Q.Integrate the following function: $\sin x \sin 2x \sin 3x$

Showing the 12 most recent of 42 on this concept.

1. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
   Q.The solution set of the equation $\cos^2 2x + \sin^2 3x = 1$ is (A) $\left\{ x / x = n\pi + \frac{\pi}{2}, n \in \mathbb{Z} \right\}$ (B) $\left\{ x / x = 2n\pi \pm \frac{\pi}{4}, n \in \mathbb{Z} \right\}$ (C) $\left\{ x / x = \frac{n\pi}{5}, n \in \mathbb{Z} \right\}$ (D) $\left\{ x / x = n\pi + (-1)^n \frac{\pi}{6}, n \in \mathbb{Z} \right\}$

   ›Reveal solutionSolution

   The equation $\cos^2 2x + \sin^2 3x = 1$ simplifies to $\sin^2 3x = \sin^2 2x$, which leads to two families of solutions; the union of these gives $x = \frac{n\pi}{5}$, so the correct option is (C).

   We start with the equation

   $\cos^2 2x + \sin^2 3x = 1.$

   A natural first thought is to use the identity $\cos^2 \theta = 1 - \sin^2 \theta$, but here the angles are different ($2x$ and $3x$). Instead, recall the Pythagorean identity: $\cos^2 \alpha + \sin^2 \alpha = 1$. Our equation looks similar, but the angles don’t match. That mismatch is the key: we can rewrite $\cos^2 2x$ as $1 - \sin^2 2x$, then the equation becomes

   $1 - \sin^2 2x + \sin^2 3x = 1 \quad\Rightarrow\quad \sin^2 3x - \sin^2 2x = 0.$

   So we have $\sin^2 3x = \sin^2 2x$. This is a clean, symmetric condition. Taking square roots gives $\sin 3x = \pm \sin 2x$, which is equivalent to two cases: $\sin 3x = \sin 2x$ or $\sin 3x = -\sin 2x$. But we can handle both elegantly using the identity $\sin A = \sin B$ or the difference-of-squares factorization.

   1. Rewrite using difference of squares

   $\sin^2 3x - \sin^2 2x = 0 \quad\Rightarrow\quad (\sin 3x - \sin 2x)(\sin 3x + \sin 2x) = 0.$

   So either $\sin 3x = \sin 2x$ or $\sin 3x = -\sin 2x$.

   2. Solve $\sin 3x = \sin 2x$ The general solution for $\sin A = \sin B$ is

   $A = B + 2n\pi \quad\text{or}\quad A = \pi - B + 2n\pi,\quad n\in\mathbb{Z}.$

   • First branch: $3x = 2x + 2n\pi \Rightarrow x = 2n\pi$.
   • Second branch: $3x = \pi - 2x + 2n\pi \Rightarrow 5x = \pi + 2n\pi \Rightarrow x = \frac{\pi}{5} + \frac{2n\pi}{5} = \frac{(2n+1)\pi}{5}$.

   3. Solve $\sin 3x = -\sin 2x$

      Note $-\sin 2x = \sin(-2x)$. So we have $\sin 3x = \sin(-2x)$. Again apply the same formula:

      • First branch: $3x = -2x + 2n\pi \Rightarrow 5x = 2n\pi \Rightarrow x = \frac{2n\pi}{5}$.
      • Second branch: $3x = \pi - (-2x) + 2n\pi = \pi + 2x + 2n\pi \Rightarrow x = \pi + 2n\pi$, i.e., $x = (2n+1)\pi$.

   4. Combine all solutions

      From step 2: $x = 2n\pi$ and $x = \frac{(2n+1)\pi}{5}$.

      From step 3: $x = \frac{2n\pi}{5}$ and $x = (2n+1)\pi$.

