Q.457. Trigonometric inequality.

Question 457.
Prove that (sin^3 A)/(sin B) + (cos^3 A)/(cos B) ≥ sec ( A - B ) for all 0 < a,b < π/2.

Answer 457.
(sin^3 A)/(sin B) + (cos^3 A)/(cos B) ≥ sec ( A - B )
<=> (sin^3 A cosB + cos^A sinB) / sinB cosB ≥ 1/cos(A - B)
<=> (sin^3 A * cosB + cos^3 A * sinB) * cos(A - B) ≥ sinB cosB
<=> (sin^2 A * 2sinA cosB + cos^2 A * 2cosA sinB) * cos(A - B) ≥ 2sinB cosB
<=> [sin^2 A {sin(A+B) + sin(A-B)} + cos^2 A {sin(A+B) - sin(A-B)}] * cos(A - B) ≥ sin2B
<=> [(sin^2 A + cos^2 A) sin(A + B) - (cos^2 A - sin^2 A) sin(A - B)] * cos(A - B) ≥ sin2B
<=> sin(A + B) cos(A - B) - cos2A sin(A - B) cos(A - B) ≥ sin2B
<=> 2sin(A + B) cos(A - B) - cos2A * 2sin(A – B) cos(A - B) ≥ 2sin2B
<=> sin2A + sin2B - cos2A * 2sin(A – B) cos(A - B) ≥ 2sin2B
<=> sin2A - sin2B - cos2A * 2sin(A – B) cos(A - B) ≥ 0
<=> 2cos(A + B) sin(A - B) - cos2A * 2sin(A - B) cos(A - B) ≥ 0
<=> sin(A - B) [cos(A + B) - cos2A cos(A - B)] ≥ 0
<=> sin(A - B) [2cos(A + B) - 2cos2A cos(A - B)] ≥ 0
<=> sin(A - B) [2cos(A + B) - cos(3A - B) - cos(A + B)] ≥ 0
<=> sin(A - B) [cos(A + B) - cos(3A - B)] ≥ 0
<=> 2sin2A sin^2 (A - B) ≥ 0
which is true
=> the given inequality is true.

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Q.455. Geometry challenge

Question 455.
A circle (radius = r), and an equilateral triangle (side = 2r), fit perfectly in a square, as shown in the diagram.
What is (length CD) divided by (height of triangle)?

Answer 455.

Trigonometric Proof:

Refer to the figure:

Let O be the center of the circle.
With r = 1, (The required ratio is independent of the value of r.)
vertical side of the square
= 1 + ODcos60° + CDcos30°
= 1 + 1/2 + (√3/2) CD
= 3 + (√3/2) CD ... ( 1 )

Horizontal side of the square
= 1 + ODcos30° + (BC - CDcos60°)
= 1 + √3/2 + 2 - CD/2
= 3 + √3/2 - (1/2) CD ... ( 2 )
Equatting ( 1 ) and ( 2 ),
3/2 + (√3/2) CD = 3 + √3/2 - (1/2) CD
=> (√3 - 1)/2 CD = (√3 - 3)/2
=> CD = √3
and Height of the triangle = 2 cos30° = √3
=> CD/Height of the triangle
= √3 / √3
= 1.
=======================================…
Proof using Co-ordinate Geometry:
Refer to the figure:

Consider the given drawn inverted as above.

