IPMAT Indore Trigonometry PYQs

Q1:ipmat indore 2022QA › TrigonometryHardSA · TITA

If

\sin \alpha+\sin \beta=\frac{\sqrt{2}}{\sqrt{3}}

and

\cos \alpha+\cos \beta=\frac{1}{\sqrt{3}}

, then the value of

\left(20 \cos \left(\frac{\alpha-\beta}{2}\right)\right)^{2}

is _________.

Enter your answer to attempt

The Setup: We are given constraint equations

\sin\alpha+\sin\beta = \frac{\sqrt{2}}{\sqrt{3}}

and

\cos\alpha+\cos\beta = \frac{1}{\sqrt{3}}

. We need to evaluate the compound trigonometric expression

\left(20\cos\left(\frac{\alpha-\beta}{2}\right)\right)^2

. Step 1: Square and add the two primary equations. Equation 1:

(\sin\alpha+\sin\beta)^2 = \left(\frac{\sqrt{2}}{\sqrt{3}}\right)^2 \implies \sin^2\alpha + \sin^2\beta + 2\sin\alpha\sin\beta = \frac{2}{3}

Equation 2:

(\cos\alpha+\cos\beta)^2 = \left(\frac{1}{\sqrt{3}}\right)^2 \implies \cos^2\alpha + \cos^2\beta + 2\cos\alpha\cos\beta = \frac{1}{3}

Summing them together:

(\sin^2\alpha + \cos^2\alpha) + (\sin^2\beta + \cos^2\beta) + 2(\cos\alpha\cos\beta + \sin\alpha\sin\beta) = \frac{2}{3} + \frac{1}{3}

Using Pythagorean and subtraction identities:

1 + 1 + 2\cos(\alpha-\beta) = 1 \implies 2 + 2\cos(\alpha-\beta) = 1

Step 2: Relate the expression to the half-angle formula. Factor out the

2

:

2(1 + \cos(\alpha-\beta)) = 1

Apply the power-reduction identity

1 + \cos\theta = 2\cos^2\left(\frac{\theta}{2}\right)

:

2 \left(2\cos^2\left(\frac{\alpha-\beta}{2}\right)\right) = 1

4\cos^2\left(\frac{\alpha-\beta}{2}\right) = 1

Step 3: Evaluate the target expression. The target is

\left(20\cos\left(\frac{\alpha-\beta}{2}\right)\right)^2

. Expand the target:

400\cos^2\left(\frac{\alpha-\beta}{2}\right)

Substitute the derived identity block:

100 \times \left[4\cos^2\left(\frac{\alpha-\beta}{2}\right)\right] = 100 \times 1 = 100

Final Answer: 100

Q2:ipmat indore 2019QA › TrigonometryHardSA · TITA

The number of pairs

(x, y)

satisfying the equation

\sin x + \sin y = \sin(x + y)

and

|x| + |y| = 1

is

Enter your answer to attempt

The Setup: A trigonometric boss phase. Factor the first equation with sum-to-product identities to expose its root conditions, then run those through the absolute-value geometry of

|x|+|y|=1

to see which survive. Step 1: Expand and factor. Sum-to-product on the left, double angle on the right:

2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)=2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x+y}{2}\right)

Bring everything to one side and pull out the common factor:

2\sin\left(\frac{x+y}{2}\right)\left[\cos\left(\frac{x-y}{2}\right)-\cos\left(\frac{x+y}{2}\right)\right]=0

Apply

\cos A-\cos B=2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{B-A}{2}\right)

to the bracket, with

A=\frac{x-y}{2}

and

B=\frac{x+y}{2}

, so that

\frac{A+B}{2}=\frac{x}{2}

and

\frac{B-A}{2}=\frac{y}{2}

:

4\sin\left(\frac{x}{2}\right)\sin\left(\frac{y}{2}\right)\sin\left(\frac{x+y}{2}\right)=0

Step 2: Identify the root scenarios. A product is zero only if a factor is, giving three parallel timelines:

\sin\left(\frac{x}{2}\right)=0\implies x=2n\pi, \qquad \sin\left(\frac{y}{2}\right)=0\implies y=2n\pi, \qquad \sin\left(\frac{x+y}{2}\right)=0\implies x+y=2n\pi

