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EAPCET Mathematics · Intermediate first year · Paper IA

Trigonometric Ratios up to Transformations

This chapter builds the trigonometric toolkit: ratios of compound, multiple and sub-multiple angles, and the transformations between sums and products.

Subject
Mathematics
Taught in
Intermediate first year · Paper IA
Mathematics in the paper
80 of 160 questions

Key formulas and results

  • sin⁡(A±B)=sin⁡Acos⁡B±cos⁡Asin⁡B\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B
  • cos⁡2A=cos⁡2A−sin⁡2A=1−2sin⁡2A=2cos⁡2A−1\cos 2A = \cos^{2}A - \sin^{2}A = 1 - 2\sin^{2}A = 2\cos^{2}A - 1
  • tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B\tan(A + B) = \dfrac{\tan A + \tan B}{1 - \tan A \tan B}
  • sin⁡3A=3sin⁡A−4sin⁡3A\sin 3A = 3\sin A - 4\sin^{3}A; cos⁡3A=4cos⁡3A−3cos⁡A\cos 3A = 4\cos^{3}A - 3\cos A
  • sin⁡C+sin⁡D=2sin⁡C+D2cos⁡C−D2\sin C + \sin D = 2\sin\dfrac{C + D}{2}\cos\dfrac{C - D}{2}
  • asin⁡θ+bcos⁡θa\sin\theta + b\cos\theta ranges over [−a2+b2,a2+b2]\left[-\sqrt{a^{2} + b^{2}}, \sqrt{a^{2} + b^{2}}\right]
  • sin⁡18∘=5−14\sin 18^\circ = \dfrac{\sqrt{5} - 1}{4} and cos⁡36∘=5+14\cos 36^\circ = \dfrac{\sqrt{5} + 1}{4}
  • tan⁡15∘=2−3\tan 15^\circ = 2 - \sqrt{3}, tan⁡75∘=2+3\tan 75^\circ = 2 + \sqrt{3}, tan⁡2212∘=2−1\tan 22\tfrac{1}{2}^\circ = \sqrt{2} - 1
  • tan⁡3A=3tan⁡A−tan⁡3A1−3tan⁡2A\tan 3A = \dfrac{3\tan A - \tan^{3}A}{1 - 3\tan^{2}A}
  • sin⁡2A=2tan⁡A1+tan⁡2A\sin 2A = \dfrac{2\tan A}{1 + \tan^{2}A} and cos⁡2A=1−tan⁡2A1+tan⁡2A\cos 2A = \dfrac{1 - \tan^{2}A}{1 + \tan^{2}A}
  • sin⁡θsin⁡(60∘−θ)sin⁡(60∘+θ)=14sin⁡3θ\sin\theta \sin(60^\circ - \theta) \sin(60^\circ + \theta) = \tfrac{1}{4}\sin 3\theta; cos⁡θcos⁡(60∘−θ)cos⁡(60∘+θ)=14cos⁡3θ\cos\theta \cos(60^\circ - \theta) \cos(60^\circ + \theta) = \tfrac{1}{4}\cos 3\theta; tan⁡θtan⁡(60∘−θ)tan⁡(60∘+θ)=tan⁡3θ\tan\theta \tan(60^\circ - \theta) \tan(60^\circ + \theta) = \tan 3\theta
  • cos⁡xcos⁡2xcos⁡4x⋯cos⁡2n−1x=sin⁡2nx2nsin⁡x\cos x \cos 2x \cos 4x \cdots \cos 2^{n-1}x = \dfrac{\sin 2^{n}x}{2^{n}\sin x}
  • sin⁡α+sin⁡(α+β)+⋯+sin⁡(α+(n−1)β)=sin⁡nβ2 sin⁡(α+(n−1)β2)sin⁡β2\sin\alpha + \sin(\alpha + \beta) + \dots + \sin(\alpha + (n - 1)\beta) = \dfrac{\sin\frac{n\beta}{2}\,\sin\left(\alpha + \frac{(n - 1)\beta}{2}\right)}{\sin\frac{\beta}{2}}; the cosine sum has cos⁡(α+(n−1)β2)\cos\left(\alpha + \frac{(n - 1)\beta}{2}\right) in place of the second sine
  • If A+B+C=πA + B + C = \pi: tan⁡A+tan⁡B+tan⁡C=tan⁡Atan⁡Btan⁡C\tan A + \tan B + \tan C = \tan A \tan B \tan C and sin⁡2A+sin⁡2B+sin⁡2C=4sin⁡Asin⁡Bsin⁡C\sin 2A + \sin 2B + \sin 2C = 4\sin A \sin B \sin C

What this chapter covers

  • Trigonometric ratios of compound angles
  • Multiple and sub-multiple angles
  • Sum-to-product and product-to-sum transformations
  • Periodicity and extreme values

How it is usually asked

  • Simplify or evaluate a trigonometric expression
  • Find the maximum or minimum of a sin θ + b cos θ
  • Evaluate products such as cos 20° cos 40° cos 80°
  • Find the period of a trigonometric function

Understanding Trigonometric Ratios up to Transformations

Everything in this chapter grows from the compound angle formulas for sin(A ± B), cos(A ± B) and tan(A ± B). Putting B = A gives the double angle formulas, such as sin 2A = 2 sin A cos A and the three forms of cos 2A, and applying them again gives triple angle formulas like sin 3A = 3 sin A − 4 sin³A. Replacing A by A/2 gives half-angle results.

