Exercise 3.1
Find the radian measures corresponding to the following degree measures:
Answer
(i) 25° (ii) −47°30′ (iii) 240° (iv) 520°
Ans. Use: Radian = (π/180) × Degree
(i) 25° = 25 × π/180 = 5π/36 radians
(ii) −47°30′ = −47.5° = −47.5 × π/180 = −19π/72 radians
(iii) 240° = 240 × π/180 = 4π/3 radians
(iv) 520° = 520 × π/180 = 26π/9 radians
Find the degree measures corresponding to the following radian measures (use π = 22/7):
Answer
(i) 11/16 (ii) −4 (iii) 5π/3 (iv) 7π/6
Ans. Use: Degree = (180/π) × Radian
(i) 11/16 × 180/π = 11/16 × 180 × 7/22 = 11 × 180 × 7 / (16 × 22) = 13860/352 = 39°22′30″
(ii) −4 × 180/π = −4 × 180 × 7/22 = −5040/22 = −229°5′27″ (approximately −229.09°)
(iii) 5π/3 × 180/π = 5 × 60 = 300°
(iv) 7π/6 × 180/π = 7 × 30 = 210°
A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?
Answer
In 1 minute = 60 seconds, the wheel makes 360 revolutions. In 1 second: 360/60 = 6 revolutions. Each revolution = 2π radians. Radians per second = 6 × 2π = 12π radians.
Find the degree measure of the angle subtended at the centre of a circle of radius 100 cm by an arc of length 22 cm (use π = 22/7).
Answer
θ = l/r = 22/100 = 11/50 radians. Degree = 11/50 × 180/π = 11 × 180 × 7 / (50 × 22) = 13860/1100 = 12°36′.
In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of the minor arc of the chord.
Answer
Radius r = 20 cm. Chord = 20 cm = r. So the triangle formed by the two radii and the chord is equilateral, giving the central angle θ = 60° = π/3 radians. Arc length = rθ = 20 × π/3 = 20π/3 cm.
If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find the ratio of their radii.
Answer
Let radii be r1 and r2. Arc length l is same. l = r1θ1 = r2θ2. θ1 = π/3, θ2 = 5π/12. So r1/r2 = θ2/θ1 = (5π/12)/(π/3) = 5/4. Ratio r1 : r2 = 5 : 4.
Find the angle in radian through which a pendulum swings if its length is 75 cm and the tip describes an arc of length: (i) 10 cm (ii) 15 cm (iii) 21 cm
Answer
θ = l/r where r = 75 cm.
(i) θ = 10/75 = 2/15 radians
(ii) θ = 15/75 = 1/5 radians
(iii) θ = 21/75 = 7/25 radians
Exercise 3.2
cos x = −1/2, x lies in the third quadrant. Find the values of other five trigonometric functions.
Answer
cos x = −1/2
⇒ sec x = −2. sin²x = 1 − cos²x = 1 − 1/4 = 3/4
⇒ sin x = ±√3/2. In Q3: sin x = −√3/2. cosec x = −2/√3 = −2√3/3. tan x = sin x/cos x = (−√3/2)/(−1/2) = √3. cot x = 1/√3 = √3/3.
sin x = 3/5, x lies in second quadrant. Find the other five trigonometric functions.
Answer
sin x = 3/5
⇒ cosec x = 5/3. cos²x = 1 − 9/25 = 16/25
⇒ cos x = ±4/5. In Q2: cos x = −4/5. sec x = −5/4. tan x = (3/5)/(−4/5) = −3/4. cot x = −4/3.
cot x = 4/3, x lies in third quadrant. Find the other five trigonometric functions.
Answer
cot x = 4/3
⇒ tan x = 3/4. sec²x = 1 + tan²x = 1 + 9/16 = 25/16
⇒ sec x = ±5/4. In Q3: sec x = −5/4, so cos x = −4/5. sin x = tan x × cos x = (3/4)(−4/5) = −3/5. cosec x = −5/3.
sec x = 13/5, x lies in fourth quadrant. Find the other five trigonometric functions.
Answer
sec x = 13/5
⇒ cos x = 5/13. sin²x = 1 − 25/169 = 144/169
⇒ sin x = ±12/13. In Q4: sin x = −12/13. cosec x = −13/12. tan x = (−12/13)/(5/13) = −12/5. cot x = −5/12.
tan x = −5/12, x lies in second quadrant. Find the other five trigonometric functions.
Answer
tan x = −5/12
⇒ cot x = −12/5. sec²x = 1 + 25/144 = 169/144
⇒ sec x = ±13/12. In Q2: sec x = −13/12, so cos x = −12/13. sin x = tan x × cos x = (−5/12)(−12/13) = 5/13. cosec x = 13/5.