Step 3: Apply the domain constraint. On

|x|+|y|=1

we have

|x|\leq 1

,

|y|\leq 1

, and by the triangle inequality

|x+y|\leq|x|+|y|=1

. Since

2\pi\approx 6.28

is far outside that, **only

n=0

survives** in all three families, collapsing them to

x=0

,

y=0

, or

x+y=0

. Step 4: Hunt the coordinates. Intersect each line with the diamond

|x|+|y|=1

: *

x=0 \implies |y|=1

: gives

(0,1)

and

(0,-1)

*

y=0 \implies |x|=1

: gives

(1,0)

and

(-1,0)

*

x=-y \implies 2|x|=1

: gives

\left(\tfrac{1}{2},-\tfrac{1}{2}\right)

and

\left(-\tfrac{1}{2},\tfrac{1}{2}\right)

Step 5: Check for double-counting before adding. The only point that could belong to two of these lines is

(0,0)

, where

x=0

and

y=0

meet - but it fails

|x|+|y|=1

, so it never enters the list. The six points above are pairwise distinct, and the count is genuinely

2+2+2

. Final Answer: 6

Q3:ipmat indore 2020QA › TrigonometryHardMCQ · MCQ

The value of

\cos^2\left(\frac{\pi}{8}\right) + \cos^2\left(\frac{3\pi}{8}\right) + \cos^2\left(\frac{5\pi}{8}\right) + \cos^2\left(\frac{7\pi}{8}\right)

is

A $1$
B $\frac{3}{2}$
C $2$
D $\frac{9}{4}$

Pick an option to attempt

The Setup: This trig expression looks intimidating, but it has zero plot armour once you spot the symmetry in the angles. We use the supplementary angle identity to fold the expression in half, then the complementary identity to finish it off. Step 1: Fold the back half. Notice that

\frac{7\pi}{8}=\pi-\frac{\pi}{8}

, and

\cos(\pi-\theta)=-\cos\theta

, so

\cos\left(\frac{7\pi}{8}\right)=-\cos\left(\frac{\pi}{8}\right)

. Squaring kills the minus sign:

\cos^2\left(\frac{7\pi}{8}\right)=\cos^2\left(\frac{\pi}{8}\right)

Identically,

\frac{5\pi}{8}=\pi-\frac{3\pi}{8}

, so

\cos^2\left(\frac{5\pi}{8}\right)=\cos^2\left(\frac{3\pi}{8}\right)

. Step 2: Simplify the sum. The four terms collapse into two identical pairs:

2\left[\cos^2\left(\frac{\pi}{8}\right)+\cos^2\left(\frac{3\pi}{8}\right)\right]

Step 3: Exploit the complementary angles. Notice that

\frac{\pi}{8}+\frac{3\pi}{8}=\frac{4\pi}{8}=\frac{\pi}{2}

. The two angles are complementary, and

\cos\left(\frac{\pi}{2}-\theta\right)=\sin\theta

, so:

\cos\left(\frac{3\pi}{8}\right)=\sin\left(\frac{\pi}{8}\right)

Step 4: Use the legendary identity. Substitute sine into the folded expression:

2\left[\cos^2\left(\frac{\pi}{8}\right)+\sin^2\left(\frac{\pi}{8}\right)\right]

By the Pythagorean identity

\cos^2\theta+\sin^2\theta=1

, the bracket collapses to 1:

2\times 1=2

Final Answer:

2

Q4:ipmat indore 2021QA › TrigonometryMediumMCQ · MCQ

If the angles

A, B, C

of a triangle are in arithmetic progression such that

\sin(2A + B) = 1/2

then

\sin(B + 2C)

is equal to

A $\frac{-1}{2}$
B $\frac{1}{2}$
C $\frac{-1}{\sqrt{2}}$
D $\frac{3}{\sqrt{2}}$

Pick an option to attempt

The Setup: Geometry meets Trigonometry. The angles of a triangle always sum to

180^\circ

. If they are in an Arithmetic Progression, the middle angle acts as the perfect average, completely locking in its value from the start. Step 1: Find angle

B

. Since

A

,

B

,

C

are in AP:

A+C=2B

Substitute this into the triangle angle sum theorem:

A+B+C=180^\circ \implies 2B+B=180^\circ \implies 3B=180^\circ \implies B=60^\circ

Step 2: Use the sine condition to find angle

A

. We know

\sin(2A+B)=1/2

. Since

A+C=120^\circ

, angle

A

is stuck below

120^\circ

, so

2A+60^\circ

never climbs past

300^\circ

- that leaves exactly two candidates,

30^\circ

and

150^\circ

. * If

2A+60^\circ=30^\circ \implies 2A=-30^\circ

(Angles in a triangle must be positive, so this is a wipe). * If

2A+60^\circ=150^\circ \implies 2A=90^\circ \implies A=45^\circ

. Step 3: Find angle

C

.

