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Solved Examples on Trigonometry

Example1: sin θ cos 7θ cos 4θ + sin 7θ cos θ cos 4θ + sin 4θ cos 7θ cos θ = sin θ sin 4θ sin 7 θ.

Solution: This question is already in sine and cosine ratios. This preferable form, but this does not seem to be of much help.

Moreover this is not in terms of one single Trigonometric Ratio i.e. either sine or cosine. Hence it is not easy to proceed in the present form. What shall we do? Divide whole thing by cos θ cos 4θ cos7θ so that we get the entire expression in terms of tangent ratios,

tan θ + tan 7θ + tan 4θ = tan θ tan 4θ tan 7θ

tan θ + tan 4θ = – tan 7θ (1 – tan θ tan 4θ).

Case I:

If 1 – tan θ tan 4θ ≠ 0, you can peacefully divide by this expression.

(tan θ + tan 4θ)/(1 – tan θ tan 4θ) = –tan 7θ

or, tan 5θ = – tan 7θ

⇒ 5θ = nπ – 7θ

⇒ θ = (nπ)/12, where n = 0, ±1, ±2 ………

Check, whether this is a solution and it satisfies the assumption 1 – tan θ tan 4θ ≠ 0.

1 – tan 4 (nπ/12). tan (nπ/12)

1 – tan (nπ/3) tan (nπ/12) ≠ 0

Case II:

1 – tan θ tan 4θ = 0, then given equation becomes

cos θ cos 4θ – sin θ sin 4θ = 0

or, (cos 3θ – cos 5θ) – (cos 5θ – cos 3θ) = 0 or,

2 (cos 3θ – cos 5θ) = 0

or, cos 3θ = cos 5θ

⇒ 5θ = 2nπ ± 3θ ……… (1) t

Taking positive sign of (1) or, 2θ = 2nπ

or, θ = nπ, where n = 0, ±1, ±2 ………

Taking negative sign of (1) or, 8θ = 2nπ

or, θ = nπ/4 where n = 0, ±1, ±2 ………

Check

Put θ = nπ in given equation 0 = 0

Put θ = nπ/4 in given equation 1/√2.1/√2 (-1) + (1/√2)(1/√2)(-1) + 0 = 0

So θ = nπ and nπ/4. Also satisfy given equation θ = nπ, nπ/4, nπ/12, where n = 0, ±1, ±2 ………

 

Example 2: Solve the equation sin4 x + cos4 x = 7/2 sin x cos x

Solution: We transform the expression sin4 x + cos4 x isolating a perfect square:

sin4 x + cos4 x = sin4 x + 2 sin2 x cos2 x + cos4 x – 2 sin2 x cos2 x

= (sin2 x + cos2 x)2 – 2 sin2 x cos2 x, which gives

sin4 x + cos4 x = 1 – 1/2 sin2 2x. … (1)

Using (1), the given equation becomes

1 – 1/2 sin2 2x = 7/4 sin 2x.

2 sin2 2x + 7sin 2x – 4 = 0

⇒ (2 sin 2x – 1) (sin 2x + 4) = 0

So, either 2 sin 2x – 1 = 0 or sin 2x + 4 = 0        

This gives 2 x = nπ + (–1)n π/6          where n = 0, ±1, ±2… …                   

or sin 2x + 4 = 0 gives no solution as –4 does not belong to range of sin 2x.

The solution of equation is x = (nπ)/2 + (–1)n π/12, where n = 0, ± 1, ± 2 …… .