The required ratio, being independent of the radius, let r = 1
=> The eqn. of the circle with its center as origin is
x^2 + y^2 = 1 ... ( 1 )

Let the length of side of the square = a
=> B = (a-1, a-1), C = (a-3, a-1) and A = (a-2, a-1-√3)
Slope of AC = - √3
Let the eqn. of the tangent AC be y = - √3x + c
=> c = r √(1 + m^2) = 2
=> eqn. of AC is y = - √3x + 2 ... ( 2 )

Solving eqn. ( 1 ) and ( 2 ) gives
x^2 + (-√3x + 2)^2 = 1
=> 4x^2 - 4√3x + 3 = 0
=> (2x - √3) = 0
=> x-coordinate of D is √3/2
Plugging in eqn. ( 2 ),
y-coordinate is 1/2
=> D = (- √3/2, 1/2)

Plugging coordinates of C in eqn. ( 2 ),
a -1 = - √3 (a - 3) + 2
=> a = (3√3 + 3) / (√3 + 1) = 3
=> C = (0, 2)

CD^2 = (0 + √3/2)^2 + (3/2)^2 = 3
=> CD = √3
and Height of the triangle = 2 cos30° = √3
=> CD/Height of the triangle
= √3 / √3
= 1.

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Q.452. Trigonometric inequality.

Question 452.
In any triangle ABC,  prove that : sin A sin B sin C  ≤  3√3/8.

Answer 452.
f = sinA sinB sinC
= sinA sinB sin[π - (A+B)]
= sinA sinB sin(A+B)
= sinA sinB (sinA cosB + cosA sinB)
= sin^2 A sinB cosB + sin^2 B sinA cosA

To find the maximum value,
δf/δA = 0
=> 2sinA cosA sinB cosB + sin^2 B * cos2A = 0
=> sin2A cosB + cos2A sinB = 0
=> sin(2A + B) = 0
=> 2A + B = π ... ( 1 ) [cannot be zero or higher multiple of π for traingle ABC]

Similarly,
δf/δB = 0
=> 2B + A = π ... ( 2 )

Solving ( 1 ) and ( 2 ),
A = B = π/3 => C = π/3
=> sinA sinB sinC = [sin(π/3)]^3 = (3√3)/8
This can be shown to be maximum and not minimum by taking some arbitrary values of A, B and C.
Taking A = π/2, B = C = π/4
=> sinA sinB sinC = 1/2
which shows that (3√3)/8 is the maximum value of sinA sinB sinC
=> sinA sinB sinC ≤ (3√3)/8.
=======================================…

For the benefit of the readers, I reproduce the elegant solution given on page 4-66 of the following link.

Click to open link.

Consider three points P(A, sinA), Q(B, sinB) and R(C, sinC)
on the curve y = sinx such that A + B + C = π
Centroid of ΔPQR = [π/3, (1/3)(sinA + sinB + sinC)]
Therefore, centroid lies on x = π/3 and is inside the triangle,
=> (1/3) (sinA + sinB + sinC) ≤ sin(π/3) = √3/2
=> (sinA sinB sinC)^(1/3) ≤ √3/2 ... [ GM ≤ AM ]
=> sinA sinB sinC ≤ (3√3)/8.

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Q.451. Challenging area problem of geometry.

Question 451.
Equilateral triangle PQR is on square ABCD, as at the picture:
For given areas of right triangles (22 and 23) find unknown trapezoidal area, exact value.

Answer 451.
Let ∠ DRP = x
=> ∠ CRQ = 120° - x
Let a = length of the side of the equilateral triangle
=> (1/2) acosx * asinx = 22
=> a^2/4 sin2x = 22 ... ( 1 )
Similarly,
a^2/4 sin(240° - 2x) = 23 ... ( 2 )

Taking ratio ( 2 ) to ( 1 ),
sin(240° - 2x) / sin2x = 23/22
=> 22 * [- (√3/2) cos2x + (1/2) sin2x] = 23 * sin2x
=> - 11√3 cos2x = 12sin2x
=> tan2x = - (11√3)/12
=> sin2x = (11√3)/13√3 = 11/13
and cos2x = sin2x/tan2x = - 12/(13√3)

Plugging sin2x in ( 1 ),
a^2 = 88 / (11/13) = 104

Area of triangle = (√3/4) a^2 = 26√3

Length of side of the square
= acosx + acos(120° - x)
= acosx - (a/2)cosx + (a√3/2) sinx
= (a/2) cosx + (a√3/2) sinx