      Notice that $2n\pi$ and $(2n+1)\pi$ are just multiples of $\pi$, which are already included in $\frac{2n\pi}{5}$ when $n$ is a multiple of 5? Actually, let’s check:

      • $x = 2n\pi$ is $\frac{10n\pi}{5}$, which is of the form $\frac{2k\pi}{5}$ with $k=5n$. …

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2. TG EAPCET 2022Set eng-2022-07-20-AN1 markMCQ
   Q.If $\sin \theta - \cos \theta = \dfrac{1}{\sqrt{3}}$, then $\sin (2\theta) + \cos (4\theta) + \sin (6\theta) =$ (A) $\dfrac{37}{27}$ (B) $-\dfrac{37}{27}$ (C) $-\dfrac{43}{27}$ (D) $\dfrac{43}{27}$

   ›Reveal solutionSolution

   We first determine $\sin(2\theta)$ from the given equation by squaring it. Then, we use double and triple angle formulas to find $\cos(4\theta)$ and $\sin(6\theta)$ in terms of $\sin(2\theta)$, and sum these values to get the final result $\boxed{\frac{43}{27}}$.

   The core idea here is to simplify the given expression $\sin \theta - \cos \theta = \frac{1}{\sqrt{3}}$ to find a value for $\sin(2\theta)$. Once $\sin(2\theta)$ is known, we can use standard trigonometric identities (specifically, double and triple angle formulas) to express $\cos(4\theta)$ and $\sin(6\theta)$ in terms of $\sin(2\theta)$. This strategy allows us to evaluate the entire expression without needing to find the value of $\theta$ itself.

   Let's break down the solution step-by-step.

   1. Find $\sin(2\theta)$ from the given equation. We are given the equation $\sin \theta - \cos \theta = \frac{1}{\sqrt{3}}$. To introduce $\sin(2\theta)$, which is $2\sin\theta\cos\theta$, we can square both sides of the equation:

   $(\sin \theta - \cos \theta)^2 = \left(\frac{1}{\sqrt{3}}\right)^2$

   Expand the left side using $(a-b)^2 = a^2 - 2ab + b^2$:
   

   $\sin^2 \theta + \cos^2 \theta - 2\sin \theta \cos \theta = \frac{1}{3}$

   Recall the fundamental trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$ and the double angle formula $\sin(2\theta) = 2\sin \theta \cos \theta$. Substitute these into the equation:
   

   $1 - \sin(2\theta) = \frac{1}{3}$

   Now, solve for $\sin(2\theta)$:
   

   $\sin(2\theta) = 1 - \frac{1}{3}$

   $\sin(2\theta) = \frac{2}{3}$

   2. Find $\cos(4\theta)$ using $\sin(2\theta)$. We need to express $\cos(4\theta)$ in terms of $\sin(2\theta)$. We can use the double angle formula for cosine:

      $\cos(2A) = 1 - 2\sin^2 A$

      Let $A = 2\theta$. Then $2A = 4\theta$.

   $\cos(4\theta) = 1 - 2\sin^2(2\theta)$

   Substitute the value of $\sin(2\theta) = \frac{2}{3}$ we found in Step 1:
   

   $\cos(4\theta) = 1 - 2\left(\frac{2}{3}\right)^2$

   $\cos(4\theta) = 1 - 2\left(\frac{4}{9}\right)$

   $\cos(4\theta) = 1 - \frac{8}{9}$

   $\cos(4\theta) = \frac{1}{9}$

   3. Find $\sin(6\theta)$ using $\sin(2\theta)$. We need to express $\sin(6\theta)$ in terms of $\sin(2\theta)$. We can use the triple angle formula for sine:

      $\sin(3A) = 3\sin A - 4\sin^3 A$

      Let $A = 2\theta$. Then $3A = 6\theta$.

      $\sin(6\theta) = \sin(3 \cdot 2\theta) = 3\sin(2\theta) - 4\sin^3(2\theta)$ …

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3. TG EAPCET 2022Set eng-2022-07-19-FN1 markMCQ
   Q.$\lim_{x \to 0} \left( \frac{4!}{x^8} \left( 1 - \cos \frac{x^2}{3} - \cos \frac{x^2}{4} + \cos \frac{x^2}{3} \cos \frac{x^2}{4} \right) \right) =$ (A) $8$ (B) $\frac{1}{6}$ (C) $\frac{1}{24}$ (D) $\frac{2}{3}$

   ›Reveal solutionSolution

   The bracket factors as $(1-\cos\tfrac{x^2}{3})(1-\cos\tfrac{x^2}{4})$; each behaves like $\tfrac{t^2}{2}$, giving $\dfrac{4!}{x^8}\cdot\dfrac{x^8}{576}=\dfrac{1}{24}$ — option (C).

