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TRIGONOMETRIC RATIOS & IDENTITIES
Page # 153
Ex.12 Find satisfying the equation, tan 15° · tan 25° · tan 35° = tan , where
Sol. LHS = tan 15° · tan (30° – 5°) · tan (30° + 5°)
let t = tan 30° and m = tan 5°
tan = tan 15° ·
(0, 15°).
t m t m
t 2 m2
3m m3 1 3m2
·
·
= tan 3(5 ) ·
=
1 tm 1 tm
1 t 2 m2
1 3m 2 3 m 2
m (3 m 2 ) (1 3m 2 )
·
=
= m = tan 5°. Hence = 5°
(1 3m 2 ) 3 m 2
tan 3 x
3 tan x tan3 x
1 3 tan2 x
tan 30º
t 2 1/ 3
t
;
Ex.13 If tan A & tan B are the roots of the quadratic equation, a x2 + b x + c = 0 then evaluate
a sin2 (A + B) + b sin (A + B) . cos (A + B) + c cos2 (A + B).
Sol.
b
c
; tan A . tan B =
a
a
tan A + tan B =
b
a = b
c a
1 ac
tan (A + B) =
Now
a b2
1
=
b2
( c a )2
1
=
E = cos2 (A + B) [a tan2 (A + B) + b tan (A + B) + c]
a )2
(c
b
2
(c
(c
b2
a)
2
c
b2 c
a)
2
=
a )2
(c
b
2
(c
b2
a)
2
c
a
a c
a
1
c
E=c
c
a )2
(c
a
c
Ex.14 Show that cos2A + cos2(A + B) + 2 cosA cos(180° + B) · cos(360° + A + B) is independent of A. Hence
find its value when B = 810°.
Sol. cos2 A + cos2 (A + B) – [2 cosA · cosB · cos (A + B)]
cos2 A + cos2(A + B) – [ {cos(A + B) + cos(A – B) } cos (A + B) ]
cos2 A + cos2 (A + B) – cos2(A + B) – (cos2 A – sin2 B)
= sin2 B which is independent of A now, sin2(810°) = sin2 (720° + 90°) = sin2 90° = 1 Ans.
Ex.15 Simplify: cos x · sin(y – z) + cos y · sin(z – x) + cos z · sin (x – y) where x, y, z R.
Sol. (1/2)[sin(y – z + x) + sin(y – z – x) + sin(z – x + y) + sin(z – x – y) + sin(x – y + z) + sin(x – y – z)] = 0
C.
MULTIPLE ANGLES AND SUB-MULTIPLE ANGLES
(a)
sin 2A = 2 sinA cosA ; sin = 2 sin
(b)
cos 2A = cos²A
cos
= cos²
2
2
cos
sin²A = 2cos²A 1 = 1
sin²
2
= 2cos²
2
1= 1
2
2 sin²A ;
2sin²
2 cos²A = 1 + cos 2A , 2sin²A = 1
cos 2A ;
2 cos²
cos
2
= 1 + cos
, 2 sin²
2
=1
2
.
.
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