Area of the square
= [(a/2) cosx + (a√3/2) sinx]^2
= a^2 [(1/4) cos^2 x + (3/4) sin^2 x + √3 sinx cosx]
= (104) * [(1/8)(1 + cos2x) + (3/8)(1 - cos2x) + (√3/2) sin2x]
= (104) * [(1/2) - (1/4) cos2x + (√3/2) sin2x]
= (104) * [(1/2) + (1/4) (12/13√3) + (√3/2) * (11/13)]
= 52 + 30√3

=> area of the trapezium
= area of the square - area of equilateral triangle - areas of the two right triangles
= 52 + 30√3 - 22 - 23 - 26√3
= 7 + 4√3.

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Q.447. Elimination of θ from two equations.

Question 447.
Eliminate θ from (x/a) cos θ + (y/b) sin θ = 1 and xsin θ - y cos θ  = √(a² sin² θ + b² cos²θ).

Answer 447.
(x/a) cosθ + (y/b) sinθ = 1
=> (x/a)^2 cos^2 θ + 2(xy/ab) sinθ cosθ + (y/b)^2 sin^2 θ = 1
=> (x/a)^2 + 2(xy/ab) tanθ + (y/b)^2 tan^2 θ = sec^2 θ
=> (x/a)^2 + 2(xy/ab) tanθ + (y/b)^2 tan^2 θ = 1 + tan^2 θ
=> [(y/b)^2 - 1] tan^2 θ + 2(xy/ab) tanθ + [(x/a)^2 - 1] = 0
=> [(y^2 - b^2)/(x^2 - a^2)] * (a/b)^2 tan^2 θ + [2xy/(x^2 - a^2)] * (a/b) tanθ + 1 = 0 ... ( 1 )

xsinθ - ycosθ =√(a² sin² θ + b² cos² θ)
=> x^2 sin^2 θ - 2xysinθ cosθ + y^2 cos^2θ = a^2 sin^2 θ + b^2 cos^2 θ
=> x^2 tan^2 θ - 2xy tanθ + y^2 = a^2 tan^2 θ + b^2
=> (x^2 - a^2) tan^2 θ - 2xy tanθ + (y^2 - b^2) = 0
=> [(x^2 - a^2)/(y^2 - b^2)] tan^2 θ - [2xy / (y^2 - b^2)] tanθ + 1 = 0 ... ( 2 )

Comparing eqns. ( 1 ) and ( 2 ),

using coefficients of tanθ and constant terms,
[2xy/(x^2 - a^2)] * (a/b) / [- 2xy / (y^2 - b^2)] = 1
=> [1/(x^2 - a^2)] * (a/b) = - 1/(y^2 - b^2)
=> - a (y^2 - b^2) = b(x^2 - a^2)
=> bx^2 + ay^2 = ba^2 + ab^2
=> x^2/a + y^2/b = a + b.

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Q.446. Trigonometric Equation.

Question 446.
Solve 2sin(10º)sin(20º + θ) = sin(θ) for 0º < θ < 90º.

Answer 446.
2sin(10º)sin(20º + θ) = sin(θ)

=> cos(θ + 10º) - cos(θ + 30º) = sinθ
=> cos(θ + 10º) = cos(θ + 30º) + cos(90º - θ)
=> cos(θ + 10º) = 2cos60º cos(θ - 30º)
=> cos(θ + 10º) = cos(θ - 30º)
=> θ + 10º = ± (θ - 30º)
=> 2θ = 20º ... [taking the -ve sign on RHS as +ve sign gives no result]
=> θ = 10º.

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Q.433. 3-D Trigonometric application.

Question 433.

The diagram shows a solid cylindrical post with a slanted elliptical face at an angle of θ to the diameter of the cylinder. Diameter DB = 10cm,

Height BC = 100cm and

0º < θ < 90º.