Find sin 765°.
Answer
765° = 2×360° + 45°. sin 765° = sin 45° = 1/√2 = √2/2.
Find cosec(−1410°).
Answer
−1410° = −4×360° + 30° (since −1410 + 1440 = 30). cosec(−1410°) = cosec(30°) = 2.
Find tan(19π/3).
Answer
19π/3 = 6π + π/3 = 3(2π) + π/3. tan(19π/3) = tan(π/3) = √3.
Find sin(−11π/3).
Answer
−11π/3 = −4π + π/3. sin(−11π/3) = sin(π/3) = √3/2.
Find cot(−15π/4).
Answer
−15π/4 = −4π + π/4. cot(−15π/4) = cot(π/4) = 1.
Exercise 3.3
Prove that: sin²(π/6) + cos²(π/3) − tan²(π/4) = −1/2
Answer
sin(π/6) = 1/2, cos(π/3) = 1/2, tan(π/4) = 1.
LHS = (1/2)² + (1/2)² − (1)² = 1/4 + 1/4 − 1 = 1/2 − 1 = −1/2 = RHS. ✓
Prove that: 2sin²(π/6) + cosec²(7π/6)cos²(π/3) = 3/2
Answer
sin(π/6) = 1/2
⇒ sin²(π/6) = 1/4. cosec(7π/6) = cosec(π + π/6) = −cosec(π/6) = −2
⇒ cosec²(7π/6) = 4. cos(π/3) = 1/2
⇒ cos²(π/3) = 1/4.
LHS = 2(1/4) + 4 × (1/4) = 1/2 + 1 = 3/2 = RHS. ✓
Prove that: cot²(π/6) + cosec(5π/6) + 3tan²(π/6) = 6
Answer
cot(π/6) = √3
⇒ cot²(π/6) = 3. cosec(5π/6) = cosec(π − π/6) = cosec(π/6) = 2. tan(π/6) = 1/√3
⇒ tan²(π/6) = 1/3.
LHS = 3 + 2 + 3(1/3) = 3 + 2 + 1 = 6 = RHS. ✓
Prove that: 2sin²(3π/4) + 2cos²(π/4) + 2sec²(π/3) = 10
Answer
sin(3π/4) = sin(π − π/4) = sin(π/4) = 1/√2
⇒ sin²(3π/4) = 1/2. cos(π/4) = 1/√2
⇒ cos²(π/4) = 1/2. sec(π/3) = 2
⇒ sec²(π/3) = 4.
LHS = 2(1/2) + 2(1/2) + 2(4) = 1 + 1 + 8 = 10 = RHS. ✓
Find the value of: (i) sin 75° (ii) tan 15°
Answer
(i) sin 75° = sin(45° + 30°) = sin45°cos30° + cos45°sin30° = (1/√2)(√3/2) + (1/√2)(1/2) = (√3+1)/(2√2) = (√6+√2)/4
(ii) tan 15° = tan(45° − 30°) = (tan45° − tan30°)/(1 + tan45°tan30°) = (1 − 1/√3)/(1 + 1/√3) = (√3−1)/(√3+1) = (√3−1)²/2 = (4−2√3)/2 = 2−√3
Prove that: cos(π/4 − x)cos(π/4 − y) − sin(π/4 − x)sin(π/4 − y) = sin(x + y)
Answer
The LHS is of the form cos(A)cos(B) − sin(A)sin(B) = cos(A + B), where A = π/4 − x and B = π/4 − y.
LHS = cos((π/4−x) + (π/4−y)) = cos(π/2 − (x+y)) = sin(x+y) = RHS. ✓
Prove that: tan(π/4 + x)/tan(π/4 − x) = ((1 + tan x)/(1 − tan x))²
Answer
tan(π/4 + x) = (1 + tan x)/(1 − tan x). tan(π/4 − x) = (1 − tan x)/(1 + tan x).
LHS = [(1+tan x)/(1−tan x)] ÷ [(1−tan x)/(1+tan x)] = [(1+tan x)/(1−tan x)]² = RHS. ✓
Prove that: [cos(π + x)cos(−x)] / [sin(π − x)cos(π/2 + x)] = cot²x
Answer
cos(π + x) = −cos x; cos(−x) = cos x; sin(π − x) = sin x; cos(π/2 + x) = −sin x.
