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Trigonometry — PYPs

13 solved Trigonometry previous year questions (PYQs) from past year papers — attempt each and check the answer.

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Q1:ipmat indore 2022QA › TrigonometryHardSA · TITA
If sin⁡α+sin⁡β=23\sin \alpha+\sin \beta=\frac{\sqrt{2}}{\sqrt{3}} and cos⁡α+cos⁡β=13\cos \alpha+\cos \beta=\frac{1}{\sqrt{3}}, then the value of (20cos⁡(α−β2))2\left(20 \cos \left(\frac{\alpha-\beta}{2}\right)\right)^{2} is _________.
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The Setup: We are given constraint equations sin⁡α+sin⁡β=23\sin\alpha+\sin\beta = \frac{\sqrt{2}}{\sqrt{3}} and cos⁡α+cos⁡β=13\cos\alpha+\cos\beta = \frac{1}{\sqrt{3}}. We need to evaluate the compound trigonometric expression (20cos⁡(α−β2))2\left(20\cos\left(\frac{\alpha-\beta}{2}\right)\right)^2. Step 1: Square and add the two primary equations. Equation 1: (sin⁡α+sin⁡β)2=(23)2  ⟹  sin⁡2α+sin⁡2β+2sin⁡αsin⁡β=23(\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⁡α+cos⁡β)2=(13)2  ⟹  cos⁡2α+cos⁡2β+2cos⁡αcos⁡β=13(\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⁡2α+cos⁡2α)+(sin⁡2β+cos⁡2β)+2(cos⁡αcos⁡β+sin⁡αsin⁡β)=23+13(\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+2cos⁡(α−β)=1  ⟹  2+2cos⁡(α−β)=11 + 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 22: 2(1+cos⁡(α−β))=12(1 + \cos(\alpha-\beta)) = 1 Apply the power-reduction identity 1+cos⁡θ=2cos⁡2(θ2)1 + \cos\theta = 2\cos^2\left(\frac{\theta}{2}\right): 2(2cos⁡2(α−β2))=12 \left(2\cos^2\left(\frac{\alpha-\beta}{2}\right)\right) = 1 4cos⁡2(α−β2)=14\cos^2\left(\frac{\alpha-\beta}{2}\right) = 1 Step 3: Evaluate the target expression. The target is (20cos⁡(α−β2))2\left(20\cos\left(\frac{\alpha-\beta}{2}\right)\right)^2. Expand the target: 400cos⁡2(α−β2)400\cos^2\left(\frac{\alpha-\beta}{2}\right) Substitute the derived identity block: 100×[4cos⁡2(α−β2)]=100×1=100100 \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)(x, y) satisfying the equation sin⁡x+sin⁡y=sin⁡(x+y)\sin x + \sin y = \sin(x + y) and ∣x∣+∣y∣=1|x| + |y| = 1 is
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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|x|+|y|=1 to see which survive. Step 1: Expand and factor. Sum-to-product on the left, double angle on the right: 2sin⁡(x+y2)cos⁡(x−y2)=2sin⁡(x+y2)cos⁡(x+y2)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: 2sin⁡(x+y2)[cos⁡(x−y2)−cos⁡(x+y2)]=02\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=2sin⁡(A+B2)sin⁡(B−A2)\cos A-\cos B=2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{B-A}{2}\right) to the bracket, with A=x−y2A=\frac{x-y}{2} and B=x+y2B=\frac{x+y}{2}, so that A+B2=x2\frac{A+B}{2}=\frac{x}{2} and B−A2=y2\frac{B-A}{2}=\frac{y}{2}: 4sin⁡(x2)sin⁡(y2)sin⁡(x+y2)=04\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⁡(x2)=0  ⟹  x=2nπ,sin⁡(y2)=0  ⟹  y=2nπ,sin⁡(x+y2)=0  ⟹  x+y=2nπ\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|x|+|y|=1 