The red route (straight lines from A to B, then B to C) and

the blue route (straight line from A to C around half the circumference of the cylinder) are the same length. 

Prove that θ = π/2 - 2tan^-1 [(π^2 - 4)/80]
and find its value.

Answer 433.

Cutting open the cylinder at BC,

AC^2 = (5π)^2 + (100 + 10tanθ)^2

(AB+BC)^2 = (10secθ + 100)^2

AB+BC = AC

=> (AB+BC)^2 = AC^2

=> (100 + 10secθ)^2 = 25π^2 + (100 + 10tanθ)^2

=> 2000secθ + 100sec^2 θ = 25π^2 + 2000tanθ + 

      100tan^2 θ

=> 2000 (secθ - tanθ) = 25π^2 - 100

=> secθ - tanθ = (π^2 - 4)/80 ... ( 1 )

secθ - tanθ = (1 - sinθ)/cosθ
                      = [cos(θ/2) - sin(θ/2)]^2 / [cos^2 (θ/2) - sin^2 (θ/2)]

                      = [cos(θ/2) - sin(θ/2)] / [cos(θ/2) + sin(θ/2)]
                      = [1 - tan(θ/2)] / [1 + tan(θ/2)]
                      =  tan(π/4 - θ/2)   ... ( 2 )

From ( 1 ) and ( 2 ),

=> tan(π/4 - θ/2) = (π^2 - 4)/80

=> θ = π/2 - 2tan^-1 [(π^2 - 4)/80]
=> θ = 81.6°

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Q.419. Hyperbolic trigo functions - proof.

Question 419.
If tanh(x/2) = tan(x/2), prove that coshx cosx = 1.

Answer 419.
tanh (x/2) = tan(x/2)
=> tan(x/2)
= sinh x / cosh x
= [(1/2) {e^(x/2) - e^(-x/2)}] / [(1/2) {e^(x/2) + e^(-x/2)}]
= [e^(x/2) - e^(-x/2)] / [e^(x/2) + e^(-x/2)] ... ( 1 )

cosx
= (1 - tan^(x/2)) / (1 + tan^(x/2))
= [1 - {e^(x/2) - e^(-x/2)}^2/{e^(x/2) + e^(-x/2)}^2] / [1 + {e^(x/2) + e^(-x/2)}^2/{e^(x/2) + e^(-x/2)}^2]
.................[plugging the value of tan(x/2) from ( 1 )]
= 4 / 2 (e^x + e^-x)
= 2 / (e^x + e^-x)
= 1 / cosh x
=> cosh x cosx = 1.

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Q.413. Trigonometry proof

Question 413.
Given : sin^4 θ/a  + cos^4  θ/b = 1/(a + b).
To Prove : sin^8  θ/a^3 + cos^8  θ/b^3 = 1/(a + b)^3.

Answer 413.
(1/a) sin^4 θ + (1/b) cos^4 θ = 1/(a + b) => (1/a) sin^4 θ + (1/b) (1 - sin^2 θ)^2 = 1/(a +b)

=> b(a + b) sin^4 θ + a (a + b) (1 - sin^2 θ)^2 = ab
=> (a + b)^2 sin^4 θ - 2a (a + b) sin^2 θ + a (a + b) = ab
=> (a + b)^2 sin^4 θ - 2a (a + b) sin^2 θ + a^2 = 0
=> [(a + b) sin^2 θ - a]^2 = 0
=> sin^2 θ = a/(a + b)
and cos^2 θ = 1 - sin^2 θ = 1 - a/(a + b) = b/(a + b)

=> (1/a^3) sin^8 θ + (1/b^3) cos^8 θ
= (1/a^3) * a^4/(a + b)^4 + (1/b^3) * b^4 / (a + b)^4
= (a + b) / (a + b)^4
= 1/(a + b)^3. (proved).