LHS = (−cos x)(cos x) / [(sin x)(−sin x)] = (−cos²x)/(−sin²x) = cos²x/sin²x = cot²x = RHS. ✓
Prove that: cos(3π/2 + x)cos(2π + x)[cot(3π/2 − x) + cot(2π + x)] = 1
Answer
cos(3π/2 + x) = sin x; cos(2π + x) = cos x; cot(3π/2 − x) = tan x; cot(2π + x) = cot x.
LHS = sin x · cos x · (tan x + cot x) = sin x · cos x · (sin x/cos x + cos x/sin x) = sin x · cos x · (sin²x + cos²x)/(sin x cos x) = 1 = RHS. ✓
Prove that: sin(n+1)x sin(n+2)x + cos(n+1)x cos(n+2)x = cos x
Answer
LHS = cos((n+1)x − (n+2)x) [using cos(A−B) = cosAcosB + sinAsinB] = cos(−x) = cos x = RHS. ✓
Prove that: cos(3π/4 + x) − cos(3π/4 − x) = −√2 sin x
Answer
Using cos A − cos B = −2sin((A+B)/2)sin((A−B)/2): A = 3π/4+x, B = 3π/4−x.
(A+B)/2 = 3π/4, (A−B)/2 = x. LHS = −2sin(3π/4)sin(x) = −2(√2/2)sin x = −√2 sin x = RHS. ✓
Prove that: sin²6x − sin²4x = sin 2x sin 10x
Answer
sin²A − sin²B = sin(A+B)sin(A−B). With A = 6x, B = 4x:
LHS = sin(10x)sin(2x) = sin 2x sin 10x = RHS. ✓
Prove that: cos²2x − cos²6x = sin 4x sin 8x
Answer
cos²A − cos²B = −sin(A+B)sin(A−B)... more directly: cos²A − cos²B = (cosA+cosB)(cosA−cosB).
cos 2x + cos 6x = 2cos4x cos2x; cos 2x − cos 6x = 2sin4x sin2x.
LHS = (2cos4x cos2x)(2sin4x sin2x) = 4sin4x cos4x sin2x cos2x... Let us use sum-to-product: cos²A − cos²B = sin(A+B)sin(B−A).
= sin(2x+6x)sin(6x−2x) = sin8x · sin4x = sin4x sin8x = RHS. ✓
Prove that: sin²x + 2sin4x + sin6x = 4cos²x sin4x
Answer
LHS = (sin²x + sin6x) + 2sin4x = 2sin4x cos2x + 2sin4x = 2sin4x(cos2x + 1) = 2sin4x · 2cos²x = 4cos²x sin4x = RHS. ✓
Prove that: cot4x(sin5x + sin3x) = cot x(sin5x − sin3x)
Answer
sin5x + sin3x = 2sin4x cosx; sin5x − sin3x = 2cos4x sinx.
LHS = cot4x · 2sin4x cosx = 2cos4x cosx. RHS = cot x · 2cos4x sinx = 2cos4x cosx. Equal. ✓
Prove that: (cos9x − cos5x)/(sin17x − sin3x) = −sin2x/cos10x
Answer
Numerator: cos9x − cos5x = −2sin7x sin2x. Denominator: sin17x − sin3x = 2cos10x sin7x.
LHS = (−2sin7x sin2x)/(2cos10x sin7x) = −sin2x/cos10x = RHS. ✓
Prove that: (sin5x + sin3x)/(cos5x + cos3x) = tan4x
Answer
sin5x + sin3x = 2sin4x cosx; cos5x + cos3x = 2cos4x cosx. LHS = 2sin4x cosx / (2cos4x cosx) = tan4x = RHS. ✓
Prove that: (sinx − siny)/(cosx + cosy) = tan((x−y)/2)
Answer
sinx − siny = 2cos((x+y)/2)sin((x−y)/2); cosx + cosy = 2cos((x+y)/2)cos((x−y)/2).
