3. Trigonometric Functions.

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1. Find the radian measures corresponding to the following degree measures:
(i) 250 (ii)-47030’ (iii)2400 (iv)5200
w
Solution:
Radians = 180 degrees
•
du
ne
ha 5
1
ad 3
.b 44
w 1
0
w
81
09
1 degree =
(i) 25° = 25 x
radians
=
Radians.
(ii) -47° 30’ = -47° 30’ x
(iii) 240° = 240 x
= - 0.2638
=
(iv) 520° = 520 x
Radians.
Radians
=
Radians
2. Find the degree measures corresponding to the following radian measures
(use π=
i)
(ii)-4 (iii)
(iv)
Solution:
Radians = 180°
∴ 1 Radian =
degrees
ii) 1 radian =
× -4 =
-4radian
= -32.72 × 7 = -229o5'24''
.in
on
= 39.375º
=
ti
ca
•
).(i)
iii) 1 radian =
=
= 3000
1
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iv) 1 radian =
= 210o
w
09
du
ne
ha 5
1
ad 3
.b 44
w 1
0
w
81
3. A wheel makes 360 revolutions in one minute. Through how many radians does
it turn is one second?
•
Solution:
One complete revolution = 2π
360 revolutions = 360 × 2π€= 720π radians / minute = 720π /60 radians/sec =
12π radians/sec.
4. In a circle of diameter 40cm. The length of a chord is 20cm. Find the length of
minor arc corresponding to the chord.
•
Solution:
ca
.
ti
5. If, in two circles, arcs of the same length subtend angles of 60o and 75o at the
centre, find the ratio of their radii.
on
•
Solution:
Let the radii be r1 and r2.
Let the angles subtend by the arcs in two circles be θ1 and θ2.
l = θ r where θ is the angle, l the length of an arc and r the radius of the circle.
=
θ2 = 75o = 75 ×
=
r1 =
radian
radian
.in
θ1 = 60o = 60 ×
r2
2
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r1 / r2 =
=
Therefore the required ratio is 5:4.
w
09
du
ne
ha 5
1
ad 3
.b 44
w 1
0
w
81
6. Find the angle in radian through which a pendulum swings if its length is 75cm
and the tip described an arcs of length
(i) 10cm
(ii) 15cm
(iii) 21cm
•
Solution:
(i) Length of the pendulum (r) = 75cm
Length of an arc (l) = 10cm
θ = l/r = 10/75radians = 2/15radians
(ii) Length of the pendulum (r) = 75cm
Length of an arc (l) = 15cm
θ = l/r = 15/75 radians = 1/5radians
(iii) Length of the pendulum (r) = 75cm
Length of an arc (l) = 21cm
θ = l/r = 21/75 radians = 7/25radians
7. Find the values of other five trigonometric functions
1. Cos x =
•
, x lies in third quadrant.
Solution:
, cosec x = -
, tanx =
, cotx =
ca
sin x =
secx = -2
8. Find the values of other five trigonometric functions
, x lies in second quadrant.
Solution:
, secx = -
9. Cot x =
•
, tanx = -
, cotx = -
,Cosecx =
, x lies in third quadrant.
.in
cos x = -
on
•
ti
(i) Sin x =
Solution:
Tanx =
, sinx = -
, Cosec x = -
, cosx = -
, secx = -
3
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10. Find the values of the other five trigonometric functions in the following
problem: sec θ = 13/5, θ lies in fourth quadrant.
•
du
ne
ha 5
1
ad 3
.b 44
w 1
0
w
81
w
Solution:
sin θ , tan θ , cosec θ and cot θ are negative in IV th quadrant sec θ and cos θ are positive in
IVth quadrant.
sec θ = 13/5
cos θ = 1/sec θ = 5/13
09
sin2 θ = 1∴ sin θ =
cosec θ =
∴ tan θ =
=
cot θ =
=
11. Tan x =
•
, x lies in second quadrant.