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Q.406. Maximum value of trigonometric function algebraically

Question 406.
Find the maximum value of cosθ (5sinθ - 4cosθ) and the corresponding value of θ.

Answer 406.
cosθ (5sinθ - 4cosθ)
= 5sinθ cosθ - 4cos^2 θ
= (5/2) sin2θ - 2(1 + cos2θ)
= (2.5) sin2θ - 2cos2θ - 2

Now, analyse
(2.5) sin2θ - 2cos2θ
= √[(2.5)^2 + 2^2] * [(2.5)/√[(2.5)^2 + 2^2] sin2θ - 2/√[(2.5)^2 + 2^2] cos2θ]
= 3.202 * [(2.5)/(3.202) sin2θ - 2/(3.202) cos2θ]

Let (2.5)/(3.202) = cosα
=> 2/(3.202) = sinα

=> (2.5) sin2θ - 2cos2θ
= 3.202 * sin(2θ - α)
maximum value of which is 3.202
=> maximum value of (2.5) sin2θ - 2cos2θ - 2
= 3.202 - 2
= 1.202.

This is the maximum value when sin(2θ - α) is maximum
=> (2θ - α) = 90°
=> 2θ = 90° + α
=> 2θ = 90° + 38.65° ... [because sinα = 2/(3.202) => α = 38.65°]
=> θ = 64.3°.

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Q.394. Quadrilateral in (3, 4, 5) triangle

Question 394.
Find the exact value of x as shown in this image.

Answer 394.
DE^2 = 7x^2 = DF^2 + EF^2
=> ∠ F is a rt. angle

Let ∠ AFD = F
=> ∠ CFE = 90° - F

Applying sine rule to Δs ADF and CEF,
sinA/x√2 = sinF/(3-x√3) and
cosA/x√5 = cosF/(4 - 2x)
=> sinF = (4/5) (3 - x√3)/x√2
and cosF = (3/5) (4 - 2x)/x√5

Squarring and adding,
(16/25) (3 - x√3)^2/(2x^2) + (9/25) (4 - 2x)^2/(5x^2) = 1
Using Wolfram Alpha,
Wolfram Alpha Link to answer
x = (6/31) [12 + 20√3 - √{10(91 + 48√3}]
≈ 0.950539. ... [The second value given is redundant as AC = 5].

I did some further Wolfram Alpha investigation to confirm the correctness of the above answer and found AF = 1.60866 and FC = 3.39134 to confirm that AF + FC = 5.
Wolfram Alpha Link

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Q. 386. Trigonometric identities.

Question 386.
Prove that
1) tan20° tan40° tan80° = √3
2) tan(a-b)/2 = √{(4-x^2-y^2)/(x^2 + y^2)} where x = sin a + sin b & y = cos a + cos b.

Answer 386.
1)
tanθ tan(60° - θ) tan (60° + θ)
=  tanθ [(tan60° - tanθ) / (1 + tan60° tanθ)] [(tan60° + tanθ) / (1 - tan60° tanθ)]
=  tanθ [(√3 - tanθ) / (1 + √3 tanθ)] [(√3 + tanθ) / (1 - √3 tanθ)]
= tanθ (3 - tan^2 θ) / (1 - √3 tan^2 θ)
= (3tanθ - tan^3 θ) / (1 - 3tan^2 θ)
= tan3θ.

Taking θ = 20°,
tan20° tan40° tan80°
= tan60°
= √3.