LHS = 2cos((x+y)/2)sin((x−y)/2) / (2cos((x+y)/2)cos((x−y)/2)) = tan((x−y)/2) = RHS. ✓
Prove that: (sinx + sin3x)/(cosx + cos3x) = tan2x
Answer
sinx + sin3x = 2sin2x cosx; cosx + cos3x = 2cos2x cosx. LHS = 2sin2x cosx / (2cos2x cosx) = tan2x = RHS. ✓
Prove that: (sinx − sin3x)/(sin²x − cos²x) = 2sinx
Answer
sinx − sin3x = −2cosx sin(−x) ... let us use: sinx − sin3x = 2cos2x sin(−x) = −2cos2x sinx. sin²x − cos²x = −cos2x. LHS = (−2cos2x sinx)/(−cos2x) = 2sinx = RHS. ✓
Prove that: (cos4x + cos3x + cos2x)/(sin4x + sin3x + sin2x) = cot3x
Answer
Numerator: cos4x + cos2x = 2cos3x cosx; add cos3x: 2cos3x cosx + cos3x = cos3x(2cosx + 1). Denominator: sin4x + sin2x = 2sin3x cosx; add sin3x: 2sin3x cosx + sin3x = sin3x(2cosx + 1). LHS = cos3x(2cosx+1)/(sin3x(2cosx+1)) = cot3x = RHS. ✓
Prove that: cot x cot 2x − cot 2x cot 3x − cot 3x cot x = 1
Answer
cot3x = cot(x + 2x) = (cotx cot2x − 1)/(cotx + cot2x). So cotx + cot2x = (cotx cot2x − 1)/cot3x, giving cot3x(cotx + cot2x) = cotx cot2x − 1. Expanding: cot3x cotx + cot3x cot2x = cotx cot2x − 1. Rearranging: cotx cot2x − cot2x cot3x − cot3x cotx = 1. ✓
Prove that: tan4x = 4tanx(1−tan²x)/(1−6tan²x+tan⁴x)
Answer
tan2x = 2tanx/(1−tan²x). tan4x = 2tan2x/(1−tan²2x) = 2·[2tanx/(1−tan²x)] / [1 − (2tanx/(1−tan²x))²] = [4tanx/(1−tan²x)] / [(1−tan²x)² − 4tan²x]/(1−tan²x)²] = 4tanx(1−tan²x)/[(1−tan²x)² − 4tan²x] = 4tanx(1−tan²x)/(1 − 2tan²x + tan⁴x − 4tan²x) = 4tanx(1−tan²x)/(1−6tan²x+tan⁴x). ✓
Prove that: cos4x = 1 − 8sin²x cos²x
Answer
cos4x = 1 − 2sin²2x = 1 − 2(2sinx cosx)² = 1 − 8sin²x cos²x. ✓
Prove that: cos6x = 32cos⁶x − 48cos⁴x + 18cos²x − 1
Answer
cos6x = cos2(3x) = 2cos²(3x) − 1. cos3x = 4cos³x − 3cosx. cos²3x = (4cos³x−3cosx)² = 16cos⁶x − 24cos⁴x + 9cos²x. cos6x = 2(16cos⁶x − 24cos⁴x + 9cos²x) − 1 = 32cos⁶x − 48cos⁴x + 18cos²x − 1. ✓
Miscellaneous Exercise on Chapter 3
Prove that: cos(π/13)cos(2π/13)cos(3π/13)cos(4π/13)cos(5π/13)cos(6π/13) = 1/64
Answer
Let θ = π/13. Use the identity: 26 sinθ · cosθ cos2θ cos4θ ... involves the product formula. Multiply both sides by 26sinθ and repeatedly use 2sinθ cosθ = sin2θ. We get 64sinθ · (product) = sin(64θ) = sin(64π/13). Since 64 = 4×13 + 12: sin(64π/13) = sin(4π + 12π/13) = sin(12π/13) = sin(π − π/13) = sin(π/13) = sinθ. So 64sinθ · product = sinθ
⇒ product = 1/64. ✓
Prove that: (sin3x + sinx)sinx + (cos3x − cosx)cosx = 0
Answer
sin3x + sinx = 2sin2x cosx; cos3x − cosx = −2sin2x sinx.
LHS = 2sin2x cosx · sinx + (−2sin2x sinx) · cosx = 2sin2x sinx cosx − 2sin2x sinx cosx = 0. ✓
Prove that: (cosx + cosy)² + (sinx − siny)² = 4cos²((x+y)/2)
Answer
cosx + cosy = 2cos((x+y)/2)cos((x−y)/2); sinx − siny = 2cos((x+y)/2)sin((x−y)/2).
LHS = [2cos((x+y)/2)cos((x−y)/2)]² + [2cos((x+y)/2)sin((x−y)/2)]²
= 4cos²((x+y)/2)[cos²((x−y)/2) + sin²((x−y)/2)] = 4cos²((x+y)/2). ✓
Prove that: (cosx − cosy)² + (sinx − siny)² = 4sin²((x−y)/2)
Answer
cosx − cosy = −2sin((x+y)/2)sin((x−y)/2); sinx − siny = 2cos((x+y)/2)sin((x−y)/2).
