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Byju's Answer
Standard IX
Mathematics
Use of Trigonometric Table
Prove that : ...
Question
Prove that :
sin
40
∘
−
cos
70
∘
=
√
3
⋅
cos
80
∘
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Solution
L.H.S. :- sin 40 - cos 70 using sin C - sin D
2
c
o
s
C
+
D
2
s
i
n
C
−
D
2
⇒
s
i
n
40
−
c
o
s
(
90
−
20
)
⇒
s
i
n
40
−
s
i
n
20
⇒
2
c
o
s
30
.
s
i
n
10
.
s
i
n
(
90
−
80
)
=
c
o
s
80
⇒
2
√
3
2
×
c
o
s
80
=
√
3
c
o
s
80
∘
R.H.S
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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.
Show that -
cos
40
∘
+
cos
50
∘
+
cos
70
∘
+
cos
80
∘
=
cos
20
∘
+
cos
10
∘
.
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.
s
i
n
70
∘
+
c
o
s
40
∘
c
o
s
70
∘
+
s
i
n
40
∘
=
Q.
The value of
sin
70
°
+
cos
40
°
cos
70
°
+
sin
40
°
is _________.
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