 

Example 3: Solve for θ, cos 2θ + cos 3θ = sin θ/2, where θ & isin [–π, π]

Solution: Using formulae for cos A + cos B and sin A + sin B, we can convert the given equation to 2cos 5θ/2 = 2sin θ cos θ/2

2 cos θ/2 [cos 5θ/2 – sin θ] = 0

either ⇒ cos θ/2 = 0 ⇒ θ/2 = (2n+1)π/2

⇒ θ = (2n+1)π where n = 0, ±1, ±2…… …… (a)

 or, cos 5θ/2 = sin θ

cos 5θ/2 = cos (π/2 – θ)

5θ/2 = 2n π ± (π/2 – θ), where n = 0, ±1, ±2 … …… (b)

Important: You have to find only those values of which θ ∈ [–π, π]. Not all the values of n & isin give the solutions in the desired range. So find such solution for which θ & isin ∈ [–π, π]. This type of solution where the values are to found in certain interval is called a particular solution of a trigonometric.

(a) Let us now examine the about found general solution θ ∈ [–π, π] = (2n+1)π/2. Only at n = 0, –1 θ lies in (–π, π)

Hence θ = – π/2, π/2 are solutions

Case I: Consider the positive sign of (b),

5θ/2 = 2nπ + (π/2- θ)

or, 7θ = (4n + 1)π

or, θ = (4n + 1)π/7 where n = 0, ±1, ±2 ……

n = 0 gives θ = + π/7, θ ∈ (–π, π)

n = ±1 ⇒ θ = – 3π/7, 5π/7 ∈ (– π, π)

n = ±2 ⇒ θ = – π, 9π/7. Now – π ∈ (–π, π) but 9π/7 does not belong to (–π, π).

Case II:

Now consider the –ve sign

5θ/2 = 2nπ – (π/2 – θ)

or, 3θ = 4nπ – π

or, θ = 4nπ/3 - π/3 = (4n–1) = π/3

n = 0, ⇒ θ = – π/3, - π/3 ∈ (–π, π)

n = 1, ⇒ θ = π, π ∈ (–π, π)

Hence solution is θ ∈ (–π, –3π/7, – π/3, – π/2, π/7, 5π/7, π, π/2}

 

Example 4: Solve cot (π cos θ) = tan (π sin θ).

Solution: Do not get confused by π cos θ or π sin θ in the arguments as they are simply some angles at this moment. Say π cos θ = B and π sin θ = A. The given equation becomes

cot (B) = tan (A)

or, tan (A) = tan (π/2 – B)

⇒ A = nπ + (π/2 – B)

or, π sin θ = nπ + (π/2 – π cos θ) (sin π + cos π) = n + 1/2.

Divide and multiple by √2 equation reduces to

√2 sin (θ + π/4) = n + 1/2. ……… (1)

Important to remember here is that is it necessary that (n+1/2)/√2 ∈ [–1, 1] i.e. for n = 0, 1/(2√2) ∈ [–1, 1]

for n = 1, 3/2√2 = 3/(2 × 1.414) = 3/2.82 does not belong to [–1, 1]

for n = –1, –1/(2√2) ∈ [–1, 1]

Hence (1) becomes

sin (θ + π /4) = +1/2√2

⇒ θ = kπ + (–1)k (± sin–11/2√2) – π /4

θ = kπ + (–1)k sin–1(1/2√2) – π /4, where n = 0, ±1, ±2 …

Note: Do not simple take up sin–1 (k) or cos–1 (k) as they are meaningful only when k domain of these inverse function.

 

Example 5: sec3 θ + cosec3 θ = (sec θ cosec θ)2(sec θ cosec θ – 1)

Solution: 1/cos3 θ + 1/sin3 θ = [1/(sin θ cos θ)3] [1 – sin θ cos θ]

⇒ (sin3 θ + cos3 θ) = 1 – sin θ cos θ

⇒ (sin θ + cos θ) (sin2 θ + cos2 θ – sin θ cos θ) = 1 – sin θ cos θ

{1-(sin 2θ)/2} {√2 cos (θ - π /4) -1} = 0

either 1- ((sin 2θ)/2 = 0 ⇒ sin 2θ = 2

⇒No solution as sin–1 2 is meaningless.

or, √2 cos (θ - π/4) – 1 = 0 ⇒ θ – π /4 = 2nπ ± π /4

⇒ θ = 2nπ + π /4 + π /4, where n = 0, ±1, ±2 ………

 

Example 6: Solve the equation

|cos x|(sin 2x-3/2 sin x + 1/2) =1.