we have ∣x∣≤1|x|\leq 1, ∣y∣≤1|y|\leq 1, and by the triangle inequality ∣x+y∣≤∣x∣+∣y∣=1|x+y|\leq|x|+|y|=1. Since 2π≈6.282\pi\approx 6.28 is far outside that, **only n=0n=0 survives** in all three families, collapsing them to x=0x=0, y=0y=0, or x+y=0x+y=0. Step 4: Hunt the coordinates. Intersect each line with the diamond ∣x∣+∣y∣=1|x|+|y|=1: * x=0  ⟹  ∣y∣=1x=0 \implies |y|=1: gives (0,1)(0,1) and (0,−1)(0,-1) * y=0  ⟹  ∣x∣=1y=0 \implies |x|=1: gives (1,0)(1,0) and (−1,0)(-1,0) * x=−y  ⟹  2∣x∣=1x=-y \implies 2|x|=1: gives (12,−12)\left(\tfrac{1}{2},-\tfrac{1}{2}\right) and (−12,12)\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)(0,0), where x=0x=0 and y=0y=0 meet - but it fails ∣x∣+∣y∣=1|x|+|y|=1, so it never enters the list. The six points above are pairwise distinct, and the count is genuinely 2+2+22+2+2. Final Answer: 6
Q3:ipmat indore 2020QA › TrigonometryHardMCQ · MCQ
The value of cos⁡2(π8)+cos⁡2(3π8)+cos⁡2(5π8)+cos⁡2(7π8)\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
  • A11
  • B32\frac{3}{2}
  • C22
  • D94\frac{9}{4}
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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 7π8=π−π8\frac{7\pi}{8}=\pi-\frac{\pi}{8}, and cos⁡(π−θ)=−cos⁡θ\cos(\pi-\theta)=-\cos\theta, so cos⁡(7π8)=−cos⁡(π8)\cos\left(\frac{7\pi}{8}\right)=-\cos\left(\frac{\pi}{8}\right). Squaring kills the minus sign: cos⁡2(7π8)=cos⁡2(π8)\cos^2\left(\frac{7\pi}{8}\right)=\cos^2\left(\frac{\pi}{8}\right) Identically, 5π8=π−3π8\frac{5\pi}{8}=\pi-\frac{3\pi}{8}, so cos⁡2(5π8)=cos⁡2(3π8)\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[cos⁡2(π8)+cos⁡2(3π8)]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 π8+3π8=4π8=π2\frac{\pi}{8}+\frac{3\pi}{8}=\frac{4\pi}{8}=\frac{\pi}{2}. The two angles are complementary, and cos⁡(π2−θ)=sin⁡θ\cos\left(\frac{\pi}{2}-\theta\right)=\sin\theta, so: cos⁡(3π8)=sin⁡(π8)\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[cos⁡2(π8)+sin⁡2(π8)]2\left[\cos^2\left(\frac{\pi}{8}\right)+\sin^2\left(\frac{\pi}{8}\right)\right] By the Pythagorean identity cos⁡2θ+sin⁡2θ=1\cos^2\theta+\sin^2\theta=1, the bracket collapses to 1: 2×1=22\times 1=2 Final Answer: 22
Q4:ipmat indore 2021QA › TrigonometryMediumMCQ · MCQ
If the angles A,B,CA, B, C of a triangle are in arithmetic progression such that sin⁡(2A+B)=1/2\sin(2A + B) = 1/2 then sin⁡(B+2C)\sin(B + 2C) is equal to
  • A−12\frac{-1}{2}
  • B12\frac{1}{2}
  • C−12\frac{-1}{\sqrt{2}}
  • D32\frac{3}{\sqrt{2}}
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The Setup: Geometry meets Trigonometry. The angles of a triangle always sum to 180∘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 BB. Since AA, BB, CC are in AP: A+C=2BA+C=2B Substitute this into the triangle angle sum theorem: A+B+C=180∘  ⟹  2B+B=180∘  ⟹  3B=180∘  ⟹  B=60∘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 AA. We know sin⁡(2A+B)=1/2\sin(2A+B)=1/2. Since A+C=120∘A+C=120^\circ, angle AA is stuck below 120∘120^\circ, so 