Exercise 3.1
Q.1: Calculate the radian measurement of the given degree measurement:
(i). 25∘
(ii). 240∘
(iii). −47∘30'
(iv). 520∘
Sol:
(i). 25∘
As we know that, 180∘ = π radian
Therefore, 25∘ = π180×25 radian = 5π36 radian
Hence, 25∘ = 5π36 radian
(ii). 240∘
As we know that 180∘ = π radian
Therefore, 240∘ = π180×240 radian = 4π3 radian
Hence, 240∘ = 4π3 radian
(iii). −47∘30'
= −4712
= −952 degree
As we know that, 180∘ = π radian
Therefore, −952 degree = π180×−952 radian = −1936×2π radian = −1972π
Hence, −47∘30' = −1972π
(iv). 520∘
As we know that, 180∘ = n radian
Therefore, 520∘ = π180×520 radian = 26π9 radian
Hence, 520∘ = 26π9 radian
Q.2: Calculate the degree measurement of the given degree measurement: [Use π = 227]
(i) 1116
(ii) -4
(iii) 5π3
(iv) 7π6
Sol:
(i). 1116:
As we know, that Π Radian = 180°
Therefore, 1116 radian = 180π×1116 degree
= 45×11π×4 degree
= 45×11×722×4 degree
= 3158 degree
= 3938 degree
= 39∘+3×608 minute [1∘ = 60’]
= 39∘+22'+12 minute
= 39∘+22'+30” [1’ = 60’’]
(ii). -4:
As we know, that Π Radian = 180°
Therefore, -4 radian = 180π×(−4) degree
= 180×7(−4)22 degree
= −252011 degree
= −229111 degree
= −229∘+1×6011 minutes [1∘ = 60’]
= −229∘+5'+511
= −229∘+5'+27” [1’ = 60’’]
(iii). 5π3
As we know, that Π Radian = 180°
Therefore, 5π3 radian = 180π×5π3 degree
= 300∘
(iv). 7π6
As we know, that Π Radian = 180°
Therefore, 7π6 radian = 180π×7π6
= 210∘
Q.3: In a minute, wheel makes 360 revolutions. Through how many radians does it turn in 1 second?
Answer:
No. of revolutions made in a minute = 360 revolutions
Therefore, no. of revolutions made in a second = 36060 = 6
In one revolution, the wheel rotates an angle of 2π radian.
Therefore, in 6 revolutions, it will turn an angle of 12π radian.
Q.4: Calculate the degree measurement of the angle subtended at the centre of a circle of radius 100 m by an arc of length 22 m.
[Use π = 227]
Answer:
As we know,
In a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then, θ=lr
Given:
r = 100 m and L = 22 m
We have,
θ=22100 radian
= 180π×22100 degree
= 180×7×2222×100 degree
= 12610 degree
= 1235 degree
= 12∘36' [1∘ = 60’]
Therefore, the req angle is 12∘36'
Q.5: In a circle of diameter 40 m, the length of the chord 20 m. Find the length of minor arc of chord.
Answer:
13
Diameter of circle = 40 m
Radius of circle = 4020 m = 20 m
Let XY be the chord (length = 20 m) of the circle.
In ΔOXY, OX = OY = radius of the circle = 20 m
XY = 20 m
Therefore,
ΔOXY is an equilateral triangle.
θ=60∘ = π3 radian
As we know,
In a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then, θ = lr
π3 = arc(AB)20 arc(AB)20 = 20π3 m
The length of the minor arc of the chord is 20π3 m.
Q.6: In two circles, arcs which has same length subtended at an angle of 60∘ and 75∘ at the center. Calculate the ratio of their radii.
Sol:
Let, the radii of two circles be r1 and r2.
Let, an arc of length l subtend an angle of 60∘ at the center of the circle of radius r1, while let an arc of length l subtend an angle of 75∘ at the center of the circle of radius r2.
60∘ = π3 radian
75∘ = 5π12 radian
In a circle of radius r unit, if an arc of length l unit subtends an angle θ then:
θ = lr
l = r θ
l = r1π3
l = r25π12 r1π3 = r25π12 r1 = r254 r1r2 = 54
Therefore, the ratio of radii is 5: 4
Q.7: Calculate the angle in radian through which a pendulum swings if the length is 75 cm and the tip describes an arc of length
(i) 10 cm
(ii) 15 cm
(iii) 21 cm
Sol:
As we know that, in a circle of radius ‘r’ unit, if an arc of length ‘l’ unit subtends an angle θ radian at the center, then:
θ=lr
Here, r = 75 cm
(i). 10 cm:
θ = 1075 radian = 215 radian
(ii). 15 cm:
θ = 1575 radian = 15 radian
(iii). 21 cm:
θ = 2175 radian = 725 radian
Exercise 3.2
Q.1: Calculate the values of five trigonometric func. if cosy = −12 and y lies in 3rd quadrant.