Solution:
, sinx =
, cosecx =
, cosx = -
ca
cotx = -
, secx = -
Solution:
sin 765o = sin
= sin
= sin
=
13. Find the value of the following trigonometric function: cosec(-1410o)
•
Solution:
cosec(-1410o) = -cosec 1410o = -cosec (8π – 30o) = cosec 30o = 2
14. Find the value of the following trigonometric function:sin (
.in
on
•
ti
12. Find the values of the trigonometric functions sin 765o
)
4
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Solution:
•
Sin
= -Sin
w
15. 2sin2
+2cos2
= -Sin(660)= - Sin (2x360 – 60) = -(-sin60) = sin60 =
+2sec2
= 10
du
ne
ha 5
1
ad 3
.b 44
w 1
0
w
81
Solution:
•
09
L.H.S = 2sin2
+2cos2
= 2sin2
= 2sin2
=
=
=
=
+2sec2
+2cos2
+ 2cos2
2×
+ 2×
1 + 1+ 8
10
R.H.S
+2sec2
+ 2sec2
+ 2× 4
16. Find the value of:(i) sin 750 (ii) tan150
•
Solution:
i)Sin (30+45)
= sin30cos45+cos30sin45 Sin(A+B)
=SinACosB+CosAsinB
=
ca
=
ti
(ii)tan15 = tan(45 – 30)
on
=
.in
=
=
17. Prove the following:Cos
cos
- sin
Sin
= sin(x+y)
5
CONTACT FOR MATHEMATICS GROUP/HOME COACHING FOR CLASS 11TH 12TH BBA,BCA,
DIPLOMA CPT CAT SSC AND OTHER EXAMS Also home tutors for other subjects in DELHI AND OTHER MAJOR CITIES
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•
Solution:
This is of the form CosACosB-SinASinB = Cos(A+B)
= Cos
09
du
ne
ha 5
1
ad 3
.b 44
w 1
0
w
81
w
∴ L H S = Cos
= Sin(x+y)
Hence proved.
= cot2x
18.
•
Solution:
=
= -cos2x/-sin2x
= cot2x
19. cos
cos(2 +x)
=1
Solution:
•
= sinxcosx
=Sinx.cosx
20. sin (n + 1)x sin (n + 2)x + cos(n + 1)xcos(n + 2)x = cosx.
21. cos
=-
sinx
.in
Solution:
L.H.S = sin (n + 1)x sin (n + 2)x + cos(n + 1)xcos(n + 2)x
= cos[(n + 1)x - (n + 2)x]
= cos (-x)
= cosx
= RH.S
on
•
ti
ca
=sinx.cosx
=sin2x+cos2x = 1
(Hence Proved)
Solution:
•
cos
cosx – sin
sinx-
6
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= cos
cosx – sin
= -2sin
sinx
sinx -
w
=-
du
ne
ha 5
1
ad 3
.b 44
w 1
0
w
81
09
22. sin2 6x – sin2 4x = sin2x sin10x
•
Solution:
L.H.S = sin2 6x – sin2 4x
[Using the formulae 2sin2x = 1 - cos2x]
=
=-
(cos12x – cos8x)
=[cos(10x + 2x) – cos(10x - 2x)]
[Using the formulae -2sinθ sinφ = cos(θ + φ ) - cos(θ - φ )]
=
× 2 sin10x sin2x
= sin10x sin2x
23. cos22x – cos26x = sin4x sin8x
•
Solution:
L.H.S = cos22x – cos26x
[Using the formulae 2cos2x = 1+ cos2x]
=-
(cos12x – cos4x)
•
.in
24. sin2x + 2sin4x + sin6x = 4cos2x sin4x
on
= × 2 sin8x sin4x
= sin4x sin8x
ti
=[cos(8x + 4x) – cos(8x - 4x)]
[Using the formulae -2sinθ sinφ = cos(θ + φ ) - cos(θ - φ )]
ca
=
Solution:
L.H.S = sin2x + 2sin4x + sin6x
=
=
=
=
=
2sin
cos
+ 2sin4x
2sin4xcos2x + 2sin4x