2A+60∘2A+60^\circ never climbs past 300∘300^\circ - that leaves exactly two candidates, 30∘30^\circ and 150∘150^\circ. * If 2A+60∘=30∘  ⟹  2A=−30∘2A+60^\circ=30^\circ \implies 2A=-30^\circ (Angles in a triangle must be positive, so this is a wipe). * If 2A+60∘=150∘  ⟹  2A=90∘  ⟹  A=45∘2A+60^\circ=150^\circ \implies 2A=90^\circ \implies A=45^\circ. Step 3: Find angle CC. C=180∘−(A+B)=180∘−(45∘+60∘)=75∘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)\sin(B+2C). B+2C=60∘+2(75∘)=60∘+150∘=210∘B+2C=60^\circ+2(75^\circ)=60^\circ+150^\circ=210^\circ sin⁡(210∘)=sin⁡(180∘+30∘)=−sin⁡(30∘)=−12\sin(210^\circ)=\sin(180^\circ+30^\circ)=-\sin(30^\circ)=-\frac{1}{2} Final Answer: −12\frac{-1}{2}
Q5:ipmat indore 2022QA › TrigonometryHardMCQ · MCQ
For 0<θ<π40\lt\theta\lt\frac{\pi}{4}, let a=((sin⁡θ)sin⁡θ)(log⁡2cos⁡θ),b=((cos⁡θ)sin⁡θ)(log⁡2sin⁡θ),c=((sin⁡θ)cos⁡θ)(log⁡2cos⁡θ)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=((sin⁡θ)sin⁡θ)(log⁡2sin⁡θ)d=\left((\sin \theta)^{\sin \theta}\right)\left(\log _{2} \sin \theta\right). Then, the median value in the sequence a,b,c,da, b, c, d is
  • Aa+b2\frac{a+b}{2}
  • Ba+d2\frac{a+d}{2}
  • Cb+c2\frac{b+c}{2}
  • Dc+d2\frac{c+d}{2}
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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<θ<π/40 < \theta < \pi/4, we can determine their strict mathematical ordering. Step 1: Analyze the components within the given domain. For θ∈(0,π/4)\theta \in (0, \pi/4), the sine and cosine functions are bounded such that: 0<sin⁡θ<12<cos⁡θ<10 < \sin\theta < \frac{1}{\sqrt{2}} < \cos\theta < 1 Because both sin⁡θ\sin\theta and cos⁡θ\cos\theta are strictly between 00 and 11, their base-22 logarithms are strictly negative: log⁡2(sin⁡θ)<log⁡2(cos⁡θ)<0\log_2(\sin\theta) < \log_2(\cos\theta) < 0 Step 2: Compare the components of expressions a,b,c,da, b, c, d. The definitions are: a=(sin⁡θ)sin⁡θlog⁡2(cos⁡θ)a = (\sin\theta)^{\sin\theta} \log_2(\cos\theta) b=(cos⁡θ)sin⁡θlog⁡2(sin⁡θ)b = (\cos\theta)^{\sin\theta} \log_2(\sin\theta) c=(sin⁡θ)cos⁡θlog⁡2(cos⁡θ)c = (\sin\theta)^{\cos\theta} \log_2(\cos\theta) d=(sin⁡θ)sin⁡θlog⁡2(sin⁡θ)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 aa and cc: Since sin⁡θ<cos⁡θ\sin\theta < \cos\theta and the base sin⁡θ∈(0,1)\sin\theta \in (0, 1), a larger exponent yields a smaller value: (sin⁡θ)cos⁡θ<(sin⁡θ)sin⁡θ(\sin\theta)^{\cos\theta} < (\sin\theta)^{\sin\theta} Because log⁡2(cos⁡θ)\log_2(\cos\theta) is negative, multiplying both sides by it reverses the inequality: (sin⁡θ)cos⁡θlog⁡2(cos⁡θ)>(sin⁡θ)sin⁡θlog⁡2(cos⁡θ)  ⟹  c>a(\sin\theta)^{\cos\theta} \log_2(\cos\theta) > (\sin\theta)^{\sin\theta} \log_2(\cos\theta) \implies c > a Second, compare bb and dd: Because sin⁡θ<cos⁡θ\sin\theta < \cos\theta, raising them to the positive power of sin⁡θ\sin\theta preserves the inequality: (sin⁡θ)sin⁡θ<(cos⁡θ)sin⁡θ(\sin\theta)^{\sin\theta} < (\cos\theta)^{\sin\theta} Because log⁡2(sin⁡θ)\log_2(\sin\theta) is negative, multiplying