C=180^\circ-(A+B)=180^\circ-(45^\circ+60^\circ)=75^\circ

Step 4: Plug the stats into the final equation

\sin(B+2C)

.

B+2C=60^\circ+2(75^\circ)=60^\circ+150^\circ=210^\circ

\sin(210^\circ)=\sin(180^\circ+30^\circ)=-\sin(30^\circ)=-\frac{1}{2}

Final Answer:

\frac{-1}{2}

Q5:ipmat indore 2022QA › TrigonometryHardMCQ · MCQ

For

0\lt\theta\lt\frac{\pi}{4}

, let

a=\left((\sin \theta)^{\sin \theta}\right)\left(\log _{2} \cos \theta\right), b=\left((\cos \theta)^{\sin \theta}\right)\left(\log _{2} \sin \theta\right), c=\left((\sin \theta)^{\cos \theta}\right)\left(\log _{2} \cos \theta\right)

and

d=\left((\sin \theta)^{\sin \theta}\right)\left(\log _{2} \sin \theta\right)

. Then, the median value in the sequence

a, b, c, d

is

A $\frac{a+b}{2}$
B $\frac{a+d}{2}$
C $\frac{b+c}{2}$
D $\frac{c+d}{2}$

Pick an option to attempt

The Setup: We are asked to order four logarithmic-exponential expressions to find their median. By analyzing the base and arguments within the interval

0 < \theta < \pi/4

, we can determine their strict mathematical ordering. Step 1: Analyze the components within the given domain. For

\theta \in (0, \pi/4)

, the sine and cosine functions are bounded such that:

0 < \sin\theta < \frac{1}{\sqrt{2}} < \cos\theta < 1

Because both

\sin\theta

and

\cos\theta

are strictly between

0

and

1

, their base-

2

logarithms are strictly negative:

\log_2(\sin\theta) < \log_2(\cos\theta) < 0

Step 2: Compare the components of expressions

a, b, c, d

. The definitions are:

a = (\sin\theta)^{\sin\theta} \log_2(\cos\theta)

b = (\cos\theta)^{\sin\theta} \log_2(\sin\theta)

c = (\sin\theta)^{\cos\theta} \log_2(\cos\theta)

d = (\sin\theta)^{\sin\theta} \log_2(\sin\theta)

Notice that all four overall expressions yield negative values (a positive exponential multiplied by a negative logarithm). Step 3: Establish the inequalities. First, compare

a

and

c

: Since

\sin\theta < \cos\theta

and the base

\sin\theta \in (0, 1)

, a larger exponent yields a smaller value:

(\sin\theta)^{\cos\theta} < (\sin\theta)^{\sin\theta}

Because

\log_2(\cos\theta)

is negative, multiplying both sides by it reverses the inequality:

(\sin\theta)^{\cos\theta} \log_2(\cos\theta) > (\sin\theta)^{\sin\theta} \log_2(\cos\theta) \implies c > a

Second, compare

b

and

d

: Because

\sin\theta < \cos\theta

, raising them to the positive power of

\sin\theta

preserves the inequality:

(\sin\theta)^{\sin\theta} < (\cos\theta)^{\sin\theta}

Because

\log_2(\sin\theta)

is negative, multiplying reverses the inequality:

(\sin\theta)^{\sin\theta} \log_2(\sin\theta) > (\cos\theta)^{\sin\theta} \log_2(\sin\theta) \implies d > b

Third, compare

a

and

d

: They share the identical positive exponential factor

(\sin\theta)^{\sin\theta}

. Since

\log_2(\sin\theta) < \log_2(\cos\theta)

, multiplying by the positive exponential preserves the inequality:

d < a

Step 4: Determine the median. Combining the inequalities gives the strict ascending order:

b < d < a < c

. For a set of 4 elements, the median is the arithmetic mean of the two central terms (

d

and

a

).

\text{Median} = \frac{a + d}{2}

Final Answer:

\frac{a+d}{2}

Q6:ipmat indore 2021QA › TrigonometryMediumMCQ · MCQ

The set of all real value of

p

for which the equation

3 \sin^2x + 12 \cos x - 3 = p

has at least one solution is

A $[-12, 12]$
B $[-12, 9]$
C $[-15, 9]$
D $[-15, 12]$

Pick an option to attempt

The Setup: We are converting a trig equation into a standard quadratic to find its absolute range (the min and max values of

p

). Since the equation involves both sine and cosine, we use the identity

\sin^2x=1-\cos^2x

to unify the variables. Step 1: Substitute and simplify the expression. Let

f(x)=p

.