LHS = 4sin²((x+y)/2)sin²((x−y)/2) + 4cos²((x+y)/2)sin²((x−y)/2) = 4sin²((x−y)/2)[sin²((x+y)/2) + cos²((x+y)/2)] = 4sin²((x−y)/2). ✓
Prove that: sinx + sin3x + sin5x + sin7x = 4cosx cos2x sin4x
Answer
Group: (sin7x + sinx) + (sin5x + sin3x) = 2sin4x cos3x + 2sin4x cosx = 2sin4x(cos3x + cosx) = 2sin4x · 2cos2x cosx = 4cosx cos2x sin4x. ✓
Prove that: (sin7x + sin5x + sin9x + sin3x)/(cos7x + cos5x + cos9x + cos3x) = tan6x
Answer
Numerator: (sin7x+sin5x) + (sin9x+sin3x) = 2sin6x cosx + 2sin6x cos3x = 2sin6x(cosx + cos3x) = 2sin6x · 2cos2x cosx. Denominator: (cos7x+cos5x) + (cos9x+cos3x) = 2cos6x cosx + 2cos6x cos3x = 2cos6x · 2cos2x cosx. LHS = (4sin6x cos2x cosx)/(4cos6x cos2x cosx) = tan6x. ✓
Prove that: sin3x + sin2x − sinx = 4sinx cos(x/2) cos(3x/2)
Answer
sin3x − sinx = 2cos2x sinx. So LHS = 2cos2x sinx + sin2x = 2cos2x sinx + 2sinx cosx = 2sinx(cos2x + cosx) = 2sinx · 2cos(3x/2)cos(x/2) = 4sinx cos(x/2)cos(3x/2). ✓
Find sin(x/2), cos(x/2) and tan(x/2) if tan x = −4/3, x in quadrant II.
Answer
tan x = −4/3
⇒ sec²x = 1 + 16/9 = 25/9
⇒ cos x = −3/5 (negative in Q2). Since x is in Q2, x/2 is in Q1 (all positive). cos(x/2) = √((1+cosx)/2) = √((1−3/5)/2) = √(1/5) = 1/√5 = √5/5. sin(x/2) = √((1−cosx)/2) = √((1+3/5)/2) = √(4/5) = 2/√5 = 2√5/5. tan(x/2) = sin(x/2)/cos(x/2) = 2.
Find sin(x/2), cos(x/2) and tan(x/2) if cos x = −1/3, x in quadrant III.
Answer
x is in Q3: π < x < 3π/2, so π/2 < x/2 < 3π/4 (Q2: sin positive, cos negative). cos(x/2) = −√((1+cosx)/2) = −√((1−1/3)/2) = −√(1/3) = −1/√3 = −√3/3. sin(x/2) = √((1−cosx)/2) = √((1+1/3)/2) = √(2/3) = √6/3. tan(x/2) = sin/cos = (√6/3)/(−√3/3) = −√2.
Find sin(x/2), cos(x/2) and tan(x/2) if sin x = 1/4, x in quadrant II.
Answer
cos²x = 1 − 1/16 = 15/16
⇒ cos x = −√15/4 (Q2). x/2 in Q1 (positive). cos(x/2) = √((1+cosx)/2) = √((1−√15/4)/2) = √((4−√15)/8). sin(x/2) = √((1−cosx)/2) = √((1+√15/4)/2) = √((4+√15)/8). tan(x/2) = √((4+√15)/(4−√15)) = √((4+√15)²/(16−15)) = 4+√15.
Important Extra Questions
What is the period of sin x and cos x?
Answer
Both sin x and cos x have a period of 2π. This means sin(x + 2π) = sin x and cos(x + 2π) = cos x for all x.
Prove that sin(A + B) + sin(A − B) = 2sinA cosB.
Answer
sin(A+B) = sinA cosB + cosA sinB; sin(A−B) = sinA cosB − cosA sinB. Adding: 2sinA cosB. ✓
Find the maximum and minimum values of sin x + cos x.
Answer
sin x + cos x = √2 sin(x + π/4). Maximum = √2 (when sin(x+π/4) = 1); Minimum = −√2 (when sin(x+π/4) = −1).
Show that tan 3x tan 2x tan x = tan 3x − tan 2x − tan x.
Answer
3x = 2x + x, so tan3x = tan(2x+x) = (tan2x + tanx)/(1 − tan2x tanx). Thus tan3x(1 − tan2x tanx) = tan2x + tanx
⇒ tan3x − tan3x tan2x tanx = tan2x + tanx
⇒ tan3x − tan2x − tanx = tan3x tan2x tanx. ✓
The value of sin 18° is (√5 − 1)/4. Use this to find cos 36°.
Answer
cos36° = 1 − 2sin²18° = 1 − 2((√5−1)/4)² = 1 − 2(6−2√5)/16 = 1 − (6−2√5)/8 = (8−6+2√5)/8 = (2+2√5)/8 = (1+√5)/4.