reverses the inequality: (sin⁡θ)sin⁡θlog⁡2(sin⁡θ)>(cos⁡θ)sin⁡θlog⁡2(sin⁡θ)  ⟹  d>b(\sin\theta)^{\sin\theta} \log_2(\sin\theta) > (\cos\theta)^{\sin\theta} \log_2(\sin\theta) \implies d > b Third, compare aa and dd: They share the identical positive exponential factor (sin⁡θ)sin⁡θ(\sin\theta)^{\sin\theta}. Since log⁡2(sin⁡θ)<log⁡2(cos⁡θ)\log_2(\sin\theta) < \log_2(\cos\theta), multiplying by the positive exponential preserves the inequality: d<ad < a Step 4: Determine the median. Combining the inequalities gives the strict ascending order: b<d<a<cb < d < a < c. For a set of 4 elements, the median is the arithmetic mean of the two central terms (dd and aa). Median=a+d2\text{Median} = \frac{a + d}{2} Final Answer: a+d2\frac{a+d}{2}
Q6:ipmat indore 2021QA › TrigonometryMediumMCQ · MCQ
The set of all real value of pp for which the equation 3sin⁡2x+12cos⁡x−3=p3 \sin^2x + 12 \cos x - 3 = p has at least one solution is
  • A[−12,12][-12, 12]
  • B[−12,9][-12, 9]
  • C[−15,9][-15, 9]
  • D[−15,12][-15, 12]
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The Setup: We are converting a trig equation into a standard quadratic to find its absolute range (the min and max values of pp). Since the equation involves both sine and cosine, we use the identity sin⁡2x=1−cos⁡2x\sin^2x=1-\cos^2x to unify the variables. Step 1: Substitute and simplify the expression. Let f(x)=pf(x)=p. 3(1−cos⁡2x)+12cos⁡x−3=p3(1-\cos^2x)+12\cos x-3=p 3−3cos⁡2x+12cos⁡x−3=p3-3\cos^2x+12\cos x-3=p −3cos⁡2x+12cos⁡x=p-3\cos^2x+12\cos x=p Step 2: Set up a boundary replacement. Let t=cos⁡xt=\cos x. Because of the constraints of the cosine function, tt is strictly bounded: t∈[−1,1]t\in[-1,1]. Our new function is g(t)=−3t2+12tg(t)=-3t^2+12t. We need the absolute min and max of this quadratic on the locked interval [−1,1][-1,1]. Step 3: Find the vertex of the parabola. For a quadratic at2+bt+cat^2+bt+c, the vertex occurs at t=−b/(2a)t=-b/(2a). t=−122(−3)=−12−6=2t=\frac{-12}{2(-3)}=\frac{-12}{-6}=2 The vertex sits at t=2t=2, which is completely outside our playable zone of [−1,1][-1,1]. Because it's a downward-opening parabola (negative aa), it is strictly increasing across our entire [−1,1][-1,1] interval. Step 4: Test the boundaries to find the range. * If t=−1t=-1 (Minimum): g(−1)=−3(−1)2+12(−1)=−3−12=−15g(-1)=-3(-1)^2+12(-1)=-3-12=-15. * If t=1t=1 (Maximum): g(1)=−3(1)2+12(1)=−3+12=9g(1)=-3(1)^2+12(1)=-3+12=9. The function seamlessly spans every value between these two extremes. Final Answer: [−15,9][-15, 9]
Q7:ipmat indore 2022QA › TrigonometryMediumMCQ · MCQ
Ayesha is standing atop a vertical tower 200m200 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∘45^{\circ}. At 11:02 AM, she observes the angle of depression of the car to be 30∘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