3(1-\cos^2x)+12\cos x-3=p

3-3\cos^2x+12\cos x-3=p

-3\cos^2x+12\cos x=p

Step 2: Set up a boundary replacement. Let

t=\cos x

. Because of the constraints of the cosine function,

t

is strictly bounded:

t\in[-1,1]

. Our new function is

g(t)=-3t^2+12t

. We need the absolute min and max of this quadratic on the locked interval

[-1,1]

. Step 3: Find the vertex of the parabola. For a quadratic

at^2+bt+c

, the vertex occurs at

t=-b/(2a)

.

t=\frac{-12}{2(-3)}=\frac{-12}{-6}=2

The vertex sits at

t=2

, which is completely outside our playable zone of

[-1,1]

. Because it's a downward-opening parabola (negative

a

), it is strictly increasing across our entire

[-1,1]

interval. Step 4: Test the boundaries to find the range. * If

t=-1

(Minimum):

g(-1)=-3(-1)^2+12(-1)=-3-12=-15

. * If

t=1

(Maximum):

g(1)=-3(1)^2+12(1)=-3+12=9

. The function seamlessly spans every value between these two extremes. Final Answer:

[-15, 9]

Q7:ipmat indore 2022QA › TrigonometryMediumMCQ · MCQ

Ayesha is standing atop a vertical tower

200 m

high and observes a car moving away from the tower on a straight, horizontal road from the foot of the tower. At 11:00 AM, she observes the angle of depression of the car to be

45^{\circ}

. At 11:02 AM, she observes the angle of depression of the car to be

30^{\circ}

. The speed at which the car is moving is approximately

A6.3 km per hour
B8.45 km per hour
C10.6 km per hour
D4.39 km per hour

Pick an option to attempt

The Setup: A car is tracked from a

200\text{m}

tower. Its angle of depression changes from

45^\circ

to

30^\circ

over exactly

2

minutes. We must use right-triangle trigonometry to find its speed in

\text{km/h}

. Step 1: Calculate the initial horizontal distance. Let the tower base be the origin. At 11:00 AM, the angle of depression is

45^\circ

. Using trigonometry,

\tan(45^\circ) = \frac{\text{Height}}{\text{Distance}_1}

.

1 = \frac{200}{d_1} \implies d_1 = 200\text{ meters}

Step 2: Calculate the subsequent horizontal distance. At 11:02 AM, the angle of depression is

30^\circ

.

\tan(30^\circ) = \frac{\text{Height}}{\text{Distance}_2}

\frac{1}{\sqrt{3}} = \frac{200}{d_2} \implies d_2 = 200\sqrt{3}\text{ meters}

Step 3: Calculate the distance traveled. The distance the car traveled in the 2-minute interval is the difference:

\Delta d = d_2 - d_1 = 200\sqrt{3} - 200 = 200(\sqrt{3} - 1)

Using the approximation

\sqrt{3} \approx 1.732

:

\Delta d \approx 200(1.732 - 1) = 200(0.732) = 146.4\text{ meters}

Step 4: Calculate the speed in km/h. The car traveled

146.4

meters in

2

minutes.

\text{Speed} = \frac{146.4\text{ meters}}{2\text{ minutes}} = 73.2\text{ meters/minute}

Convert to kilometers per hour (

\times 60 \text{ minutes/hour}, \div 1000 \text{ meters/km}

):

\text{Speed} = \frac{73.2 \times 60}{1000} = \frac{4392}{1000} = 4.392\text{ km/h}

Final Answer: 4.39 km per hour

Q8:ipmat indore 2024QA › TrigonometryEasyMCQ · MCQ

The angle of elevation of the top of a pole from a point A on the ground is 30. The angle of elevation changes to 45, after moving 20 meters towards the base of the pole. Then the height of the pole, in meters, is

A $15(\sqrt{5} + 1)$
B $20(\sqrt{3} + 1)$
C $30$
D $10(\sqrt{3} + 1)$

Pick an option to attempt

The Setup: This problem models two angles of elevation to the top of a pole from different horizontal distances. We use right-triangle trigonometry to build a system of linear equations mapping distance to height. Step 1: Establish the initial trigonometric relation. Let

h

be the height of the pole, and

x

be the initial horizontal distance from point

A

to the pole's base.