Adding and subtracting the compound angle formulas produces the transformation formulas. Sum-to-product formulas such as sin C + sin D = 2 sin((C + D)/2) cos((C − D)/2) turn a sum into a product, which is easier to simplify or set to zero, and the product-to-sum formulas do the reverse. Choosing the direction that makes terms cancel is the main skill.

Two further ideas appear often. The expression a sin θ + b cos θ can be written as √(a² + b²) sin(θ + α), so its values lie between −√(a² + b²) and √(a² + b²). Periods matter too: sin x and cos x repeat every 2π, tan x every π. EAPCET asks for exact values, simplified products of cosines, extreme values and periods.

Worked example

EAPCET-style question

The value of sin⁡20∘sin⁡40∘sin⁡80∘\sin 20^\circ \sin 40^\circ \sin 80^\circ is

  1. (1)38\dfrac{\sqrt{3}}{8}
  2. (2)18\dfrac{1}{8}
  3. (3)34\dfrac{\sqrt{3}}{4}
  4. (4)38\dfrac{3}{8}

Solution

  1. sin⁡40∘sin⁡80∘=12[cos⁡40∘−cos⁡120∘]=12[cos⁡40∘+12]\sin 40^\circ \sin 80^\circ = \tfrac{1}{2}[\cos 40^\circ - \cos 120^\circ] = \tfrac{1}{2}\left[\cos 40^\circ + \tfrac{1}{2}\right].
  2. So the product =12sin⁡20∘cos⁡40∘+14sin⁡20∘= \tfrac{1}{2}\sin 20^\circ \cos 40^\circ + \tfrac{1}{4}\sin 20^\circ.
  3. sin⁡20∘cos⁡40∘=12[sin⁡60∘+sin⁡(−20∘)]=12[32−sin⁡20∘]\sin 20^\circ \cos 40^\circ = \tfrac{1}{2}[\sin 60^\circ + \sin(-20^\circ)] = \tfrac{1}{2}\left[\tfrac{\sqrt{3}}{2} - \sin 20^\circ\right].
  4. Substitute: 14(32−sin⁡20∘)+14sin⁡20∘=38\tfrac{1}{4}\left(\tfrac{\sqrt{3}}{2} - \sin 20^\circ\right) + \tfrac{1}{4}\sin 20^\circ = \tfrac{\sqrt{3}}{8}.
  5. This matches the identity sin⁡θsin⁡(60∘−θ)sin⁡(60∘+θ)=14sin⁡3θ\sin\theta \sin(60^\circ - \theta) \sin(60^\circ + \theta) = \tfrac{1}{4}\sin 3\theta with θ=20∘\theta = 20^\circ.

Answer: 38\dfrac{\sqrt{3}}{8}

Shortcuts and results to remember

  • sin⁡4θ+cos⁡4θ=1−12sin⁡22θ\sin^{4}\theta + \cos^{4}\theta = 1 - \tfrac{1}{2}\sin^{2}2\theta and sin⁡6θ+cos⁡6θ=1−34sin⁡22θ\sin^{6}\theta + \cos^{6}\theta = 1 - \tfrac{3}{4}\sin^{2}2\theta, so their ranges are [12,1]\left[\tfrac{1}{2}, 1\right] and [14,1]\left[\tfrac{1}{4}, 1\right].
  • cot⁡θ−tan⁡θ=2cot⁡2θ\cot\theta - \tan\theta = 2\cot 2\theta; repeated use gives tan⁡A+2tan⁡2A+4tan⁡4A+⋯+2n−1tan⁡2n−1A+2ncot⁡2nA=cot⁡A\tan A + 2\tan 2A + 4\tan 4A + \dots + 2^{n-1}\tan 2^{n-1}A + 2^{n}\cot 2^{n}A = \cot A.
  • 2+2+⋯+2+2cos⁡θ\sqrt{2 + \sqrt{2 + \dots + \sqrt{2 + 2\cos\theta}}} with nn square roots equals 2cos⁡θ2n2\cos\dfrac{\theta}{2^{n}}.
  • To lower powers: sin⁡3θ=3sin⁡θ−sin⁡3θ4\sin^{3}\theta = \dfrac{3\sin\theta - \sin 3\theta}{4} and cos⁡3θ=3cos⁡θ+cos⁡3θ4\cos^{3}\theta = \dfrac{3\cos\theta + \cos 3\theta}{4}.
  • acos⁡x+bsin⁡x+ca\cos x + b\sin x + c ranges from c−a2+b2c - \sqrt{a^{2} + b^{2}} to c+a2+b2c + \sqrt{a^{2} + b^{2}}.

