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Byju's Answer
Standard XII
Mathematics
Trigonometric Ratios of Allied Angles
Prove that:i ...
Question
Prove that:
(i) sin 38° + sin 22° = sin 82°
(ii) cos 100° + cos 20° = cos 40°
(iii) sin 50° + sin 10° = cos 20°
(iv) sin 23° + sin 37° = cos 7°
(v) sin 105° + cos 105° = cos 45°
(vi) sin 40° + sin 20° = cos 10°
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Solution
(i)
Consider
LHS
:
s
in
38
°
+
sin
22
°
=
2
sin
38
°
+
22
°
2
cos
38
°
-
22
°
2
∵
sin
A
+
sin
B
=
2
sin
A
+
B
2
cos
A
-
B
2
=
2
sin
30
°
cos
8
°
=
2
×
1
2
cos
(
90
°
-
8
°
)
=
sin
82
°
=
RHS
Hence
,
LHS
=
RHS
.
(ii)
Consider
LHS
:
cos
100
°
+
cos
20
°
=
2
cos
100
°
+
20
°
2
cos
100
°
-
20
°
2
∵
cos
A
+
cos
B
=
2
cos
A
+
B
2
cos
A
-
B
2
=
2
cos
60
°
cos
40
°
=
2
×
1
2
cos
40
°
=
cos
40
°
Hence
,
LHS
=
RHS
.
(iii)
Consider
LHS
:
sin
50
°
+
sin
10
°
=
2
sin
50
°
+
10
°
2
cos
50
°
-
10
°
2
∵
sin
A
+
sin
B
=
2
sin
A
+
B
2
cos
A
-
B
2
=
2
sin
30
°
cos
20
°
=
2
×
1
2
cos
20
°
=
cos
20
°
Hence
,
LHS
=
RHS
.
(iv)
Consider
LHS
:
s
in
23
°
+
sin
37
°
=
2
sin
23
°
+
37
°
2
cos
23
°
-
37
°
2
∵
sin
A
+
sin
B
=
2
si
n
A
+
B
2
cos
A
-
B
2
=
2
sin
30
°
cos
-
7
°
=
2
sin
30
°
cos
7
°
=
2
×
1
2
cos
7
°
=
cos
7
°
Hence
,
LHS
=
RHS
.
(v)
Consider
LHS
:
s
in
105
°
+
cos
105
°
=
sin
105
°
+
cos
90
°
+
15
°
=
sin
105
°
-
sin
15
°
=
2
si
n
105
°
-
15
°
2
cos
105
°
+
15
°
2
∵
sin
A
+
sin
B
=
2
sin
A
-
B
2
cos
A
+
B
2
=
2
sin
45
°
cos
60
°
=
2
sin
90
°
-
45
°
cos
60
°
=
2
×
1
2
cos
45
°
=
cos
45
°
Hence
,
LHS
=
RHS
.
(vi)
Consider
LHS
:
sin
40
°
+
sin
20
°
=
2
sin
40
°
+
20
°
2
cos
40
°
-
20
°
2
∵
sin
A
+
sin
B
=
2
sin
A
+
B
2
cos
A
-
B
2
=
2
sin
30
°
cos
10
°
=
2
×
1
2
cos
10
°
=
cos
10
°
Hence
,
LHS
=
RHS
.
Suggest Corrections
0
Similar questions
Q.
Prove that:
(i) cos 10° cos 30° cos 50° cos 70° =
3
16
(ii) cos 40° cos 80° cos 160° =
-
1
8
(iii) sin 20° sin 40° sin 80° =
3
8
(iv) cos 20° cos 40° cos 80° =
1
8
(v) tan 20° tan 40° tan 60° tan 80° = 3
(vi) tan 20° tan 30° tan 40° tan 80° = 1
(vii) sin 10° sin 50° sin 60° sin 70° =
3
16
(viii) sin 20° sin 40° sin 60° sin 80° =
3
16
Q.
Prove that:
(i) cos 55° + cos 65° + cos 175° = 0
(ii) sin 50° − sin 70° + sin 10° = 0
(iii) cos 80° + cos 40° − cos 20° = 0
(iv) cos 20° + cos 100° + cos 140° = 0
(v)
sin
5
π
18
-
cos
4
π
9
=
3
sin
π
9
(vi)
cos
π
12
-
sin
π
12
=
1
2
(vii) sin 80° − cos 70° = cos 50°
(viii) sin 51° + cos 81° = cos 21°
Q.
Prove that
sin
10
∘
+
sin
20
∘
+
sin
40
∘
+
sin
50
∘
=
sin
70
∘
+
sin
80
∘
.
Q.
Prove that
cos
20
∘
+
sin
20
∘
cos
20
∘
−
sin
20
∘
=
tan
65
∘
Q.
Show that
sin
10
∘
+
sin
20
∘
+
sin
40
∘
+
sin
50
∘
=
sin
70
∘
+
sin
80
∘
.
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