2sin4x(cos2x + 1)
4sin4xcos2x
R.H.S
7
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25. cot4x(sin5x + sin3x) = cotx(sin5x – sin3x)
•
Solution:
L.H.S = cot4x(sin5x + sin3x)
w
[Using the formula sin5x + sin3x = 2 sin
cos
]
du
ne
ha 5
1
ad 3
.b 44
w 1
0
w
81
= 2cot4x sin
cos
09
= 2cot4x sin
cos
= 2cot4x sin4x cosx
= 2cos4x cosx
R.H.S = cotx(sin5x – sin3x)
[Using the formula sin5x - sin3x = 2 cos
= 2cotx cos
sin
]
sin
= 2cotx cos
sin
= 2cos4x cosx
∴ L.H.S = R.H.S
26. Prove that
•
=
Solution:
L.H.S =
ca
=
ti
on
=
=
.in
=
= R.H.S
27. Prove that
•
= tan4x
Solution:
L.H.S =
8
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=
w
=
du
ne
ha 5
1
ad 3
.b 44
w 1
0
w
81
09
=
= tan4x
= R.H.S
28. Prove that
= tan
Solution:
•
L.H.S =
=
=
= tan
= R.H.S
ca
29.
ti
Solution:
•
on
=
.in
=
= tan2x.
30. Prove that
•
=2sinx
Solution:
L.H.S =
9
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CONTACT FOR COACHING MATHEMATICS FOR 11TH 12TH NDA DIPLOMA SSC CAT SAT CPT
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=
=
w
[Using the identity cos2x = cos2x – sin2x and sinx – siny = 2cos
sin
]
du
ne
ha 5
1
ad 3
.b 44
w 1
0
w
81
09
=
= 2sinx
= R.H.S
31. Prove that
Solution:
•
=
=
cot3x
=
R.H.S
Hence proved
•
Solution:
Cot2x(cotx – cot 3x)- cot 3x.cotx
ti
ca
32. cot x cot 2x – cot 2x cot3x – cot3x cotx =1
= cot 3x.cotx
.in
= cot2x
- cot3x.cotx
on
= cot 2x
=
=
10
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DIPLOMA CPT CAT SSC AND OTHER EXAMS Also home tutors for other subjects in DELHI AND OTHER MAJOR CITIES
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CONTACT FOR COACHING MATHEMATICS FOR 11TH 12TH NDA DIPLOMA SSC CAT SAT CPT
CONTACT FOR ADMISSION GUIDANCE B.TECH BBA BCA, MCA MBA DIPLOMA AND OTHER COURSES 09810144315
=
du
ne
ha 5
1
ad 3
.b 44
w 1
0
w
81
w
=
09
=
=
=
=
=
ca
=
ti
=
=
.in
on
=
= 1.
33. tan4x =
Solution:
•
L.H.S
= tan4x
11
CONTACT FOR MATHEMATICS GROUP/HOME COACHING FOR CLASS 11TH 12TH BBA,BCA,
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= tan(3x+x)
=
du
ne
ha 5
1
ad 3
.b 44
w 1
0
w
81
w
09
=
=
=
=
L.H.S = R.H.S
34. Prove that cos4x = 1 - 8sin2xcos2x.
ti
Solution:
= cos4x = 1-2sin22x
= 1-2 (2sinx cosx)2
= 1-8 sin2x cos2x = R.H.S
Hence proved.
ca
•
•
= 4(2cos2x-1)3 - 3(2cos2x-1)
= 4[(2cos2x)3 - 3×(2cos2x)2×1 + 3 × 2 cos2x×1 - 13]-6cos2x + 3
.in
Solution:
cos6x = 4 cos32x - 3 cos2x
on
35. Prove that cos6x = 32cos6 x- 48cos4x + 18cos2x-1
= 32cos6x - 48cos4x + 18cos2x -1
= R.H.S
Hence proved
12
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