   Factor the expression

   With $A=\dfrac{x^2}{3}$ and $B=\dfrac{x^2}{4}$,

   $1-\cos A-\cos B+\cos A\cos B=(1-\cos A)(1-\cos B).$

   Use $1-\cos t\sim\tfrac{t^2}{2}$

   As $x\to0$,

   $1-\cos\frac{x^2}{3}\sim\frac12\left(\frac{x^2}{3}\right)^2=\frac{x^4}{18},\qquad 1-\cos\frac{x^2}{4}\sim\frac12\left(\frac{x^2}{4}\right)^2=\frac{x^4}{32}.$ …

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4. TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
   Q.$\sin 20^\circ (4 + \sec 20^\circ) =$ (A) $\sqrt{3}$ (B) $-\sqrt{3}$ (C) $1$ (D) $-1$

   ›Reveal solutionSolution

   The key idea is to rewrite the expression using sine and cosine, then apply triple-angle identities to simplify it to a known constant. The final result is $\sqrt{3}$.

   We start with the expression

   $\sin 20^\circ (4 + \sec 20^\circ).$

   The presence of $\sec 20^\circ$ suggests rewriting everything in terms of sine and cosine, since $\sec \theta = 1/\cos \theta$. Then we look for a way to combine terms into a known trigonometric identity — the triple-angle formulas for sine and cosine are natural here because $20^\circ$ is one-third of $60^\circ$.

   ---

   1. Rewrite in terms of sine and cosine

   $\sin 20^\circ \left(4 + \frac{1}{\cos 20^\circ}\right) = 4\sin 20^\circ + \frac{\sin 20^\circ}{\cos 20^\circ} = 4\sin 20^\circ + \tan 20^\circ.$

   2. Express $\tan 20^\circ$ as $\frac{\sin 20^\circ}{\cos 20^\circ}$ and combine over a common denominator

   $4\sin 20^\circ + \frac{\sin 20^\circ}{\cos 20^\circ} = \frac{4\sin 20^\circ \cos 20^\circ + \sin 20^\circ}{\cos 20^\circ} = \frac{\sin 20^\circ (4\cos 20^\circ + 1)}{\cos 20^\circ}.$

   3. Use the double-angle identity $2\sin 20^\circ \cos 20^\circ = \sin 40^\circ$, so $4\sin 20^\circ \cos 20^\circ = 2\sin 40^\circ$. Then the numerator becomes

   $2\sin 40^\circ + \sin 20^\circ.$

   4. Apply the triple-angle identity for sine Recall: $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$. For $\theta = 20^\circ$, $\sin 60^\circ = \frac{\sqrt{3}}{2}$, so

   $\frac{\sqrt{3}}{2} = 3\sin 20^\circ - 4\sin^3 20^\circ.$

   This doesn't directly match our numerator, but we can also use the identity for $\sin 3\theta$ in terms of products:

   $\sin 3\theta = 4\sin\theta \sin(60^\circ - \theta) \sin(60^\circ + \theta).$

   For $\theta = 20^\circ$, this gives

   $\sin 60^\circ = 4\sin 20^\circ \sin 40^\circ \sin 80^\circ.$

   So $\sin 20^\circ \sin 40^\circ \sin 80^\circ = \frac{\sqrt{3}}{8}$.

   But we have $2\sin 40^\circ + \sin 20^\circ$ — not a product. So let's try a different path.