Sol:
(i) sec y :
Since, cos y = 12
Therefore, sec y = 1cosy = 1(−12)
Hence, sec y = -2
(ii) sin y :
Since, sin2y+cos2y=1
Therefore, sin2y=1−cos2y ⇒ sin2y=1−(−12)2 ⇒ sin2y=1−14 ⇒ sin2y=34 ⇒ siny=±3√2
Since, y lies in the third quadrant, the value of sin y will be negative.
Therefore, sin y = 3√2
(iii) cosec y = 1siny = 1(−3√2) = −23√
Therefore, cosec y = −23√
(iv) tan y = sinycosy = tany=(−3√2)(−12) = 3–√
Therefore, tan y = 3–√
(v) cot y = 1tany = 13√
Therefore, cot y = 13√
Q.2: Calculate the other five trigonometric function if we are given the values for sin y = 35, where y lies in second quadrant.
Sol:
sin y = 35
Therefore, cosec y = 1Siny = 135=53
Since, sin2y+cos2y=1 ⇒ cos2y=1–sin2y ⇒ cos2y=1–(35)2 ⇒ cos2y=1–925 ⇒ cos2y=1625 ⇒ cos y = ±45
Since, y lies in the 2nd quadrant, the value of cos y will be negative,
Therefore, cos y = – 45
⇒ sec y = 1cosy=1(−45)=–54
⇒ tan y = sinycosy=(35)(−45)=−34
⇒ cot y = 1tany=−43
Q.3: Find the values of other five trigonometric functions if coty=34, where y lies in the third quadrant.
Sol:
cot y = 34
Since, tan y = 1coty=134=43
Since, 1+tan2y=sec2y
⇒ 1+(43)2=sec2y ⇒ 1+169=sec2y ⇒ 259=sec2y ⇒ sec y = ±53
Since, y lies in the 3rd quadrant, the value of sec y will be negative.
Therefore, sec y = −53
cos y = 1secy=1(−53)=–35
Since, tan y = sinycosy ⇒ 43=siny(−35) ⇒ sin y = (43)×(−35)=−45 ⇒ cosec y = 1siny=−54
Q.4: Find the values of other five trigonometric if secy=135, where y lies in the fourth quadrant.
Sol:
sec y = 135
cos y = 1secy=1(135)=513
Since, sin2y+cos2y=1
⇒ sin2y=1–cos2y ⇒ sin2y=1–(513)2 ⇒ sin2y=1–25169=144169 ⇒ sin y = ±1213
Since, y lies in the 4th quadrant, the value of sin y will be negative.
Therefore, sin y = −1213
⇒ cosec y = 1siny=1(−1213)=−1312
⇒ tan y = sinycosy=(−1213)(513)=−125
⇒ cot y = 1tany=1(−125)=−512
Q.5: Find the values of other five trigonometric function if tan y = −512 and y lies in second quadrant.
Sol:
tan y = −512 [Given]
And, cot y = 1tany=1(−512)=–125
Since, 1+tan2y=sec2y
Therefore, 1+(–512)2=sec2y ⇒ sec2y=1+(25144) ⇒ sec2y=169144 ⇒ sec y = ±1312
Since, y lies in the 2nd quadrant, the value of sec y will be negative.
Therefore, sec y = −1312
cos y = 1secy=1(−1312)=(−1213)
Since, tany=sinycosy
⇒ −512=siny(−1213) ⇒ sin y = (−512)×(−1213)=513
⇒ cosec y = 1siny=1(513)=135
Q.6: Calculate the value of trigonometric function sin 765°.
Sol:
The values of sin y repeat after an interval of 360° or 2n.
Therefore, sin 765° = sin ( 2 × 360° + 45° ) = sin 45° = 12√
Q.7: Calculate the value of trigonometric function cosec [-1410°]
Sol:
It is known that the values of tan y repeat after an interval of 360° or 2n.
Therefore, cosec [-1410°] = cosec [-1410° + 4 × 360°] = cosec [ -1410° + 1440°] = cosec 30° = 2
Q.8: Calculate the value of the trigonometric function tan19π3.
Sol:
It is known that the values of tan y repeat after an interval of 360° or 2n.
Therefore, tan19π3 = tan613π = tan(6π+π3) = tanπ3 = tan 60° = 3–√
Q.9: Calculate the value of the trigonometric function sin(−11π3).
Answer:
It is known that the values of tan y repeat after an interval of 360° or 2n.
Therefore, sin(−11π3) = sin(−11π3+2×2π) = sinπ3 = 3√2
Q.10: Calculate the value of the trigonometric function cot(−15π4).
Sol:
It is known that the values of tany repeat after an interval of 180∘ or n.