 2)
x = sina + sinb and y = cosa + cosb
Squarring and adding the given eqns.,
(sina + sinb)^2 + (cosa + cosb)^2 = x^2 + y^2
=> (sin^2 a + cos^2 a) + (sin^2 b + cos^2 b) + 2 (cosa cosb + sina sinb) = x^2 + y^2
=> 2 [1 + cos(a - b)] = x^2 + y^2
=> cos(a - b) = (x^2 + y^2)/2 - 1
=> [1 - tan^2 (a-b)/2] / [1 + tan^2 (a-b)/2] = (x^2 + y^2 - 2)/2
=> [1 + tan^2 (a-b)/2 - 1 + tan^2 (a-b)/2] / [1 + tan^2 (a-b)/2 + 1 - tan^2 (a-b)/2]
......................= (2 - x^2 - y^2 + 2) / (2 + x^2 + y^2 - 2)
=> tan^2 (a-b)/2 = (4 - x^2 - y^2) / (x^2 + y^2)
=> tan(a - b)/2 = √ [(4 - x^2 - y^2) / (x^2 + y^2)].

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Q.378. Trigonometric derivation.

Question 378.
Express tan5A in terms of tanA.

Answer 378.
tan5A
= tan(2A + 3A)
= (tan2A + tan3A) / (1 - tan2A tan3A) ... ( 1 )

tan2A + tan3A
= 2tanA / (1 - tan^2 A) + (3tanA - tan^3 A) / (1 - 3tan^2 A)
= (2tanA - 6tan^3 A + 3tanA - 3tan^3 A - tan^3 A + tan^5 A)
    diided by [(1 - tan^2 A)(1 - 3tan^2A]
= (tan^5 A - 10tan^3 A + 5tanA) / [(1 - tan^2 A)(1 - 3tan^2A] ... ( 2 )

1 - tan2A tan3A
= 1 - [2tanA (3tanA - tan^3 A)] / [(1 - tan^2 A)(1 - 3tan^2A]
= (1 - 4tan^2 A + 3tan^4 A - 6tan^2 A + 2tan^4 A) / [(1 - tan^2 A)(1 - 3tan^2A]
= (1 - 10tan^2 A + 5tan^4 A) / [(1 - tan^2 A)(1 - 3tan^2A] ... ( 3 )

Putting results ( 2 ) and ( 3 ) in ( 1 ),
tan5A
= (tan^5 A - 10tan^3 A + 5tanA) / (1 - 10tan^2 A + 5tan^4 A).

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Q.360. Solving the triangle given two sides and an unincluded angle.

Question 360.
Find all sides and angles of a triangle given b = 125, c = 162 and  angle B = 40°.

Answer 360.
This is a case of two sides and a non-included angle.
b, c and angle B are given.
(i) c sin B = 162 sin40° = 104 < b ( = 125) => solution exists.
(ii) b > csinB and b < c => there are two distinctsolutions.
 sinC
= (c/b) sinB
= (162/125) sin40°
= 0.8331
=> C = 56.4° or C = 180° - 56.4° = 123.6°
A = 180° - (B + C) = 180° - (40° + 56.4°) = 83.6° OR
A = 180° - (40° + 123.6°) = 16.4°

If A = 83.6°,
a = b * (sinA/sinB) = 125 * (sin83.6°/sin40°) = 193.3

If A = 16.4°,
a = 125 * (sin16.4°/sin40°) = 54.9

Answers:
1) a = 193.3, A = 83.6°, C = 56.4°
OR
2) a = 54.9, A = 16.4°, C = 123.6°.

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Q.357. Trigonometric identity

Question 357.
Prove that -64sin^7 θ = sin7θ - 7sin5θ + 21sin3θ - 35sinθ.

Answer 357.
Though I had not posted my answer on Yahoo, I have given it a place in my blog with my solution as the question is good and challenging.

- 64 sin^7 x
= - 64 sin^4 x sin^3 x
= 4 (2 sin^2 x)^2 * (- 4 sin^3 x)
= 4 (1 – cos2x)^2 * (sin3x – 3sinx)
= 4 (1 – 2cos2x + cos^2 2x) * (sin3x – 3sinx)
= (2 – 4cos2x + 2cos^2 2x) * (2sin3x – 6sinx)
= (2 – 4cos2x + 1 + cos4x) * (2sin3x – 6sinx)
= (3 – 4cos2x + cos4x) * (2sin3x – 6sinx)
= 6sin3x – 18sinx – 8sin3x cos2x + 24sinx cos2x + 2sin3x cos4x - 6sinx cos4x
= 6sin3x – 18sinx – 4sin5x – 4sinx + 12sin3x – 12sinx + sin7x – sinx – 3sin5x + 3sin3x
= sin7x – 7sin5x + 21sin3x – 35sinx.

