
0
⇒
3
(
sin
θ
−
cos
θ
)
−
4
(
sin
3
θ
−
cos
3
θ
)
−
(
sin
2
θ
+
cos
2
θ
−
2
sin
θ
cos
θ
)
=
0
⇒
3
(
sin
θ
−
cos
θ
)
−
4
(
sin
θ
−
cos
θ
)
(
sin
2
θ
+
cos
2
θ
+
sin
θ
cos
θ
)
−
(
sin
2
θ
+
cos
2
θ
−
2
sin
θ
cos
θ
)
=
0
⇒
3
(
sin
θ
−
cos
θ
)
−
4
(
sin
θ
−
cos
θ
)
(
1
+
sin
θ
cos
θ
)
−
(
sin
θ
−
cos
θ
)
2
=
0
⇒
(
sin
θ
−
cos
θ
)
(
3
−
4
−
4
sin
θ
cos
θ
−
sin
θ
+
cos
θ
)
=
0
⇒
(
sin
θ
−
cos
θ
)
(
cos
θ
−
1
−
4
sin
θ
cos
θ
−
sin
θ
)
=
0
So
sin
θ
−
cos
θ
=
0
⇒
sin
θ
=
cos
θ
⇒
tan
θ
=
1
→
option A possible
Checking opotion B,
cos
θ
=
0
or
θ
=
90
for 2nd equation
cos
θ
−
1
−
4
sin
θ
cos
θ
−
sin
θ
=
cos
90
−
1
−
4
⋅
sin
90
⋅
cos
90
−
sin
90
=
0
−
1
−
0
−
1
=
−
2
So option B is not possible
Checking opotion C
tan
(
θ
2
)
=
−
1
or
θ
2
=
−
45
or
θ
=
−
90
for 2nd equation
cos
θ
−
1
−
4
sin
θ
cos
θ
−
sin
θ
=
cos
(
−
90
)
−
1
−
4
⋅
sin
(
−
90
)
⋅
cos
(
−
90
)
−
sin
(
−
90
)
=
0
−
1
−
0
+
1
=
0
So option C is possible
Checking opotion D,
cos
(
θ
2
)
=
0
or
θ
=
180
∘
for 2nd equation
cos
θ
−
1
−
4
sin
θ
cos
θ
−
sin
θ
=
cos
180
−
1
−
4
⋅
sin
180
⋅
cos
180
−
sin
180
=
−
1
−
1
−
0
−
0
=
−
2
So option D is not possible
Answer link
Answer:
√3/2
Let 30°= A
putting it in the equation
we, get
sinA•cosA+cosA•sinA
according to trigonometric ratios of compound angles we get
sin2A=2sinA•cosA
and
sinA=1/2 A=30°
cosA=√3/2 A=30°
= 2×1/2×√3/2
=√3/2 (and)