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The Setup: A car is tracked from a 200m200\text{m} tower. Its angle of depression changes from 45∘45^\circ to 30∘30^\circ over exactly 22 minutes. We must use right-triangle trigonometry to find its speed in km/h\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∘45^\circ. Using trigonometry, tan⁡(45∘)=HeightDistance1\tan(45^\circ) = \frac{\text{Height}}{\text{Distance}_1}. 1=200d1  ⟹  d1=200 meters1 = \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∘30^\circ. tan⁡(30∘)=HeightDistance2\tan(30^\circ) = \frac{\text{Height}}{\text{Distance}_2} 13=200d2  ⟹  d2=2003 meters\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: Δd=d2−d1=2003−200=200(3−1)\Delta d = d_2 - d_1 = 200\sqrt{3} - 200 = 200(\sqrt{3} - 1) Using the approximation 3≈1.732\sqrt{3} \approx 1.732: Δd≈200(1.732−1)=200(0.732)=146.4 meters\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 146.4146.4 meters in 22 minutes. Speed=146.4 meters2 minutes=73.2 meters/minute\text{Speed} = \frac{146.4\text{ meters}}{2\text{ minutes}} = 73.2\text{ meters/minute} Convert to kilometers per hour (×60 minutes/hour,÷1000 meters/km\times 60 \text{ minutes/hour}, \div 1000 \text{ meters/km}): Speed=73.2×601000=43921000=4.392 km/h\text{Speed} = \frac{73.2 \times 60}{1000} = \frac{4392}{1000} = 4.392\text{ km/h} Final Answer: 4.39 km per hour
Q8:jipmat 2025QA › TrigonometryMediumQA · MCQ
Two ships are sailing in the sea on the two sides of a lighthouse. The angles of elevation of the top of the lighthouse as observed from two ships are 30 deg and 45 deg respectively. If the lighthouse is 100 m high, the distance between two ships is approximately:
  • A173 m
  • B200 m
  • C273 m
  • D300 m
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The Setup: We are dealing with two right-angled triangles sharing the same vertical height (the lighthouse). The ships are on *opposite* sides, meaning the total distance between them is the sum of their individual horizontal distances to the base of the lighthouse. Step 1: Calculate distance for the first ship (d1d_1). Using the tangent ratio (opposite/adjacent): tan⁡(30∘)=100d1\tan(30^\circ)=\frac{100}{d_1} 13=100d1  ⟹  d1=1003\frac{1}{\sqrt{3}}=\frac{100}{d_1} \implies d_1=100\sqrt{3} Since 3≈1.732\sqrt{3} \approx 1.732: d1=100(1.732)=173.2 md_1=100(1.732)=173.2\text{ m} Step 2: Calculate distance for the second ship (d2d_2). tan⁡(45∘)=100d2\tan(45^\circ)=\frac{100}{d_2} 1=100d2  ⟹  d2=100 m1=\frac{100}{d_2} \implies d_2=100\text{ m} Step 3: Add them together for the total distance. Total Distance=d1+d2=173.2+100=273.2 m\text{Total Distance}=d_1+d_2=173.2+100=273.2\text{ m} Rounding to the nearest whole number gives us 273273. Final Answer: 273
Q9:jipmat 2025QA › TrigonometryHardQA · MCQ
If sin⁡θ+cos⁡θsin⁡θ−cos⁡θ=3\frac{\sin\theta + \cos\theta}{\sin\theta - \cos\theta} = 3, then value of sin⁡4θ−cos⁡4θ\sin^4\theta - \cos^4\theta is
  • A15\frac{1}{5}
  • B25\frac{2}{5}
  • C35\frac{3}{5}
  • D45\frac{4}{5}