Solution: This can be solved in two ways. One is logical argument and another is procedural method.

Method 1: Procedural Method

Take log on both sides

(sin 2x - 3/2 sin x + 1/2) log |cos x| = 0

either 2 sin 2x – 3sin x + 1 = 0         or log |cos x| = 0

Considering this first 2sin 2x – 3sin x + 1 = 0                                   

or, 2sin 2x – 2sin x – sin x + 1 = 0                                            

This gives sin x = 1/2 or sin x = 1                                   

The general solution x = nπ + (-1)n π/6      

or x = nπ + (-1)nπ/2, where n = 0, ±1, ±2 ……

But this is wrong as at x = π/2 given equation becomes 0o form.

Hence no solution exists for x = nπ + (-1)n π/2.

Now consider the case when log |cos x| = 0

Then |cos x| = 1

Hence, cos x = ± 1. This gives x = nπ, where n = 0, ±1, ±2 ……

Hence, the solution of the given equation is x = nπ, nπ + (-1)n π/6, where n = 0, ±1, ±2 ………

Method II:

Logical Method

We know (Any number)0 = 1

⇒ sin2 x – 3/2 sin x + 1 = 0

⇒x = nπ + (-1)n π/3

or, x = nπ + (-1)n π/3

and also (1)(any number) = 0

⇒ |cos x| = 1 ⇒ x = nπ

⇒ The solution of given equation

x = nπ, nπ + (-1)n π/3, where n = 0, ±1, ±2 ………

 

Example 7: Solve for x and y.

x2 + 2x sin(xy) +1 = 0

Solution:

The given equation can be written as

(x + sin (xy))2 + 1 – sin2 xy = 0

(x + sin xy)2 + cos2 xy = 0

⇒ x + sin xy = 0 …. (a) and

cos xy = 0 …. (b)

cos xy = 0 ⇒ sin xy = 1, –1

Then (x + 1) = 0 or (x – 1) = 0

⇒ x = –1, then from (a), sin xy = 1 which means sin y = –1

also, if x = 1 , sin xy = –1 ⇒ sin y = –1

⇒ y = 2nπ + 3π/2

⇒ (x, y) = (±1, 2nπ + 3π/2), where n = 0, ±1, ±2 ……

Note: If (a – b)2 + (b – c)2 + (c – a)2 = 0. Then each of (a – b)2, (b – c)2 and (c –a)2 must be zero ⇒ a = b, b = c, c = a.

 

Example 8: Solve the system –

sin x + cos x = 1/√2 + sin y – cos y … (a)

2 sin 2x = 3/2 + sin2 y … (b)

Solution: We put u = sin x + cos x, v = sin y – cos y

Thus sin2 x = u2 – 1 and sin2 y = 1 – v2

So now we will have u = 1/√2 + v …… (i)

2u2 + v2 = 9/2 …… (ii)

As a reduced or transformed system –

This possesses two solutions u1 = -4/(3√2) v1 = -7/(3√2)

and v2 = 1/2, u2 = √2

Hence we see that the original system is equivalent to collection of two systems of equations –

(a) (sin x + cos x = -4/(3√2))

     (sin y - cos y = -7/(3√2))

(b) (sin x + cos x = -√2)

      (sin y - cos y = -1/√2)

System (a) has no solution since sin y – cos y = 7/(3√2).

⇒ sin (y - π /4) = 7/(3√2×√2) = 7/6, 7/6>1 but sin (y- π /4) cannot be greater than 1.

Hence, we solve system (b) by multiplying 1/√2

sin (x + π/4) = 1 ⇒ x = nπ + (-1)n π/2 - π/4, where n = 0, ±1, ±2 ……

sin (y + π/4) = 1/2 ⇒ y = mπ + (-1)m π/6 - π/4, where m = 0, ±1, ±2 ……

These pairs of values of x and y constitute the set of all solutions of the original system.