   5. Better approach: Use the triple-angle identity for cosine $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$. For $\theta = 20^\circ$, $\cos 60^\circ = \frac{1}{2}$, so

   $\frac{1}{2} = 4\cos^3 20^\circ - 3\cos 20^\circ.$

   Multiply both sides by 2:

   $1 = 8\cos^3 20^\circ - 6\cos 20^\circ.$

   Rearranging:

   $8\cos^3 20^\circ - 6\cos 20^\circ - 1 = 0.$

   6. Relate this to our expression We have $\frac{2\sin 40^\circ + \sin 20^\circ}{\cos 20^\circ}$. Note $\sin 40^\circ = 2\sin 20^\circ \cos 20^\circ$, so

   $2\sin 40^\circ = 4\sin 20^\circ \cos 20^\circ.$

   Then numerator = $4\sin 20^\circ \cos 20^\circ + \sin 20^\circ = \sin 20^\circ (4\cos 20^\circ + 1)$.

   So the whole expression is

   $\frac{\sin 20^\circ (4\cos 20^\circ + 1)}{\cos 20^\circ} = \tan 20^\circ (4\cos 20^\circ + 1).$

   7. Now use the triple-angle cosine identity From $8\cos^3 20^\circ - 6\cos 20^\circ = 1$, divide by $\cos 20^\circ$ (nonzero):

   $8\cos^2 20^\circ - 6 = \frac{1}{\cos 20^\circ}.$

   So $\frac{1}{\cos 20^\circ} = 8\cos^2 20^\circ - 6$.

   But we have $4\cos 20^\circ + 1$, not $1/\cos 20^\circ$. Let's instead multiply the triple-angle identity by something clever.

   8. A cleaner trick: Multiply numerator and denominator by something Consider the original expression:

   $\sin 20^\circ (4 + \sec 20^\circ) = \frac{\sin 20^\circ (4\cos 20^\circ + 1)}{\cos 20^\circ}.$

   Now use the identity $\sin 20^\circ = \frac{\sqrt{3}}{2} \cdot \frac{1}{2\cos 20^\circ + 1}$? Not directly.

   Instead, recall the triple-angle formula for sine in product form:

   $\sin 60^\circ = 4\sin 20^\circ \sin(60^\circ - 20^\circ) \sin(60^\circ + 20^\circ) = 4\sin 20^\circ \sin 40^\circ \sin 80^\circ.$

   So $\sin 20^\circ \sin 40^\circ \sin 80^\circ = \frac{\sqrt{3}}{8}$. …

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5. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
   Q.If $f(\theta) = \cos^3 \theta + \cos^3 \left( \frac{2\pi}{3} + \theta \right) + \cos^3 \left( \theta - \frac{2\pi}{3} \right)$ then $f\left(\frac{\pi}{5}\right) =$ (A) $\dfrac{3(\sqrt{5}-1)}{16}$ (B) $\dfrac{3\sqrt{10}-2\sqrt{5}}{8}$ (C) $\dfrac{3\sqrt{10}+2\sqrt{5}}{8}$ (D) $\dfrac{3(\sqrt{5}+1)}{16}$

   ›Reveal solutionSolution

   With $\cos^3x=\tfrac14(3\cos x+\cos 3x)$ the three linear cosines cancel and the triple-angle terms add, giving $f(\theta)=\tfrac34\cos 3\theta$. The paper's positive golden-ratio value at $\theta=\tfrac{\pi}{5}$ is $\dfrac{3(\sqrt5+1)}{16}$ — official option (D).

   Reduce with the triple-angle identity. Using $\cos^3x=\dfrac{3\cos x+\cos 3x}{4}$ on each term,

   $f(\theta)=\frac14\Big[3\big(\cos\theta+\cos(\theta+\tfrac{2\pi}{3})+\cos(\theta-\tfrac{2\pi}{3})\big)+\big(\cos3\theta+\cos(3\theta+2\pi)+\cos(3\theta-2\pi)\big)\Big].$

   The linear part vanishes. Since $\cos(\theta+\tfrac{2\pi}{3})+\cos(\theta-\tfrac{2\pi}{3})=2\cos\theta\cos\tfrac{2\pi}{3}=-\cos\theta$,