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Q.355. Trigonometry proofs

Question 355.
If tan x/2 = tanh y/2, then  prove that
(a) sinh y = tan x and
(b) y=log tan(π/4+x/2).

Answer 355.
(a)
tanx
= 2tan(x/2) / [1 - tan^2 (x/2)]
= 2 tanh (y/2) / [1 - tanh^2 (y/2)]
= 2 [{(e^(y/2) - e^(-y/2)} / {e^(y/2) + e^(-y/2)}] / [1 - [{e^(y2) - e^(-y/2)} / {e^(y/2) + e^(y/2)}]^2 ]
= 2 [{e^(y/2) - e^(-y/2)} * {e^(y/2) + e^(-y/2)}] / [{e^(y/2) + e^(-y/2)}^2 - {e^(y/2) - e^(-y/2)}^2]
= 2 (e^y - e^-y) / 4
= ((e^y - e^-y) / 2
= sinh y

(b)
tan(x/2) = tanh (y/2)

tan (π/4+x/2).
= [1 + tan(x/2)] / [1 - tan(x/2)]
= [1 + tanh (y/2)] / [1 - tanh (y/2)]
= [1 + {e^(y/2) - e^(-y/2)} / {e^(y/2) + e^(-y/2)}] / [1 - {e^(y/2) - e^(-y/2)} / {e^(y/2) + e^(-y/2)}]
= [2e^(y/2) / {e^(y/2) + e^(-y/2)}] / [2e^(-y/2) / {e^(y/2) + e^(-y/2)}]
= e^y
=> tan(π/4+x/2) = e^y
=> y = log tan (π/4+x/2).

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Q.352. Trigonometric identities.

Question 352.
1) Expand sin^3θ cos^4θ in terms of sines of multiples of θ.
2) Prove that - 64sin^7x = sin7x - 7sin5x + 21sin3x - 35sinx.

Answer 352.
1)
sin^3 θ cos^4 θ
= (1/8) (2sinθ cosθ)^3 cosθ
= (1/8) sin^3 (2θ) cosθ
= (1/32) [3sin(2θ) - sin(6θ)] cosθ
= (1/32) [3sin(2θ) cosθ - sin(6θ) cosθ]
= (1/64) [3 (sin3θ + sinθ) - (sin7θ + sin5θ)]
= (1/64) (3sinθ + 3sin3θ - sin5θ - sin7θ).
2)
- 64 sin^7 x
= - 64 sin^4 x sin^3 x
= 4 (2 sin^2 x)^2 * (- 4 sin^3 x)
= 4 (1 – cos2x)^2 * (sin3x – 3sinx)
= 4 (1 – 2cos2x + cos^2 2x) * (sin3x – 3sinx)
= (2 – 4cos2x + 2cos^2 2x) * (2sin3x – 6sinx)
= (2 – 4cos2x + 1 + cos4x) * (2sin3x – 6sinx)
= (3 – 4cos2x + cos4x) * (2sin3x – 6sinx)
= 6sin3x – 18sinx – 8sin3x cos2x + 24sinx cos2x + 2sin3x cos4x - 6sinx cos4x
= 6sin3x – 18sinx – 4sin5x – 4sinx + 12sin3x – 12sinx + sin7x – sinx – 3sin5x + 3sin3x
= sin7x – 7sin5x + 21sin3x – 35sinx.

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Q.344. Trigonometric equations

Question 344.
Find general solution of sin(5x) = tan(x).