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The Setup: This looks like a trig nightmare, but we can easily finesse it by isolating a single trig ratio (like tan⁡θ\tan\theta) first, then substituting it back into a simplified version of the target expression. Step 1: Cross-multiply the initial equation to find the relationship between sine and cosine. sin⁡θ+cos⁡θsin⁡θ−cos⁡θ=31\frac{\sin\theta + \cos\theta}{\sin\theta - \cos\theta} = \frac{3}{1} sin⁡θ+cos⁡θ=3(sin⁡θ−cos⁡θ)\sin\theta + \cos\theta = 3(\sin\theta - \cos\theta) sin⁡θ+cos⁡θ=3sin⁡θ−3cos⁡θ\sin\theta + \cos\theta = 3\sin\theta - 3\cos\theta Group the terms together: 4cos⁡θ=2sin⁡θ4\cos\theta = 2\sin\theta Divide by 2cos⁡θ2\cos\theta to get tangent: sin⁡θcos⁡θ=42  ⟹  tan⁡θ=2\frac{\sin\theta}{\cos\theta} = \frac{4}{2} \implies \tan\theta = 2 Step 2: Build a right-angled triangle. Since tan⁡θ=OppositeAdjacent=21\tan\theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{2}{1}, we can find the hypotenuse using Pythagoras. Hypotenuse2=22+12=5  ⟹  Hypotenuse=5\text{Hypotenuse}^2 = 2^2 + 1^2 = 5 \implies \text{Hypotenuse} = \sqrt{5} This means sin⁡θ=25\sin\theta = \frac{2}{\sqrt{5}} and cos⁡θ=15\cos\theta = \frac{1}{\sqrt{5}}. Step 3: Simplify the target expression using the difference of squares identity: a2−b2=(a−b)(a+b)a^2 - b^2 = (a - b)(a + b). sin⁡4θ−cos⁡4θ=(sin⁡2θ−cos⁡2θ)(sin⁡2θ+cos⁡2θ)\sin^4\theta - \cos^4\theta = (\sin^2\theta - \cos^2\theta)(\sin^2\theta + \cos^2\theta) Since sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 (the most basic trig fact), the expression simply becomes: sin⁡2θ−cos⁡2θ\sin^2\theta - \cos^2\theta Step 4: Plug in our values from the triangle. sin⁡2θ=(25)2=45\sin^2\theta = (\frac{2}{\sqrt{5}})^2 = \frac{4}{5} cos⁡2θ=(15)2=15\cos^2\theta = (\frac{1}{\sqrt{5}})^2 = \frac{1}{5} Final Value=45−15=35\text{Final Value} = \frac{4}{5} - \frac{1}{5} = \frac{3}{5} Final Answer: 3/5
Q10: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
  • A15(5+1)15(\sqrt{5} + 1)
  • B20(3+1)20(\sqrt{3} + 1)
  • C3030
  • D10(3+1)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 hh be the height of the pole, and xx be the initial horizontal distance from point AA to the pole's base. tan⁡(30∘)=hx  ⟹  13=hx  ⟹  x=h3\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 2020 meters closer, the new distance is x−20x - 20. tan⁡(45∘)=hx−20\tan(45^\circ) = \frac{h}{x - 20} Since tan⁡(45∘)=1\tan(45^\circ) = 1: 1=hx−20  ⟹  h=x−201 = \frac{h}{x - 20} \implies h = x - 20 Step 3: Substitute and solve for hh. Replace xx with h3h\sqrt{3}: h=h3−20h = h\sqrt{3} - 20 20=h3−h=h(3−1)20 = h\sqrt{3} - h = h(\sqrt{3} - 1) h=203−1h = \frac{20}{\sqrt{3} - 1} Step 4: Rationalize the denominator. Multiply the numerator and denominator by the conjugate (3+1)(\sqrt{3} + 1): h=20(3+1)(3−1)(3+1)=20(3+1)3−1h = \frac{20(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{20(\sqrt{3} + 1)}{3 - 1} h=20(3+1)2=10(3+1)h = \frac{20(\sqrt{3} + 1)}{2} = 10(\sqrt{3} + 1) Final Answer: 10(3+1)10(\sqrt{3}+1)
Q11:ipmat indore 2019QA › TrigonometryHardMCQ · MCQ