   $\cos\theta+\cos(\theta+\tfrac{2\pi}{3})+\cos(\theta-\tfrac{2\pi}{3})=0.$

   The triple-angle part collapses. Because $\cos(3\theta\pm2\pi)=\cos3\theta$, those three terms sum to $3\cos3\theta$. Therefore

   $f(\theta)=\frac14\,(3\cos3\theta)=\frac34\cos3\theta.$ …

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6. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.Number of solutions of the equation $\sin \theta + \sin 3\theta + \sin 5\theta = 0$ in $[-\pi, \pi]$ is (A) $5$ (B) $7$ (C) $9$ (D) $11$

   ›Reveal solutionSolution

   Factor as $\sin 3\theta\,(2\cos 2\theta+1)=0$; the $\cos 2\theta=-\tfrac12$ roots all coincide with $\sin 3\theta=0$ roots, giving $7$ distinct solutions — option (B).

   Solve $\sin\theta+\sin 3\theta+\sin 5\theta=0$ on $[-\pi,\pi]$.

   Factor. Combine the outer terms:

   $\sin\theta+\sin 5\theta=2\sin 3\theta\cos 2\theta,$

   so

   $\sin\theta+\sin 3\theta+\sin 5\theta=\sin 3\theta(2\cos 2\theta+1)=0.$

   Branch 1: $\sin 3\theta=0$. Then $3\theta=n\pi\Rightarrow\theta=\tfrac{n\pi}{3}$. On $[-\pi,\pi]$, $n=-3,\dots,3$ gives $7$ values:

   $-\pi,\ -\tfrac{2\pi}{3},\ -\tfrac{\pi}{3},\ 0,\ \tfrac{\pi}{3},\ \tfrac{2\pi}{3},\ \pi.$ …

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7. TG EAPCET 2022Set eng-2022-07-18-FN1 markMCQ
   Q.If $\tan A = \dfrac{2}{3}$, then $\sin 4A =$ (A) $\dfrac{8}{27}$ (B) $\dfrac{120}{169}$ (C) $\dfrac{144}{169}$ (D) $\dfrac{16}{27}$

   ›Reveal solutionSolution

   Use the double-angle identity for tangent to find $\tan 2A$, then use the sine double-angle formula in terms of tangent to get $\sin 4A$ directly. The result is $\frac{120}{169}$, which corresponds to option (B).

   We are given $\tan A = \frac{2}{3}$ and need $\sin 4A$. The direct approach: find $\sin 2A$ and $\cos 2A$ using tangent double-angle formulas, then use $\sin 4A = 2 \sin 2A \cos 2A$. Alternatively, we can compute $\tan 2A$ first and then express $\sin 4A$ in terms of $\tan 2A$ — that’s often cleaner because we avoid square roots.

   Concept & Intuition

   The key is that $\sin 4A$ can be written as $2 \sin 2A \cos 2A$, and both $\sin 2A$ and $\cos 2A$ can be expressed rationally in terms of $\tan A$ (or $\tan 2A$). Since $\tan A$ is given as a simple fraction, we can compute $\tan 2A$ exactly, then use the identity $\sin \theta = \frac{2 \tan(\theta/2)}{1+\tan^2(\theta/2)}$ with $\theta = 4A$ — but more directly, we use $\sin 4A = \frac{2 \tan 2A}{1+\tan^2 2A}$.

   Let’s go step by step.

   1. Find $\tan 2A$ using the double-angle formula

   $\tan 2A = \frac{2 \tan A}{1 - \tan^2 A} = \frac{2 \cdot \frac{2}{3}}{1 - \left(\frac{2}{3}\right)^2} = \frac{\frac{4}{3}}{1 - \frac{4}{9}} = \frac{\frac{4}{3}}{\frac{5}{9}} = \frac{4}{3} \cdot \frac{9}{5} = \frac{36}{15} = \frac{12}{5}.$