Therefore, cot(−15π4) = cot(−15π4+4π) = cotπ4 = 1
Exercise 3.3
Q.1: Prove:
sin2π6+cos2π3–tan2π4=−12
Sol:
Now, taking L.H.S.
sin2π6+cos2π3–tan2π4:
= (12)2+(12)2–(1)2
= 14+14–1 = −12
= R.H.S.
Q.2: Prove:
2sin2π6+cosec27π6cos2π3=32
Sol:
Now, taking L.H.S.
2sin2π6+cosec27π6cos2π3 :
= 2(12)2+cosec2(π+π6)(12)2
= 2×14+(−cosecπ6)2(14)
= 12+(−2)2(14)
= 12+44
= 12+1
= 32
= R.H.S.
Q.3: Prove:
cot2π6+cosec5π6+3tan2π6=6
Sol:
Taking L.H.S.
cot2π6+cosec5π6+3tan2π6 :
= (3–√)2+cosec(π–π6)+3(13√)2
= 3+cosecπ6+3×13
= 3 + 2 + 1 = 6
= R.H.S.
Q.4: Prove:
2sin23π4+2cos2π4+2sec2π3=10
Sol:
Now, taking L.H.S.
2sin23π4+2cos2π4+2sec2π3 :
= 2{sin(π–π4)}2+2(12√)2+2(2)2
= 2{sinπ4}2+2×12+8
= 2(12√)2 + 1 + 8
= 1 + 1 + 8 = 10
= R.H.S.
Q.5: Calculate the value of:
(i). sin75∘
(ii). tan15∘
Sol:
(i). sin75∘:
= sin(45∘+30∘)
= sin45∘cos30∘+cos45∘sin30∘
Since, [sin (x + y) = sin x cos y + cos x sin y]
= (12√)(3√2)+(12√)(12)
= 3√22√+122√
= 3√+122√
(ii). tan15∘:
= tan(45∘–30∘)
= tan45∘–tan30∘1+tan45∘tan30∘
Since, [tan (x – y) = tanx–tany1+tanxtany]
= 1–13√1+1(13√)
= 3√–13√3+1√3√
= 3√–13√+1
= (3√–1)2(3√+1)(3√–1)
= 3+1–23√(3√)2–(1)2
= 4–23√3–1
= 2–3–√
Q.6:Prove:
cos(π4–x)cos(π4–y)–sin(π4–x)sin(π4–y)=sin(x+y)
Sol:
Now, taking L.H.S.
cos(π4–x)cos(π4–y)–sin(π4–x)sin(π4–y):
= 12[2cos(π4–x)cos(π4–y)]+12[−2sin(π4–x)sin(π4–y)] =12[cos{(π4–x)+(π4–y)}+cos{(π4–x)–(π4–y)}]+12[cos{(π4–x)+(π4–y)}–cos{(π4–x)–(π4–y)}]
Since, [2cos A cos B = cos (A + B) + cos (A – B)]
And, [2sin A sin B = cos (A + B) – cos (A – B)]
= 2×12[cos{(π4–x)+(π4–y)}]
= cos[π2–(x+y)]
= sin (x + y)
= R.H.S.
Q.7: Prove:
tan(π4+x)tan(π4–x)=(1+tanx1–tanx)2
Sol:
Since, tan (A + B)= tanA+tanB1–tanAtanB
And, tan (A – B) = tanA–tanB1+tanAtanB
Now, taking L.H.S.
tan(π4+x)tan(π4–x):
= (tanπ4+tanx1–tanπ4tanx)(tanπ4–tanx1+tanπ4tanx)
= (1+tanx1–tanx)(1–tanx1+tanx)
= (1+tanx1–tanx)2
= R.H.S.
Q.8: Prove:
cos(π+x)cos(−x)sin(π–x)cos(π2+x)=cot2x
Sol:
Now, taking L.H.S.
cos(π+x)cos(−x)sin(π–x)cos(π2+x):
= [−cosx][cosx](sinx)(−sinx)
= −cos2x−sin2x = cot2x
= R.H.S.
Q.9: Prove:
cos(3π2+x)cos(2π+x)[cot(3π2–x)+cot(2π+x)]=1
Sol:
Now, taking L.H.S.
cos(3π2+x)cos(2π+x)[cot(3π2–x)+cot(2π+x)]:
= sinxcosx[tanx+cotx]
= sinxcosx(sinxcosx+cosxsinx)
= (sinxcosx)[sin2x+cos2xsinxcosx]
= sin2x+cos2x = 1
= R.H.S.
Q.10: Prove:
sin(n+1)xsin(n+2)x+cos(n+1)xcos(n+2)x=cosx
Sol:
Now, taking L.H.S.
sin(n+1)xsin(n+2)x+cos(n+1)xcos(n+2)x:
=12[2sin(n+1)xsin(n+2)x+2cos(n+1)xcos(n+2)x]
=12[cos{(n+1)x–(n+2)x}–cos{(n+1)x+(n+2)x}+cos{(n+1)x+(n+2)x}+cos{(n+1)x–(n+2)x}]
Since, [2sin A sin B = cos (A + B) – cos (A – B)]
And, [2cos A cos B = cos (A + B) + cos (A – B)]
= 12×2cos{(n+1)x–(n+2)x}
= cos(−x) = cos x
= R.H.S.