Given that cos⁡x+cos⁡y=1\cos x + \cos y = 1, the range of sin⁡x−sin⁡y\sin x - \sin y is
  • A[−1,1][-1, 1]
  • B[−2,2][-2, 2]
  • C[0,3][0, \sqrt{3}]
  • D[−3,3][-\sqrt{3}, \sqrt{3}]
Pick an option to attempt
The Setup: One equation constrains the pair (x,y)(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⁡yS=\sin x-\sin y, given cos⁡x+cos⁡y=1\cos x+\cos y=1. Step 2: Square both and add. (cos⁡x+cos⁡y)2=1  ⟹  cos⁡2x+2cos⁡xcos⁡y+cos⁡2y=1(\cos x+\cos y)^2=1 \implies \cos^2x+2\cos x\cos y+\cos^2y=1 (sin⁡x−sin⁡y)2=S2  ⟹  sin⁡2x−2sin⁡xsin⁡y+sin⁡2y=S2(\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\sin^2+\cos^2=1 for each of xx and yy: 1+1+2(cos⁡xcos⁡y−sin⁡xsin⁡y)=1+S21+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)\cos(x+y), so: 2+2cos⁡(x+y)=1+S2  ⟹  S2=1+2cos⁡(x+y)2+2\cos(x+y)=1+S^2 \implies S^2=1+2\cos(x+y) **Step 3: Bound cos⁡(x+y)\cos(x+y) using the constraint.** Sum-to-product on the given equation, writing u=x+y2u=\frac{x+y}{2} and v=x−y2v=\frac{x-y}{2}: 2cos⁡ucos⁡v=1  ⟹  cos⁡ucos⁡v=122\cos u\cos v=1 \implies \cos u\cos v=\frac{1}{2} Since ∣cos⁡v∣≤1|\cos v|\leq 1, we need ∣cos⁡u∣≥12|\cos u|\geq\frac{1}{2}, hence cos⁡2u≥14\cos^2u\geq\frac{1}{4}. The double-angle formula then gives: cos⁡(x+y)=2cos⁡2u−1≥2⋅14−1=−12\cos(x+y)=2\cos^2u-1 \geq 2\cdot\frac{1}{4}-1=-\frac{1}{2} and of course cos⁡(x+y)≤1\cos(x+y)\leq 1 always. Substituting into S2=1+2cos⁡(x+y)S^2=1+2\cos(x+y): S2∈[ 1−1, 1+2 ]=[0, 3]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\cos x+\cos y=1: * x=π3, y=−π3x=\frac{\pi}{3},\ y=-\frac{\pi}{3}: then cos⁡x+cos⁡y=12+12=1\cos x+\cos y=\frac{1}{2}+\frac{1}{2}=1 ✓ and S=32−(−32)=3S=\frac{\sqrt{3}}{2}-\left(-\frac{\sqrt{3}}{2}\right)=\sqrt{3}, the maximum. * x=y=π3x=y=\frac{\pi}{3}: then cos⁡x+cos⁡y=1\cos x+\cos y=1 ✓ and S=0S=0, the minimum of ∣S∣|S|. Swapping xx and yy negates SS, so −3-\sqrt{3} is reached too, and since SS varies continuously it sweeps everything between: S∈[−3, 3]S\in\left[-\sqrt{3},\ \sqrt{3}\right] Final Answer: [−3,3][-\sqrt{3}, \sqrt{3}]
Q12:ipmat indore 2019QA › TrigonometryHardMCQ · MCQ
If sinθ+cosθ=msin \theta + cos \theta = m, then sin6θ+cos6θsin^6 \theta + cos^6 \theta equals
  • A3(m2+1)4\frac{3(m^2+1)}{4}
  • B3(m2−1)4\frac{3(m^2-1)}{4}
  • C1−3(m2−1)41-\frac{3(m^2-1)}{4}
  • D1−3(m2−1)241-\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⁡θcos⁡θ\sin\theta\cos\theta, which squaring the given equation hands us immediately. Step 1: Extract the cross-term. Square both sides: (sin⁡θ+cos⁡θ)2=m2  ⟹  sin⁡2θ+cos⁡2θ+2sin⁡θcos⁡θ=m2(\sin\theta+\cos\theta)^2=m^2 \implies \sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta=m^2 Since sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1: 1+2sin⁡θcos⁡θ=m2  ⟹  sin⁡θcos⁡θ=m2−121+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 a3+b3=(a+b)(a2−ab+b2)a^3+b^3=(a+b)\left(a^2-ab+b^2\right) with a=sin⁡2θa=\sin^2\theta and b=cos⁡2θb=\cos^2\theta: sin⁡6θ+cos⁡6θ=(sin⁡2θ+cos⁡2θ)(sin⁡4θ−sin⁡2θcos⁡2θ+cos⁡4θ)\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 