 

Example 9: Solve the following equations for x and y

x cos3 y + 3x cos y sin2 y = 14 …… (1)

x sin3 y + 3x cos2 y sin y = 13 …… (2)

Solution: This does not seem to be a purely trigonometric equation as it involves two variables x and y in Trigo-Algebraic combinations. most of the students get panicky on seeing such problems, but just remember that there’s always a trick involved in solving such questions.

Case I: If x ≠ 0 Divide (1) by (2)

(cos3 y + 3 cos y.sin2 y)/(sin3 y + 3 cos2 y.sin y) = 14/13 Applying componendo and dividendo, we get

((cos y + sin y) / (cos y - sin y))3 = (14 + 13)/(14 - 13) = 27 = 33

⇒ (cos y + sin y) / (cos y - sin y) or (1 + tan y)/(1 - tan y) = 3/1

Again using componendo and Dividendo we get

2/2 tan y = 4/(2)

4 tan y = 2

⇒ tan y = 1/2 [This is possible in 1st & 3rd quadrant]

In Ist quadrant,

sin y = 1/√5, cos y = 2/√5

Hence, putting value of y in equation (1)

x [8/(5√5) + 3.2/√5 1/5] = 14 x = 5√5

When y is in 3rd quadrant, sin y = -1/√5 and cos y = -2/√5

Putting value of above in equation (1)

Hence, the value of x is x [-8/(5√5) + 3(-2/√5) 1/√5] = 14

⇒ x = – 5√5

Hence, y = tan–1 1/2, x = 5√5, where 2nπ < y < 2nπ + π/2 y = tan–1 1/2, x = –5√5, where 2nπ + π < y < 2nπ + 3π/2.

 

Example 10: If cot-1{tan (sin-1 x)} = π/4, determine x.

Solution: Since sin-1 x = tan-1 (x/√(1-x2)), – 1 < x < 1

The given equation is equivalent to

cot-1{tan tan-1 (x/√(1-x2))} = π/4, x ∈ (-1, 1).

This gives cot-1{x/√(1-x2)} = π/4, x ∈ (-1, 1).

If 0 < x < 1, tan-1{√(1-x2)/ x}= π/4

Hence, this gives √(1-x2)/x = 1.

This yields (1 - x2) = x2

or 2x2 = 1 ⇒ x = +1/√2

For –1 < x ≤ 0, equation (1) reduces to

π + tan-1{√(1-x2)/x} = π/4

⇒ tan-1{√(1-x2)/x} = –3π/4

{√(1-x2)/x} = 1

⇒ x = (1-x2)

⇒ 2x2 = 1 ⇒ x = –1/√2.

Hence x = {–1/√2, 1/√2}.

 

Example 11: Simplify

E = a3/2 cosec2 (tan-1 a/ß) + ß3/2 sec2 (tan-1 a/ß) when a, ß > 0.

Solution: Let tan-1 a/ß = θ ⇒ tan θ = a/ß

Now we have E = (a3cosec2 θ + ß3 sec2 θ)/2

or, E = (a3cos2 θ + ß3sin2 θ )/(2sin2 θ cos2 θ ) …… (1)

Now, sin θ = a/√(a2 + ß2) and cos θ = ß/√(a22)

Putting values of sin θ and cos θ in (1)

E = (a + ß) (a2 + ß2)/ 2.

 

Example 12: Find the sum

cot-1 2 + cot-1 8 + cot-1 18 +………… to infinity.