Q.11 Prove:
cos(3π4+x)–cos(3π4–x)=−2–√sinx
Sol:
Since, cos A – cos B = −2sin(A+B2)sin(A–B2)
Now, taking L.H.S.
cos(3π4+x)–cos(3π4–x):
= −2sin{(3π4+x)+(3π4–x)2}sin{(3π4+x)–(3π4–x)2}
= −2sin(3π4)sinx
= −2sin(π–π4)sinx
= −2sinπ4sinx
= −2×12√×sinx
= −2–√sinx
= R.H.S.
Q.12: Prove:
sin26x–sin24x=sin2xsin10x
Sol:
Since, sin A + sin B = 2sin(A+B2)cos(A–B2)
And, sin A – sin B = 2cos(A+B2)sin(A–B2)
Now, taking L.H.S.
sin26x–sin24x:
= (sin6x+sin4x)(sin6x–sin4x)
= [2sin(6x+4x2)cos(6x–4x2)][2cos(6x+4x2)sin(6x–4x2)]
= (2sin 5x cos x) (2cos 5x sin x)
= (2 sin 5x cos 5x) (2 cos x sin x)
= sin 10x sin 2x
= R.H.S.
Q.13: Prove:
cos22x–cos26x=sin4xsin8x
Sol:
Since, cos A + cos B = 2cos(A+B2)cos(A–B2)
And, cos A – cos B = 2sin(A+B2)sin(A–B2)
Now, taking L.H.S.
cos22x–cos26x:
= (cos 2x + cos 6x) (cos 2x – cos 6x)
= [2cos(2x+6x2)cos(2x–6x2)][−2sin(2x+6x2)sin(2x–6x2)]
= [2cos 4x cos (-2x)] [ -2sin 4x sin ( -2x)]
= [2cos 4x cos 2x] [-2sin 4x ( -sin2x)]
= [2sin 4x cos 4x] [2sin 2x cos 2x]
= sin 8x sin 4x
= R.H.S.
Q.14:Prove:
sin2x+2sin4x+sin6x=4cos2xsin4x
Sol:
Now, taking L.H.S.
sin 2x + 2sin 4x + sin 6x:
= [sin 2x + sin 6x] + 2 sin 4x
= [2sin(2x+6x2)(2x–6x2)]+2sin4x
Since, sin A + sin B = 2sin(A+B2)cos(A–B2)
= 2sin 4x cos(-2x) + 2sin 4x
= 2sin 4x cos 2x + 2sin 4x
= 2sin 4x (cos 2x + 1)
= 2sin4x(2cos2x–1+1)
= 2sin4x(2cos2x)
= 4cos2xsin4x
= R.H.S.
Q.15: Prove:
cot4x(sin5x+sin3x)=cotx(sin5x–sin3x)
Sol:
Now, taking L.H.S.
cot 4x (sin 5x + sin 3x):
= cos4xsin4x[2sin(5x+3x2)cos(5x–3x2)]
Since, sin A – sin B = 2cos(A+B2)sin(A–B2)
= (cos4xsin4x)[2sin4xcosx]
=2cos 4x cos x . . . . . . . . . . . . . . . (1)
Now, taking R.H.S.
cot x (sin 5x – sin 3x):
= cosxsinx[2cos(5x+3x2)sin(5x–3x2)] sinA–sinB=2cos(A+B2)sin(A–B2)
= cosxsinx[2cos4xsinx]
= 2 cos 4x cos x . . . . . . . . . . . . . . . . . . . . (2)
From equation (1) and (2):
L.H.S. = R.H.S.
Q.16: Prove:
cos9x–cos5xsin17x–sin3x=−sin2xcos10x
Sol:
Since, cos A – cos B = 2sin(A+B2)sin(A–B2)
And, sin A – sin B = 2cos(A+B2)sin(A–B2)
Now, taking L.H.S.
cos9x–cos5xsin17x–sin3x:
= −2sin(9x+5x2)sin(9x–5x2)2cos(17x+3x2)sin(17x–3x2)
= −2sin7xsin2x2cos10xsin7x
= −sin2xcos10x
= R.H.S.
Q.17: Prove:
sin5x+sin3xcos5x+cos3x=tan4x
Sol:
Since, sin A + sin B = 2sin(A+B2)cos(A–B2)
And, cos A + cos B = 2cos(A+B2)cos(A–B2)
Now, taking L.H.S.
sin5x+sin3xcos5x+cos3x:
= 2sin(5x+3x2)cos(5x–3x2)2cos(5x+3x2)cos(5x–3x2)
= 2sin4xcosx2cos4xcosx = tan 4x
=R.H.S.
