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TRIGONOMETRIC RATIOS & IDENTITIES
Page # 157
Ex.26 In a kite ABCD, AB = AD and CB = CD. If
A = 108° and
C = 36° then the ratio of the area of ABD to
2
the area of CBD can be written in the form
Sol.
a b tan 36
where a, b and c are relatively prime positive
c
integers. Determine the ordered triple (a, b, c).
Since the triangles ABD and CBD have a common base,
hence the ratio of their areas equals the ratio of their heights.
h
, then h = x tan 36°.
x
k
|||ly tan 72° = then k = x tan 72°.
x
Since tan 36° =
Hence,
h x tan 36
=
=
k x tan 72
tan 36
2 tan 36
1 tan 2 36
=
1 tan 2 36
2
Then ordered triple (a, b, c) is (1,1, 2)
Ex.27 If , , and be the roots of the equation, 2 cos 2
2 cos + 1 = 0, all lying in the interval [0, 2 ]
then find the value of the product, cos . cos . cos . cos .
Sol.
4 cos2
cos
2 cos
5 1
or cos
4
=
=
1=0
or
5
cos
5 1
=
4
=
=
sin
2
10
4
8
= cos
16
=
5
10
1
5
4
10
= cos
6
10
9
3
7
;
or
5
5
5
Hence P = cos
5
cos
3
7
9
1
cos
cos
=
5
5
5
16
Ex.28 If sin x, sin22x and cos x · sin 4x form an increasing geometric sequence, find the numerical value of
Sol.
Given sin x, sin22x and cos x · sin 4x are in G.P.
(r > 1 as G.P. is increasing)
sin4 2x = (sin x) (cos x) (sin 4x)
16 sin4x cos4 x = sin x cos x sin 4x
3
3
16 sin x cos x = sin 4x
(sin x 0, cos x 0)
16(sin x cos x)3 = 2 sin 2x · cos 2x
(sin 2x)3 = sin 2x · cos 2x
2
2
sin 2x = cos 2x
(sin 2x 0), 1 – cos 2x = cos 2x, y2 + y – 1 = 0
1
cos 2x =
5
2
;
cos 2x cannot be
1
1 cos 2x
sin x =
=
2
5 1
2
cos 2x =
r=
5 1
2
·
r=
5 1
hence rejected
2
cos 2x =
1
5
2
5 1
5 1
3
5
2
=
=
2 2
2
2
sin 2 2 x
= 4 sin x cos2x = 2 sin x(1 + cos 2x)
sin x
4
5 1
=
= 2
2 2
2
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