Solution: Let tn denote the nth term of the series

Then tn = cot-1 2n2

or, tn = cot-1 ((4n2-1+1)/2)

= cot-1 [((2n-1) (2n+1) + 1) / ((2n+1) - (2n-1))]

= cot-1 (2n – 1) – cot-1 (2n + 1) …… (1)

putting n = 1, 2, 3, ………… etc. in (1), we get

t1 = cot-1 1 – cot-1 3 = cot-1 2

t2 = cot-1 3 – cot-1 5 = cot-1 8

t3 = cot-1 5 – cot-1 7 = cot-1 18

……………………………………

……………………………………

tn = cot-1 (2n – 1) – cot-1 (2n + 1) = cot-1 2n2

adding Sn = cot-1 1 – cot-1 (2n + 1)

lim n→∞Sn = lim n→∞ (cot-11 - cot-1 (2n+1) )

= π /4 – lim n n→∞ cot-1 (2n+1)

= π /4 – 0 = π /4.

Hence the required sum = π /4.

 

Example 13: tan-1 (a/s) + tan-1 (b/s) + tan-1 (c/s) = π [where s = (a + b + c)/2]

show that abc = 2s3.

Solution: Let’s consider first two terms

tan-1 (a/s) + tan-1 (b/s)

(a/s). (b/s) = a. b/s2

We know b + c > a

⇒ 2a/(a + b + c) ≤ 2a/(a + a)

⇒ a/s < 1

⇒ a.b/s2 < 1 (since b/s also < 1)

so, tan-1 a/s + tan-1 b/s = tan-1 [(a/s + b/s)/(1-ab/s2 )]

= tan-1 s (a+b)/(s2-ab)

So given equation becomes tan-1 s(a + b)/(s2 - ab) = π – tan-1 (c/s)

taking tan of both the sides we get

tan tan-1 s(a+b)/(s2 - ab) = tan { π - tan-1 (c/s) }

Hence LHS = tan tan-1 s(a+b)/(s2-ab) = s(a+b)/(s2-ab) …… (1)

and RHS = tan {π - tan-1(c/s) } = – tan {tan-1 c/s} = - c/s …… (2)

From (1) and (2), we get,

⇒ s(a + b)/ (s2 – ab) = –c/s

⇒ s2(a + b) = – cs2 + abc

⇒ s2(a + b) + cs2 = abc

⇒ s2(a + b + c) = abc

⇒ 2s3 = abc.         Hence proved.

 

Example 14: If tan-1 (√(1+x2) - √(1-x2)) / (√(1+x2) + √(1-x2) = a, then prove that x2 = sin2 a.

Solution: We will have to start with the given result.

tan-1 (√(1+x2) - √(1-x2)) / (√(1+x2) + √(1-x2) = a

(√(1+x2) - √(1-x2)) / (√(1+x2) + √(1-x2) = tan a

Now, using componendo and dividendo, we have

(√(1 + x2) / √(1 - x2)) = (1 + tan a)/ (1 - tan a)

This gives √(1 + x2)/√(1 - x2) = (cos a + sin a)/ (cos a - sin a)

Squaring both sides, we get

(1 + x2) / (1 - x2) = (cos2a + sin2a + 2cos a. sin a)/ (cos2a + sin2a - 2cos a. sin a)

This can be written as

(1 + x2) / (1 - x2) = (1 + 2cos a. sin a)/ (1 - 2cos a. sin a)

⇒ (1 + x2) / (1 - x2) = (1 + sin 2a)/ (1 - sin 2a)

Again applying componendo and dividendo,

x2 /1 = sin 2a/1

Hence we get x2 = sin 2a.

 

Example 15:

If a ≠ b, solve for x.

sec-1 x/a - sec-1 x/b = sec-1 b - sec-1 a

Solution: We know that

sec-1 y = cos-1 1/y, if y ∈ (-∞, -1] ∪ [1, ∞)

           = Not defined    otherwise

Hence, the given equation becomes

cos-1 a/x - cos-1 b/x = cos-1 1/b - cos-1 1/a

⇒ cos-1 a/x + cos-1 1/a = cos-1 1/b + cos-1 b/x

⇒ cos–1 [a/x.1/a - √(1 - (a/x)2). √(1 - 1/a2)] = cos-1 [b/x.1/b - √(1 - (b/x)2 ). √(1-1/b2)]

taking cos both the sides as we know cos cos–1 x = x

1/x - √((1 - (a/x)2)(1-1/a2)) = 1/x-√((1-(b/x)2 )(1-1/b2)) …… (1)