Q.18: Prove:
sinx–sinycosx+cosy=tanx–y2
Sol:
Since, cos A + cos B = 2cos(A+B2)cos(A–B2)
And, sin A – sin B = 2cos(A+B2)sin(A–B2)
Now taking L.H.S.
sinx–sinycosx+cosy:
= 2cos(x+y2)sin(x–y2)2cos(x+y2)cos(x–y2)
= sin(x–y2)cos(x–y2)
= tan(x–y2)
= R.H.S.
Q.19: Prove:
sinx+sin3xcosx+cos3x=tan2x
Sol::
Since, sin A + sin B = 2sin(A+B2)cos(A–B2)
And, cos A + cos B = 2cos(A+B2)cos(A–B2)
Now, taking L.H.S.
sinx+sin3xcosx+cos3x:
= 2sin(x+3x2)cos(x–3x2)2cos(x+3x2)cos(x–3x2)
= sin2xcos2x = tan 2x
= R.H.S.
Q.20: Prove:
sinx–sin3xsin2x–cos2x=2sinx
Answer:
Since, sin A – sin B = 2cos(A+B2)sin(A–B2)
cos2A–sin2A=cos2A
Now, taking L.H.S.
sinx–sin3xsin2x–cos2x:
= 2cos(x+3x2)sin(x–3x2)–cos2x
= 2cos2xsin(−x)−cos2x
= −2x(−sinx) = 2 sin x
= R.H.S.
Q.21: Prove:
cos4x+cos3x+cos2xsin4x+sin3x+sin2x=cot3x
Sol:
Taking L.H.S.
cos4x+cos3x+cos2xsin4x+sin3x+sin2x:
= (cos4x+cos2x)+cos3x(sin4x+sin2x)+sin3x
= 2cos(4x+2x2)cos(4x–2x2)+cos3x2sin(4x+2x2)cos(4x–2x2)+sin3x ⎡⎣⎢sinA+sinB=2sin(A+B2)cos(A–B2)∗cosA+cosB=2cos(A+B2)cos(A–B2)⎤⎦⎥
= 2cos3xcosx+cos3x2sin3xcosx+sin3x
= cos3x(2cosx+1)sin3x(2cosx+1) = cot 3x
= R.H.S.
Q.22: Prove:
cotxcot2x–cot2xcot3x–cot3xcotx=1
Sol:
Now, taking L.H.S.
cot x cot 2x – cot 2x cot 3x – cot 3x cot x :
= cotxcot2x–cot3x(cot2x+cotx)
= cotxcot2x–cot(2x+x)(cot2x+cotx)
= cotxcot2x–[cot2xcotx–1cotx+cot2x](cot2x+cotx)
= [cot(A+B)=cotAcotB–1cotA+cotB]
= cotxcot2x–(cot2xcotx–1) = 1
= R.H.S
Q.23: Prove:
tan4x=4tanx(1–tan2x)1–6tan2x+tan4x
Sol:
Since, tan 2A = 2tanA1–tan2A
Now, taking L.H.S.
tan 4x :
= tan2(2x)
= 2tan2x1–tan2(2x)
= 2(2tanx1–tan2x)1–(2tanx1–tan2x)2
= (4tanx1–tan2x)1–⎛⎝4tan2x(1–tan2x)2⎞⎠
= (4tanx1–tan2x)⎡⎣(1–tan2x)2–4tan2x(1–tan2x)2⎤⎦
= 4tanx(1–tan2x)(1–tan2x)2–4tan2x
= 4tanx(1–tan2x)1+tan4x–2tan2x–4tan2x
= 4tanx(1–tan2x)1–6tan2x+tan4x
= R.H.S.
Q.24: Prove:
cos4x=1–8sin2xcos2x
Sol:
Now, taking L.H.S.
cos 4x:
= cos2(2x)
= 1–2sin22x[cos2A=1–2sin2A]
= 1–2(2sinxcosx)2[sin2A=2sinAcosA]
= 1–8sin2xcos2x
= R.H.S.
Q.25: Prove:
cos6x=32cos6x–48cos4x+18cos2x−1
Sol:
Now, taking L.H.S.
cos 3 (2x ):
= 4cos32x–3cos2x[cos3A=4cos3A–3cosA]
= 4(2cos2x–1)3–3(2cos2x−1)[cos2x=2cos2x−1]
= 4[(2cos2x)3–(1)3–3(2cos2x)2+3(2cos2x)]–6cos2x+3
= 4[8cos6x–1–12cos4x+6cos2x]–6cos2x+3
= 32cos6x–4–48cos4x+24cos2x–6cos2x+3
= 32cos6x–48cos4x+18cos2x−1
= R.H.S.