\tan(30^\circ) = \frac{h}{x} \implies \frac{1}{\sqrt{3}} = \frac{h}{x} \implies x = h\sqrt{3}

Step 2: Establish the secondary trigonometric relation. After moving

20

meters closer, the new distance is

x - 20

.

\tan(45^\circ) = \frac{h}{x - 20}

Since

\tan(45^\circ) = 1

:

1 = \frac{h}{x - 20} \implies h = x - 20

Step 3: Substitute and solve for

h

. Replace

x

with

h\sqrt{3}

:

h = h\sqrt{3} - 20

20 = h\sqrt{3} - h = h(\sqrt{3} - 1)

h = \frac{20}{\sqrt{3} - 1}

Step 4: Rationalize the denominator. Multiply the numerator and denominator by the conjugate

(\sqrt{3} + 1)

:

h = \frac{20(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{20(\sqrt{3} + 1)}{3 - 1}

h = \frac{20(\sqrt{3} + 1)}{2} = 10(\sqrt{3} + 1)

Final Answer:

10(\sqrt{3}+1)

Q9:ipmat indore 2019QA › TrigonometryHardMCQ · MCQ

Given that

\cos x + \cos y = 1

, the range of

\sin x - \sin y

is

A $[-1, 1]$
B $[-2, 2]$
C $[0, \sqrt{3}]$
D $[-\sqrt{3}, \sqrt{3}]$

Pick an option to attempt

The Setup: One equation constrains the pair

(x,y)

, and we want the range of a second expression. Squaring both and adding is the standard move, because the Pythagorean identity then collapses most of the terms. Step 1: Name the target. Let

S=\sin x-\sin y

, given

\cos x+\cos y=1

. Step 2: Square both and add.

(\cos x+\cos y)^2=1 \implies \cos^2x+2\cos x\cos y+\cos^2y=1

(\sin x-\sin y)^2=S^2 \implies \sin^2x-2\sin x\sin y+\sin^2y=S^2

Adding, and grouping

\sin^2+\cos^2=1

for each of

x

and

y

:

1+1+2\left(\cos x\cos y-\sin x\sin y\right)=1+S^2

The bracket is exactly the expansion of

\cos(x+y)

, so:

2+2\cos(x+y)=1+S^2 \implies S^2=1+2\cos(x+y)

**Step 3: Bound

\cos(x+y)

using the constraint.** Sum-to-product on the given equation, writing

u=\frac{x+y}{2}

and

v=\frac{x-y}{2}

:

2\cos u\cos v=1 \implies \cos u\cos v=\frac{1}{2}

Since

|\cos v|\leq 1

, we need

|\cos u|\geq\frac{1}{2}

, hence

\cos^2u\geq\frac{1}{4}

. The double-angle formula then gives:

\cos(x+y)=2\cos^2u-1 \geq 2\cdot\frac{1}{4}-1=-\frac{1}{2}

and of course

\cos(x+y)\leq 1

always. Substituting into

S^2=1+2\cos(x+y)

:

S^2\in[\,1-1,\ 1+2\,]=[0,\ 3]

Step 4: Confirm both extremes are actually reachable. Bounds are worthless unless attained, so produce explicit angles satisfying

\cos x+\cos y=1

: *

x=\frac{\pi}{3},\ y=-\frac{\pi}{3}

: then

\cos x+\cos y=\frac{1}{2}+\frac{1}{2}=1

✓ and

S=\frac{\sqrt{3}}{2}-\left(-\frac{\sqrt{3}}{2}\right)=\sqrt{3}

, the maximum. *

x=y=\frac{\pi}{3}

: then

\cos x+\cos y=1

✓ and

S=0

, the minimum of

|S|

. Swapping

x

and

y

negates

S

, so

-\sqrt{3}

is reached too, and since

S

varies continuously it sweeps everything between:

S\in\left[-\sqrt{3},\ \sqrt{3}\right]

Final Answer:

[-\sqrt{3}, \sqrt{3}]

Q10:ipmat indore 2019QA › TrigonometryHardMCQ · MCQ

If

sin \theta + cos \theta = m

, then

sin^6 \theta + cos^6 \theta

equals

A $\frac{3(m^2+1)}{4}$
B $\frac{3(m^2-1)}{4}$
C $1-\frac{3(m^2-1)}{4}$
D $1-\frac{3(m^2-1)^2}{4}$

Pick an option to attempt

The Setup: An algebraic identity wearing a trigonometry costume. We must climb from first powers to sixth powers, and the bridge is the single quantity