Squaring both sides we get

1/x2 - (a2/x + 1/a2) = 1/x2 (b2/x + 1/b2)

⇒ ((a2-b2))/x2 =1/b2 -1/a2 = (a2-b2)/(ab)2

⇒ (a2 – b2) [1/x2 -1/(ab)2] = 0

⇒ either a = ± b ⇒ the equation is satisfied ∀ x ∈ R …… (2)

or, 1/x2 =1/(ab)2 ⇒ x = ± ab.

We squared the equation (1) of the solution i.e. we distorted the original equation. Hence there is every chance of some additional solutions to appear. We, therefore, must check a = + b.

If we take a = b, the equation becomes an identity and always satisfied. But what if we take a = ±b.

⇒ sec-1 (–y) = π – sec-1 (y)

⇒ sec-1 (-x/b) – sec-1 (x/b) = sec-1 b – sec-1 (–b)

⇒ Given equation becomes π – 2 sec-1(x/b) = π – 2 sec-1 b

⇒ sec-1 (x/b) = sec-1 (b)

⇒ x = b2

x = –ab given equation becomes

sec-1 (–b) – sec-1 (–a) = sec-1 b – sec-1 a

⇒2 π = 0 which is not possible

⇒ no solution for x = – ab

So x = ab, b2 are the solution of the given equation.

 

Example 16:

Prove that cos–1 ((cos θ + cos φ)/(1 + cos θ cos φ)) = 2 tan-1 (tan θ/2 tan φ/2)

Solution:

R.H.S. = 2 tan-1 (tan θ/2 tan φ/2)

= cos-1 [(1 - tan2 θ/2 tan2 φ/2) / (1 + tan2 θ/2 tan2 φ/2)

= cos-1 [{1 – (sin2 θ/2 sin2 φ/2)/(cos2 θ/2 cos2 φ/2)}/ {1 + (sin2 θ/2 sin2 φ/2)/(cos2 θ/2 cos2 φ/2)}]

On simplifying this term we obtain

cos-1 [(cos θ +  cos φ)/(1 + cos (θ - φ) – sin θ sin φ]

= cos-1 [(cos θ + cos φ)/ (1 + cos θ cos φ)]

= L.H.S

 

Example 17: Prove that

tan-1 (1/2 tan 2A) + tan-1 (cot A) + tan-1 (cot3A) = 0 if π /4 < A < π /2

                                                                                   = π if 0 < A < π /4.

Solution: We consider two cases

Case I: When π /4 < A < π /2

Then we have 0 < cot A < 1 and 0 < cot3 A < 1.

⇒ cot A . cot3 A < 1.

Hence, tan-1 (cot A) + tan-1 (cot3 A)

= tan-1 [(cot A + cot3 A)/(1 - cot A .cot3 A)]

[Since tan-1 x + tan-1 y = tan-1 ((x+y)/(1-xy)) if x > 0, y > 0 and xy < 1]

Now, proceeding on similar lines as in the previous example, we can obtain

= – tan-1 (1/2 tan 2 A).

= tan-1 (1/2 tan 2 A) + tan-1 (cot A) + tan-1 (cot3 A) = 0.

Case II. When 0 < A < π /4

cot A > 1 and cot3 A > 1.

⇒ cot A . cot3 A > 1

Hence, tan-1 (cot A) + tan-1 (cot3 A)

= π + tan-1 [(cot A + cot3 A)/(1 - cotA. cot3 A)]

[Since tan-1 x + tan-1 y = π + tan-1 ((x + 1)/(1- xy))         if x > 0, y > 0 and xy > 1]

= π – tan-1 (1/2 tan 2 A) [as shown in case I]      

⇒ tan-1 (1/2 tan 2 A) + tan-1 (cot A) + tan-1 (cot3 A) = π.

Hence the result.

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