Exercise 3.4
Q.1: Find general solutions and the principle solutions of the given equation: tan x = 3–√
Sol:
tan x = 3–√ [Given]
As we know that, tanπ3=3–√
And, tan4π3 = tan(π+π3) = tan(π3) = 3–√
Therefore, the principle solutions are x=π3 and 4π3
Now, tanx=tanπ3 x=nπ+π3, where n ∈ Z
Therefore, the general solution is x=nπ+π3, where n ∈ Z.
Q.2: Find general solutions and the principle solutions of the given equation: sec x = 2
Sol:
sec x = 2 [Given]
As we know that, secπ3=2
And, sec5π3 = sec(2π–π3) = secπ3 = 2
Therefore, the principle solutions are x=π3 and 5π3.
Now, secx=secπ3
And, cosx=cosπ3[secx=1cosx] x=2nπ±π3, where n ∈ Z
Therefore, the general solution is x=2nπ±π3, where n ∈ Z.
Q.3: Find general solutions and the principle solutions of the given equation:
cot = −3–√
Sol:
cot = −3–√ [Given]
As we know that, cotπ6=3–√⇒cot(π–π6)=−cotπ6=−3–√
And, cot(2π–π6) = −cotπ6 = −3–√
That is cot5π6=−3–√ and cot11π6=−3–√
Therefore, the principle solutions are x=5π6 and 11π6.
Now,cotx=cot5π6
And, tanx=tan5π6[cotx=1tanx] x=nπ+5π6, where n ∈ Z
Therefore, the general solution is x=nπ+5π6, where n ∈ Z.
Q.4: Find general solutions and the principle solutions of the given equation: cosec x = -2
Sol:
cosec x = -2 [Given]
As we know that, cosecπ6=2
Hence, cosec(π+π6) = −cosecπ6 = -2
And, cosec(2π–π6) = −cosecπ6 = -2
That is cosec7π6=−2 and cosec11π6=−2.
Therefore, the principle solutions are x=7π6 and 11π6.
Now,cosecx=cosec7π6
And, sinx=sin7π6[cosecx=1sinx] x=nπ+(−1)n7π6, where n ∈ Z
Therefore, the general solution is x=nπ+(−1)n7π6, where n ∈ Z.
Q.5: Find the general solution of the given equation: cos 4x = cos 2x
Sol:
cos 4x = cos 2x [Given]
i.e. cos 4x – cos 2x = 0
−2sin(4x+2x2)sin(4x–2x2)=0
Since, cos A – cos B = 2sin(A+B2)sin(A–B2)
(sin 3x) (sin x) = 0
sin 3x = 0 or sin x = 0
sin 3x = 0
3x=nπ
x=nπ3, where n ∈ Z
sin x = 0
⇒ x=nπ, where n ∈ Z
Q.6: Find the general solution of the given equation: cos 3x + cos x – cos 2x = 0
Sol:
cos 3x + cos x – cos 2x = 0 [Given]
2cos(3x+x2)cos(3x–x2)–cos2x=0
Since, cos A + cos B = 2cos(A+B2)cos(A–B2)
2cos 2x cos x – cos 2x = 0
cos 2x (2cos x – 1) = 0
cos 2x = 0 or 2cos x -1 = 0
cos 2x = 0
2x=(2n+1)π2, where n ∈ Z
x=(2n+1)π4, where n ∈ Z
2cos x -1 = 0
cosx=12 cosx=cosπ3, where n ∈ Z
x=2nπ±π3, where n ∈ Z
Q.7: Find the general solution of the given equation: sin 2x + cos x = 0
Sol:
sin 2x + cos x = 0 [Given]
2sin x cos x + cos x = 0
cos x (2sin x + 1) = 0
cos x = 0 or 2sin x + 1 = 0
cos x = 0
cosx=(2n+1)π2, where n ∈ Z
2sin x + 1 = 0
= −sinπ6 = sin(π+π6) = sin7π6 ⇒ x=nπ+(−1)n7π6, where n ∈ Z
Therefore, the general solution is (2n+1)π2 or nπ+(−1)n7π6, where n ∈ Z.
Q.8: Find the general solution of the given equation:
sec22x=1–tan2x
Sol:
sec22x=1–tan2x [Given]
1+tan22x=1–tan2x tan22x+tan2x=0
tan 2x ( tan 2x + 1) = 0
tan 2x = 0 or tan 2x + 1 = 0
tan 2x = 0
tan 2x = tan 0
2x = n∏ + 0, where n ∈ Z
x=nπ2, where n ∈ Z
tan 2x + 1 = 0
tan 2x = -1= −tanπ4= tan(π–π4)= tan3π4 ⇒ 2x=nπ+3π4, where n ∈ Z
⇒ x=nπ2+3π8, where n ∈ Z
Therefore, the general solution is nπ2 or nπ2+3π8, where n ∈ Z.