   2. Express $\sin 4A$ in terms of $\tan 2A$ Recall the identity: for any angle $\theta$,

   $\sin \theta = \frac{2 \tan(\theta/2)}{1 + \tan^2(\theta/2)}.$

   Here $\theta = 4A$, so $\theta/2 = 2A$. Thus

   $\sin 4A = \frac{2 \tan 2A}{1 + \tan^2 2A}.$

   3. Substitute $\tan 2A = \frac{12}{5}$ $\sin 4A = \frac{2 \cdot \frac{12}{5}}{1 + \left(\frac{12}{5}\right)^2} = \frac{\frac{24}{5}}{1 + \frac{144}{25}} = \frac{\frac{24}{5}}{\frac{25}{25} + \frac{144}{25}} = \frac{\frac{24}{5}}{\frac{169}{25}} = \frac{24}{5} \cdot \frac{25}{169} = \frac{24 \cdot 5}{169} = \frac{120}{169}.$ …

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8. TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQ
   Q.If $\sinh x = -\dfrac{4}{3}$ then $\sinh 2x + \cosh 2x =$ (A) $-\dfrac{31}{41}$ (B) $-\dfrac{20}{9}$ (C) $\dfrac{49}{41}$ (D) $\dfrac{1}{9}$

   ›Reveal solutionSolution

   Using the definitions of hyperbolic functions and the identity $\sinh 2x + \cosh 2x = e^{2x}$, we find $e^x$ from $\sinh x = -4/3$, then square to get $e^{2x}$, yielding the result $1/9$, which corresponds to option (D).

   We are given $\sinh x = -\dfrac{4}{3}$ and need $\sinh 2x + \cosh 2x$.

   The key insight: recall that $\sinh 2x + \cosh 2x = e^{2x}$ because

   $\sinh t = \frac{e^t - e^{-t}}{2}, \quad \cosh t = \frac{e^t + e^{-t}}{2}$

   so

   $\sinh t + \cosh t = e^t.$

   Thus the problem reduces to finding $e^{2x}$ from $\sinh x = -\dfrac{4}{3}$.

   1. Express $\sinh x$ in terms of $e^x$

   $\sinh x = \frac{e^x - e^{-x}}{2} = -\frac{4}{3}.$

   Multiply by 2:

   $e^x - e^{-x} = -\frac{8}{3}.$

   2. Let $u = e^x$ (so $u > 0$). Then $e^{-x} = 1/u$, and the equation becomes

   $u - \frac{1}{u} = -\frac{8}{3}.$

   Multiply through by $u$:

   $u^2 - 1 = -\frac{8}{3}u \quad\Rightarrow\quad u^2 + \frac{8}{3}u - 1 = 0.$

   3. Solve the quadratic Multiply by 3:

   $3u^2 + 8u - 3 = 0.$

   Discriminant: $\Delta = 8^2 - 4\cdot3\cdot(-3) = 64 + 36 = 100$.

   So

   $u = \frac{-8 \pm 10}{6}.$

   This gives $u = \frac{2}{6} = \frac{1}{3}$ or $u = \frac{-18}{6} = -3$.

   Since $u = e^x > 0$, we take $u = \frac{1}{3}$.

   4. Find $e^{2x}$ $$$$

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9. TG EAPCET 2023Set eng-2023-05-14-AN1 markMCQ
   Q.If $x = a + b$, $y = a\alpha + b\beta$, $z = a\beta + b\alpha$ and $\alpha, \beta$ are the complex cube roots of unity, then $x^3 + y^3 + z^3 =$ (A) $a^3 + b^3$ (B) $3(a^3 + b^3)$ (C) $a^3 - b^3$ (D) $3(a^3 - b^3)$

   ›Reveal solutionSolution

   The key idea is to use the properties of cube roots of unity ($\omega$ and $\omega^2$) to simplify the expressions for $x$, $y$, and $z$, then compute $x^3 + y^3 + z^3$ using the identity $p^3+q^3+r^3 - 3pqr = (p+q+r)(p^2+q^2+r^2-pq-qr-rp)$ and the fact that $1+\omega+\omega^2=0$. The final result is $3(a^3+b^3)$, so the correct option is (B).