a2+b2=(a+b)2−2aba^2+b^2=(a+b)^2-2ab on sin⁡4θ+cos⁡4θ\sin^4\theta+\cos^4\theta, which turns a2−ab+b2a^2-ab+b^2 into (a+b)2−3ab(a+b)^2-3ab: sin⁡4θ−sin⁡2θcos⁡2θ+cos⁡4θ=(sin⁡2θ+cos⁡2θ)2−3sin⁡2θcos⁡2θ=1−3(sin⁡θcos⁡θ)2\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⁡6θ+cos⁡6θ=1−3(m2−12)2=1−3(m2−1)24\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 1−3(m2−1)41-\frac{3(m^2-1)}{4} - the difference is one exponent, so check it before committing. Spot-check at θ=0\theta=0: then m=sin⁡0+cos⁡0=1m=\sin 0+\cos 0=1, and the formula gives 1−3(1−1)24=11-\frac{3(1-1)^2}{4}=1, matching sin⁡60+cos⁡60=0+1=1\sin^6 0+\cos^6 0=0+1=1. Final Answer: 1−3(m2−1)241-\frac{3(m^2-1)^2}{4}
Q13:ipmat indore 2023QA › TrigonometryMediumMCQ · MCQ
If cosαcos \alpha + cosβcos \beta = 1 then the maximum value of sinα−sinβsin \alpha - sin \beta is
  • A2\sqrt{2}
  • B22
  • C3\sqrt{3}
  • D11
Pick an option to attempt
The Setup: Given the constraint cos⁡α+cos⁡β=1\cos\alpha + \cos\beta = 1, we must find the maximum value of sin⁡α−sin⁡β\sin\alpha - \sin\beta. Step 1: Rewrite both expressions with the sum-to-product identities. Let u=α+β2u = \frac{\alpha+\beta}{2} and v=α−β2v = \frac{\alpha-\beta}{2}. The constraint and the target expression become: cos⁡α+cos⁡β=2cos⁡ucos⁡v=1\cos\alpha + \cos\beta = 2\cos u \cos v = 1 sin⁡α−sin⁡β=2cos⁡usin⁡v\sin\alpha - \sin\beta = 2\cos u \sin v Both share the same factor 2cos⁡u2\cos u, which is what makes the constraint substitutable. Step 2: Eliminate uu using the constraint. The constraint gives 2cos⁡u=1cos⁡v2\cos u = \frac{1}{\cos v} directly (note cos⁡v≠0\cos v \neq 0). Substitute this into the target: sin⁡α−sin⁡β=(1cos⁡v)sin⁡v=tan⁡v\sin\alpha - \sin\beta = \left(\frac{1}{\cos v}\right)\sin v = \tan v The whole problem has collapsed to maximising tan⁡v\tan v. Step 3: Find the admissible range of vv. cos⁡u\cos u is a cosine, so it is bounded by ∣cos⁡u∣≤1|\cos u| \le 1: ∣12cos⁡v∣≤1  ⟹  ∣cos⁡v∣≥12\left|\frac{1}{2\cos v}\right| \le 1 \implies |\cos v| \ge \frac{1}{2} On the branch where tan⁡v\tan v is positive and increasing, this restricts vv to 0≤v≤π30 \le v \le \frac{\pi}{3}. Step 4: Maximise on that range. tan⁡v\tan v increases throughout [0,π3]\left[0, \frac{\pi}{3}\right], so the maximum sits at the right endpoint: Maximum=tan⁡π3=3\text{Maximum} = \tan\frac{\pi}{3} = \sqrt{3} Step 5: Confirm the maximum is actually attained. At v=π3v = \frac{\pi}{3} the constraint forces cos⁡u=12cos⁡(π/3)=1\cos u = \frac{1}{2\cos(\pi/3)} = 1, so u=0u = 0, giving α=π3\alpha = \frac{\pi}{3} and β=−π3\beta = -\frac{\pi}{3}. Check both conditions: cos⁡π3+cos⁡(−π3)=12+12=1\cos\frac{\pi}{3} + \cos\left(-\frac{\pi}{3}\right) = \frac{1}{2} + \frac{1}{2} = 1 sin⁡π3−sin⁡(−π3)=32+32=3\sin\frac{\pi}{3} - \sin\left(-\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} = \sqrt{3} Both hold, so 3\sqrt{3} is genuinely achieved and is the maximum. Final Answer: 3\sqrt{3}

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