More practice questions

Practice question 1

The value of cos⁡20∘cos⁡40∘cos⁡80∘\cos 20^\circ \cos 40^\circ \cos 80^\circ is

  1. (1)18\dfrac{1}{8}
  2. (2)14\dfrac{1}{4}
  3. (3)116\dfrac{1}{16}
  4. (4)38\dfrac{\sqrt{3}}{8}

Solution

  1. Use cos⁡θcos⁡(60∘−θ)cos⁡(60∘+θ)=14cos⁡3θ\cos\theta \cos(60^\circ - \theta) \cos(60^\circ + \theta) = \tfrac{1}{4}\cos 3\theta
  2. Take θ=20∘\theta = 20^\circ: 60∘−θ=40∘60^\circ - \theta = 40^\circ and 60∘+θ=80∘60^\circ + \theta = 80^\circ
  3. Product =14cos⁡60∘= \tfrac{1}{4}\cos 60^\circ
  4. =14×12=18= \tfrac{1}{4} \times \tfrac{1}{2} = \tfrac{1}{8}

Answer: 18\dfrac{1}{8}

Practice question 2

The value of cos⁡π7cos⁡2π7cos⁡4π7\cos\dfrac{\pi}{7}\cos\dfrac{2\pi}{7}\cos\dfrac{4\pi}{7} is

  1. (1)−18-\dfrac{1}{8}
  2. (2)18\dfrac{1}{8}
  3. (3)14\dfrac{1}{4}
  4. (4)−14-\dfrac{1}{4}

Solution

  1. Use cos⁡xcos⁡2xcos⁡4x=sin⁡8x8sin⁡x\cos x \cos 2x \cos 4x = \dfrac{\sin 8x}{8\sin x} with x=π7x = \tfrac{\pi}{7}
  2. Product =sin⁡(8π/7)8sin⁡(π/7)= \dfrac{\sin(8\pi/7)}{8\sin(\pi/7)}
  3. sin⁡8π7=sin⁡(π+π7)=−sin⁡π7\sin\tfrac{8\pi}{7} = \sin\left(\pi + \tfrac{\pi}{7}\right) = -\sin\tfrac{\pi}{7}
  4. Product =−sin⁡(π/7)8sin⁡(π/7)=−18= \dfrac{-\sin(\pi/7)}{8\sin(\pi/7)} = -\dfrac{1}{8}

Answer: −18-\dfrac{1}{8}

Common mistakes to avoid

  • Writing cos(A + B) = cos A cos B + sin A sin B; the sum formula for cosine has a minus sign.
  • Using 1 + tan A tan B in the denominator of tan(A + B); for the sum it is 1 − tan A tan B.
  • Taking a + b as the maximum of a sin θ + b cos θ; the maximum is √(a² + b²), plus any constant added to the expression.
  • Giving 2π as the period of sin²x or |sin x|; squaring or taking modulus halves it to π.

How to prepare it

Know the exact values of the standard angles, including 15°, 18° and 36° — for example sin 18° = (√5 − 1)/4.

When an expression looks heavy, try converting sums to products (or the reverse) before anything else.

Once the formulas feel familiar, test them the way the exam does: practise Trigonometric Ratios up to Transformations questions chapter-wise, then attempt a full previous year paper to see whether it holds up under a 3-hour timer.

Related Mathematics chapters

See every chapter in the EAPCET Mathematics syllabus.

Practise Trigonometric Ratios up to Transformations chapter-wise

Chapter-wise quizzes show exactly which topics in this chapter you still get wrong — before a full paper does.

Frequently asked questions

Is Trigonometric Ratios up to Transformations part of the EAPCET Mathematics syllabus?
Yes. Trigonometric Ratios up to Transformations is taught in Intermediate first year · Paper IA Mathematics, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.
How should I prepare Trigonometric Ratios up to Transformations for EAPCET?
Know the exact values of the standard angles, including 15°, 18° and 36° — for example sin 18° = (√5 − 1)/4. When an expression looks heavy, try converting sums to products (or the reverse) before anything else.
How many Mathematics questions are there in EAPCET?
80 of the 160 questions in the Engineering paper are Mathematics, spread across 37 chapters.
What mistakes do students make in Trigonometric Ratios up to Transformations?
Writing cos(A + B) = cos A cos B + sin A sin B; the sum formula for cosine has a minus sign. Using 1 + tan A tan B in the denominator of tan(A + B); for the sum it is 1 − tan A tan B. Taking a + b as the maximum of a sin θ + b cos θ; the maximum is √(a² + b²), plus any constant added to the expression. Giving 2π as the period of sin²x or |sin x|; squaring or taking modulus halves it to π.