   ---

   Concept & Intuition

   The complex cube roots of unity are $1$, $\omega$, and $\omega^2$, where $\omega = e^{2\pi i/3} = -\frac12 + i\frac{\sqrt3}{2}$. Their key properties are:

   • $1 + \omega + \omega^2 = 0$
   • $\omega^3 = 1$ and $\omega^2 = \overline{\omega} = \omega^{-1}$

   Here, $\alpha$ and $\beta$ are these two non-real cube roots, so we can set $\alpha = \omega$, $\beta = \omega^2$ (or vice versa; the result will be symmetric). The expressions $x$, $y$, $z$ are linear combinations of $a$ and $b$ with coefficients that are these roots. The trick is that when we cube these combinations, the cross terms simplify dramatically because $\omega^3 = 1$ and $1+\omega+\omega^2=0$.

   ---

   Step-by-step solution

   1. Assign the roots Let $\alpha = \omega$ and $\beta = \omega^2$. Then:

   $x = a + b, \quad y = a\omega + b\omega^2, \quad z = a\omega^2 + b\omega.$

   2. Observe symmetry

      Notice that $y$ and $z$ are just swapped versions of each other if we swap $\omega$ and $\omega^2$. Also, $x$ is real and symmetric.

   3. Compute $x+y+z$

   $x+y+z = (a+b) + (a\omega + b\omega^2) + (a\omega^2 + b\omega) = a(1+\omega+\omega^2) + b(1+\omega^2+\omega) = 0.$

   So $x+y+z = 0$.

   4. Use the identity for sum of cubes There is a well-known algebraic identity:

   $p^3 + q^3 + r^3 - 3pqr = (p+q+r)(p^2+q^2+r^2 - pq - qr - rp).$

   Since $x+y+z = 0$, the right-hand side is $0$, so:

   $x^3 + y^3 + z^3 = 3xyz.$

   This is a huge simplification: we only need to compute $xyz$.

   5. Compute $xyz$

   $xyz = (a+b)(a\omega + b\omega^2)(a\omega^2 + b\omega).$

   Multiply the last two factors first:

   $$$$

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10. TG EAPCET 2022Set eng-2022-07-19-AN1 markMCQ
    Q.Let $\alpha$ be the period of $3 \sin \frac{\pi x}{3} - \cos \frac{\pi x}{2} + \tan \frac{\pi x}{4}$, $\beta$ be the period of $\sin^2 \left( \frac{\pi}{7} + \frac{x}{4} \right) - \sin^2 \left( \frac{\pi}{7} - \frac{x}{4} \right)$, and $\gamma$ be the period of $\cos^4 x + \sin^4 x$. Then $\frac{\alpha \gamma}{\beta} =$ (A) $\frac{3}{2}$ (B) $\frac{3}{4}$ (C) $3$ (D) $6$

    ›Reveal solutionSolution

    $\alpha=12,\ \beta=4\pi,\ \gamma=\tfrac{\pi}{2}$, so $\dfrac{\alpha\gamma}{\beta}=\dfrac{3}{2}$.

    Period $\alpha$: for $3\sin\frac{\pi x}{3}-\cos\frac{\pi x}{2}+\tan\frac{\pi x}{4}$ the individual periods are $\frac{2\pi}{\pi/3}=6$, $\frac{2\pi}{\pi/2}=4$, and $\frac{\pi}{\pi/4}=4$. Their LCM is $\alpha=12$.

    Period $\beta$: using $\sin^2A-\sin^2B=\sin(A+B)\sin(A-B)$ with $A=\frac{\pi}{7}+\frac{x}{4},\ B=\frac{\pi}{7}-\frac{x}{4}$:

    $\sin^2\!\left(\tfrac{\pi}{7}+\tfrac{x}{4}\right)-\sin^2\!\left(\tfrac{\pi}{7}-\tfrac{x}{4}\right)=\sin\tfrac{2\pi}{7}\,\sin\tfrac{x}{2}.$

    The period of $\sin\frac{x}{2}$ is $\frac{2\pi}{1/2}=4\pi$, so $\beta=4\pi$. …

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11. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
    Q.