\sin\theta\cos\theta

, which squaring the given equation hands us immediately. Step 1: Extract the cross-term. Square both sides:

(\sin\theta+\cos\theta)^2=m^2 \implies \sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta=m^2

Since

\sin^2\theta+\cos^2\theta=1

:

1+2\sin\theta\cos\theta=m^2 \implies \sin\theta\cos\theta=\frac{m^2-1}{2}

Step 2: Reduce the sixth powers to that one quantity. Use the sum of cubes

a^3+b^3=(a+b)\left(a^2-ab+b^2\right)

with

a=\sin^2\theta

and

b=\cos^2\theta

:

\sin^6\theta+\cos^6\theta=\left(\sin^2\theta+\cos^2\theta\right)\left(\sin^4\theta-\sin^2\theta\cos^2\theta+\cos^4\theta\right)

The first bracket is 1. For the second, use the identity

a^2+b^2=(a+b)^2-2ab

on

\sin^4\theta+\cos^4\theta

, which turns

a^2-ab+b^2

into

(a+b)^2-3ab

:

\sin^4\theta-\sin^2\theta\cos^2\theta+\cos^4\theta=\left(\sin^2\theta+\cos^2\theta\right)^2-3\sin^2\theta\cos^2\theta=1-3\left(\sin\theta\cos\theta\right)^2

Step 3: Substitute. Note the cross-term gets squared, so the bracket from Step 1 is squared whole:

\sin^6\theta+\cos^6\theta=1-3\left(\frac{m^2-1}{2}\right)^2=1-\frac{3\left(m^2-1\right)^2}{4}

Option 3 drops that squaring and reads

1-\frac{3(m^2-1)}{4}

- the difference is one exponent, so check it before committing. Spot-check at

\theta=0

: then

m=\sin 0+\cos 0=1

, and the formula gives

1-\frac{3(1-1)^2}{4}=1

, matching

\sin^6 0+\cos^6 0=0+1=1

. Final Answer:

1-\frac{3(m^2-1)^2}{4}

Q11:ipmat indore 2023QA › TrigonometryMediumMCQ · MCQ

If

cos \alpha

+

cos \beta

= 1 then the maximum value of

sin \alpha - sin \beta

is

A $\sqrt{2}$
B $2$
C $\sqrt{3}$
D $1$

Pick an option to attempt

The Setup: Given the constraint

\cos\alpha + \cos\beta = 1

, we must find the maximum value of

\sin\alpha - \sin\beta

. Step 1: Rewrite both expressions with the sum-to-product identities. Let

u = \frac{\alpha+\beta}{2}

and

v = \frac{\alpha-\beta}{2}

. The constraint and the target expression become:

\cos\alpha + \cos\beta = 2\cos u \cos v = 1

\sin\alpha - \sin\beta = 2\cos u \sin v

Both share the same factor

2\cos u

, which is what makes the constraint substitutable. Step 2: Eliminate

u

using the constraint. The constraint gives

2\cos u = \frac{1}{\cos v}

directly (note

\cos v \neq 0

). Substitute this into the target:

\sin\alpha - \sin\beta = \left(\frac{1}{\cos v}\right)\sin v = \tan v

The whole problem has collapsed to maximising

\tan v

. Step 3: Find the admissible range of

v

.

\cos u

is a cosine, so it is bounded by

|\cos u| \le 1

:

\left|\frac{1}{2\cos v}\right| \le 1 \implies |\cos v| \ge \frac{1}{2}

On the branch where

\tan v

is positive and increasing, this restricts

v

to

0 \le v \le \frac{\pi}{3}

. Step 4: Maximise on that range.

\tan v

increases throughout

\left[0, \frac{\pi}{3}\right]

, so the maximum sits at the right endpoint:

\text{Maximum} = \tan\frac{\pi}{3} = \sqrt{3}

Step 5: Confirm the maximum is actually attained. At

v = \frac{\pi}{3}

the constraint forces

\cos u = \frac{1}{2\cos(\pi/3)} = 1

, so

u = 0

, giving

\alpha = \frac{\pi}{3}

and

\beta = -\frac{\pi}{3}

. Check both conditions:

\cos\frac{\pi}{3} + \cos\left(-\frac{\pi}{3}\right) = \frac{1}{2} + \frac{1}{2} = 1

\sin\frac{\pi}{3} - \sin\left(-\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} = \sqrt{3}

Both hold, so

\sqrt{3}

is genuinely achieved and is the maximum. Final Answer:

\sqrt{3}

IPMAT Indore Trigonometry — PYPs

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