Q.9: Find the general solution of the given equation: sin x + sin 3x + sin 5x = 0
Sol:
sin x + sin 3x + sin 5x = 0 [Given]
(sin x + sin 5x) + sin 3x = 0
[2sin(x+5x2)cos(x–5x2)]+sin3x=0
Since, sin A + sin B = 2sin(A+B2)cos(A–B2)
2sin 3x cos (-2x) + \sin 3x = 0
2sin 3x cos 2x + sin 3x = 0
sin 3x (2cos 2x + 1) = 0
sin 3x = 0 or 2cos 2x + 1 = 0
Now,
sin 3x = 0
3x=nπ, where n ∈ Z
x=nπ3, where n ∈ Z
2cos 2x + 1 = 0
cos2x=−12 = −cosπ3 = cos(π–π3) = cos2π3 2x=2nπ±2π3, where n ∈ Z
x=nπ±π3, where n ∈ Z
Therefore, the general solution is nπ3 or nπ±π3, where n ∈ Z.
Miscellaneous Exercise
Q.1: Prove that:
2cosπ13cos9π13+cos3π13+cos5π13=0
Sol:
Taking L.H.S.
2cosπ13cos9π13+cos3π13+cos5π13 :
= 2cosπ13cos9π13+2cos(3π13+5π132)cos(3π13–5π132)
Since, [cosx+cosy=2cos(x+y2)cos(x–y2)]
= 2cosπ13cos9π13+2cos4π13cos(−π13)
= 2cosπ13cos9π13+2cos4π13cosπ13
= 2cosπ13[cos9π13+cos4π13]
= 2cosπ13[2cos(9π13+4π132)cos(9π13–4π132)]
= 2cosπ13[2cosπ2cos5π26]
= 2cosπ13×2×0×cos5π26
= 0
= R.H.S.
Therefore, 2cosπ13cos9π13+cos3π13+cos5π13=0
Q.2: Prove that:
(sin3x+sinx)sinx+(cos3x–cosx)cosx=0
Sol:
Taking L.H.S.
(sin3x+sinx)sinx+(cos3x–cosx)cosx :
= sin3xsinx+sin2x+cos3xcosx–cos2x
= cos3xcosx+sin3xsinx–(cos2x–sin2x)
= cos(3x–x)–cos2x
Since, [cos(A−B)=cosAcosB+sinAsinB]
= cos2x–cos2x
= 0
= R.H.S.
Therefore, (sin3x+sinx)sinx+(cos3x–cosx)cosx=0
Q-3: Prove that:
(cosx+cosy)2+(sinx–siny)2=4cos2x+y2
Sol:
Taking L.H.S.
(cosx+cosy)2+(sinx–siny)2 :
= cos2x+cos2y+2cosxcosy+sin2x+sin2y–2sinxsiny
= (cos2x+sin2x)+(cos2y+sin2y)+2(cosxcosy–sinxsiny)
= 1+1+2cos(x+y)
Since, [cos(A+B)=cosAcosB–sinAsinB]
= 2+2cos(x+y)
= 2[1+cos(x+y)]
= 2[1+2cos2(x+y2)–1]
Since, [cos2A=2cos2A–1]
= 4cos2(x+y2)
= R.H.S.
Therefore, (cosx+cosy)2+(sinx–siny)2=4cos2x+y2
Q-4: Prove that:
(cosx–cosy)2+(sinx–siny)2=4sin2x–y2
Sol:
Taking L.H.S.
(cosx–cosy)2+(sinx–siny)2 :
= cos2x+cos2y–2cosxcosy+sin2x+sin2y–2sinxsiny
= (cos2x+sin2x)+(cos2y+sin2y)–2(cosxcosy–sinxsiny)
= 1+1–2[cos(x–y)]
Since, [cos(A−B)=cosAcosB+sinAsinB]
= 2[1–cos(x–y)]
= 2[1–{1–2sin2(x–y2)}]
Since, [cos2A=1–2sin2A]
= 4sin2x–y2
= R.H.S.
Therefore, (cosx–cosy)2+(sinx–siny)2=4sin2x–y2
Q-5: Prove that:
sinx+sin3x+sin5x+sin7x=4cosxcos2xcos4x
Sol:
Taking L.H.S.
sinx+sin3x+sin5x+sin7x :
= (sinx+sin5x)+(sin3x+sin7x)
= 2sin(x+5x2)cos(x–5x2)+2sin(3x+7x2)cos(3x+7x2)
Since, [sinA+sinB=2sin(A+B2).cos(A–B2)]
= 2sin3xcos(−2x)+2sin5xcos(−2x)
= 2sin3xcos2x+2sin5xcos2x
= 2cos2x[sin3x+sin5x]
= 2cos2x[2sin(3x+5x2).cos(3x–5x2)]
= 2cos2x[2sin4x.cos(−x)]
= 4cos2xsin4xcosx
= R.H.S.