Answer 344.
sin5x
= sin5x - sinx + sinx
= 2cos3x sin2x + sinx
= 2(4cos^3 x - 3cosx) * 2sinx cosx + sinx
= sinx (16cos^4 x - 12cos^2 x + 1)

Hence,
sin5x = tanx
=> sin5x - sinx/cosx = 0
=> sin5x cosx - sinx = 0
=> sinx (16cos^5 x - 12cos^3 x + cosx - 1) = 0
=> sinx = 0
=> x = kπ
or 16cos^5 x - 12cos^3 x + cosx - 1 = 0
As this is not easily factorizable, using Wolfram Alpha (link as under)
=> x = 2kπ ± 0.509694
=> x = {kπ, 2kπ ± 0.509694, k ∈ Z} is the general solution.
Source(s):
http://www.wolframalpha.com/input/?i=16cos%5E5+x+-+12cos%5E3+x+%2B+cosx+-+1+%3D+0

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Q.336. Trigonometric proof.

Question 336.

If sin2x = 22 - 8√7,  prove that sinx + cosx + tanx + cosecx + secx + cotx = 7, given that angle x is acute.

Answer 336.

I had not answered this question, but while voting, I found this question to be good and have a better simpler answer posted here.

sin2x = 22 - 8√7

=> 2sinx cosx = 22 - 8√7

=> sinx cosx = 11 - 4√7   ...   ( 1 )

1 + sin2x = 23 - 8√7

=> (sinx + cosx)^2 = 23 - 2√(112) = (4 - √7)^2

=> sinx + cosx = 4 - √7   ...   ( 2 )

sinx + cosx + tanx + cosecx + secx + cotx

= (sinx + cosx) + (cosecx + secx) + (tanx + cotx)

= (sinx + cosx) + (1/sinx + 1/cosx) + (sinx/cosx + cosx/sinx)

= (sinx + cosx) + (sinx + cosx) / sinx cosx + (sin^2 x + cos^2 x) / sinx cosx

= (4 - √7) + (4 - √7) / (11 - 4√7) + 1 / (11 - 4√7)

= [(4 - √7) * (11 - 4√7) + (4 - √7) + 1] / (11 - 4√7)

= (44 + 28 - 27√7 + 5 - √7) / (11 - 4√7)

= 7 * (11 - 4√7) / (11 - 4√7)

= 7.

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Q.328. Application of De Moivere's Theorem.

Question 328.
Using the relationship e^(ix) = cos x + i sin x, express cos5x in term of cosx.
Hence show that x = cos(π/10) is a root of the equation 16x^4 - 20x^2  + 5 = 0.

Answer 328.
cos x + i sin x = e^(ix)
=> (cosx + isinx)^5 = e^(5ix)
=> cos^5 x + 5C1 cos^4 x * (isinx) + 5C2 cos^3 x (isinx)^2 + 5C3 cos^2 x (isinx)^3 + 5C4 cosx (isinx)^4 + 5C5 (isinx)^5 = cos5x + isin5x

Comparing real parts on both sides,
cos5x
= cos^5 x + 5C2 cos^3 x (isinx)^2 + 5C4 cosx (isinx)^4
= cos^5 x - 10cos^3 x sin^2 x + 5cosx sin^4 x
= cos^5 x - 10cos^3 x (1 - cos^2 x) + 5cosx (1 - cos^2 x)^2
= cos^5 x - 10cos^3 x + 10cos^5 x + 5cosx - 10cos^3 x + 5cos^5 x
= 16cos^5 x - 20cos^3 x + 5cosx

Plugging cos(π/10) in 16x^4 - 20x^2 + 5, we get
16cos^4 (π/10) - 20cos^2 (π/10) + 5
= [16cos^5 (π/10) - 20cos^3 (π/10) + 5cos(π/10)] / cos(π/10)
= cos [5(π/10)] / cos(π/10)
= cos(π/2) / cos(π/10)
= 0. [because cos(π/2) = 0].

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