Therefore, sinx+sin3x+sin5x+sin7x=4cosxcos2xcos4x
Q-6: Prove that:
(sin7x+sin5x)+(sin9x+sin3x)(cos7x+cos5x)+(cos9x+cos3x)=tan6x
Sol:
Since, sinA+sinB=2sin(A+B2).cos(A−B2)
And, cosA+cosB=2cos(A+B2).cos(A−B2)
Taking L.H.S.
(sin7x+sin5x)+(sin9x+sin3x)(cos7x+cos5x)+(cos9x+cos3x) :
= [2sin(7x+5x2).cos(7x–5x2)]+[2sin(9x+3x2).cos(9x–3x2)][2cos(7x+5x2).cos(7x–5x2)]+[2cos(9x+3x2).cos(9x–3x2)]
= [2sin6x.cosx]+[2sin6x.cos3x][2cos6x.cosx]+[2cos6x.cos3x]
= 2sin6x[cosx+cos3x]2cos6x[cosx+cos3x]
= tan6x
= R.H.S.
Therefore, (sin7x+sin5x)+(sin9x+sin3x)(cos7x+cos5x)+(cos9x+cos3x)=tan6x
Q-7: Show that: sin3y+sin2y–siny=4sinycosy2cos3y2
Sol:
Here, L.H.S = sin3y+sin2y–siny
= sin3y+(sin2y−siny)=sin3y+[2cos(2y+y2)sin(2y–y2)]
Since, (sinx–siny)=[2cos(x+y2)sin(x–y2)]
= sin3y+[2cos(3y2)sin(y2)]
=[2sin(3y2)cos(3y2)]+[2cos(3y2)sin(y2)]
Since, sin 2x = 2 sin x cos x
=2cos3y2⋅[sin 3y2+siny2]
=2cos3y2⋅[sin (3y2+y2)2][cos (3y2–y2)2]
Since, sin x + sin y = 2sin(x+y2)cos(x–y2)
=2cos(3y2)2sinycos(y2) =4sinycos(y2)cos(3y2)
= R.H.S
Q-8: The value of tany=−43 where y in in 2nd quadrant then find out the values of siny2,cosy2andtany2.
Sol:
Here, y is in 2nd quadrant.
So, π2<y<π⇒π4<y2<π2
Thus, siny2,cosy2andtany2 lies in 1st quadrant.
Now,
tany=−43 sec2y=1+tan2y=1+(43)2=1+169=259
So, cos2y=925 ⇒cosy=±35
As y is in 2nd quadrant, cos y is negative.
cosy=−35
So, cos y = 2cos2y2–1
⇒−35=2cos2y2–1⇒2cos2y2=1–−35⇒2cos2y2=25⇒2cos2y2=15⇒2cos2y2=35√ [Since, cosy2 is positive]
⇒cosy2=5√5sin2y2+cos2y2=1⇒sin2y2+cos215√=1⇒sin2y2=1–15=45⇒sin2y2=25√ [Since, siny2 is positive]
⇒sin2y2=25√5tany2=siny2cosy2=25√15√=2
Q-9: The value of cosy=−13 where y in in 3rd quadrant then find out the values of siny2,cosy2andtany2.
Sol:
Here, y is in 3rd quadrant.
So, π<y<3π2⇒π2<y2<3π4
Thus, cosy2andtany2 are negative and siny2 is positive.
Now, cosy=−13 [Given]
cosy=1–2sin2y2⇒sin2y2=1–cosy2⇒sin2y2=1–−132=1+132=23 ⇒siny2=2√3√×3√3√=6√3 [Since, siny2 is positive]
Now, cosy=2cos2y2–1 2cos2y2=1+cosy2=1–132=13 ⇒ cosy2=−13√=−13√×−3√3=−13√
Therefore, tany2=siny2cosy2=2√3√−13√=−2–√
Q-10: The value of siny=14 where y in in 2nd quadrant then find out the values of siny2,cosy2andtany2.
Sol:
Here, y is in 2nd quadrant.
So, π2<y<π⇒π4<y2<π2
Thus, siny2,cosy2andtany2 lies in 1st quadrant.
Now, siny=14 cos2y=1–2sin2y=1–(14)2=1–116=1516⇒cosy=−15√4 [Since, cosy is negative]
sin2y2=1–(−15√4)2=4+15√8 ⇒ siny2=4+15√8−−−−−√ [ Since, siny2 is positive ]
=4+15√8×22−−−−−−−−√=8+215√16−−−−−−√=8+215√√4 cos2y=1+cosy2=1+(−15√4)2=4–15√8
Therefore, cosy2=4–15√8−−−−−√ = 4–15√8×22−−−−−−−−√
= 8–215√16−−−−−√=8–215√√4
Now, tany2=siny2cosy2=(8+215√√4)(8−215√√4)
= 8+215√√8–215√√=8+215√√8–215√√×8+215√√8+215√√
= (8+215√)2√64−60√=8+215√2=4+15−−√
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