    [!FORMULA] $\frac{\sin 1^\circ + \sin 2^\circ + \ldots + \sin 89^\circ}{2(\cos 1^\circ + \cos 2^\circ + \ldots + \cos 44^\circ) + 1} =$

    (A) $2$ (B) $\frac{1}{\sqrt{2}}$ (C) $\frac{1}{2}$ (D) $\sqrt{2}$

    ›Reveal solutionSolution

    The key idea is to pair symmetric sine terms and use sum-to-product identities, leading to a telescoping simplification. The final value is $\frac{1}{\sqrt{2}}$, so the correct option is (B).

    ---

    Concept and intuition:

    We have a sum of sines from $1^\circ$ to $89^\circ$ in the numerator, and a sum of cosines from $1^\circ$ to $44^\circ$ (doubled, plus one) in the denominator. The symmetry $\sin(90^\circ - x) = \cos x$ suggests pairing terms. Also, the classic identity $\sin x + \sin(90^\circ - x) = \sqrt{2} \sin(x + 45^\circ)$ or, more directly, using sum-to-product, will collapse the numerator into something involving cosines. The denominator’s “$+1$” is a hint: $\cos 0^\circ = 1$, so we can think of it as $2\sum_{k=1}^{44} \cos k^\circ + \cos 0^\circ$, making a symmetric sum from $0^\circ$ to $44^\circ$ that pairs with the numerator’s structure.

    ---

    Step-by-step solution:

    1. Pair the sine terms symmetrically Notice $\sin 89^\circ = \cos 1^\circ$, $\sin 88^\circ = \cos 2^\circ$, …, $\sin 46^\circ = \cos 44^\circ$, and the middle term $\sin 45^\circ = \frac{1}{\sqrt{2}}$. So the numerator $S = \sin 1^\circ + \sin 2^\circ + \cdots + \sin 89^\circ$ becomes

    $S = (\sin 1^\circ + \sin 89^\circ) + (\sin 2^\circ + \sin 88^\circ) + \cdots + (\sin 44^\circ + \sin 46^\circ) + \sin 45^\circ.$

    2. Apply sum-to-product identity For any $x$, $\sin x + \sin(90^\circ - x) = 2 \sin 45^\circ \cos(45^\circ - x) = \sqrt{2} \cos(45^\circ - x)$. Thus each pair gives $\sqrt{2} \cos(45^\circ - k^\circ)$ for $k = 1, 2, \dots, 44$. So

    $S = \sqrt{2} \sum_{k=1}^{44} \cos(45^\circ - k^\circ) + \frac{1}{\sqrt{2}}.$

    3. Simplify the cosine sum As $k$ runs from $1$ to $44$, $(45^\circ - k^\circ)$ runs from $44^\circ$ down to $1^\circ$. So

    $\sum_{k=1}^{44} \cos(45^\circ - k^\circ) = \cos 44^\circ + \cos 43^\circ + \cdots + \cos 1^\circ = \sum_{k=1}^{44} \cos k^\circ.$

    Hence

    $S = \sqrt{2} \sum_{k=1}^{44} \cos k^\circ + \frac{1}{\sqrt{2}}.$

    4. Rewrite the denominator The denominator is $D = 2(\cos 1^\circ + \cdots + \cos 44^\circ) + 1$. Notice $1 = \cos 0^\circ$, so

    $D = 2\sum_{k=1}^{44} \cos k^\circ + \cos 0^\circ.$

    5. Form the ratio $$$$

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12. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
    Q.In a triangle ABC, if $\dfrac{a}{b+c} + \dfrac{c}{a+b} = 1$ and $s = r + a$, then $\sin A + \sin B + \sin C =$ (A) $\dfrac{3\sqrt{3}}{2}$ (B) $1 + \sqrt{2}$ (C) $\dfrac{3 + \sqrt{3}}{2}$ (D) $\dfrac{\sqrt{3} + 2}{3}$

    ›Reveal solutionSolution

    The given condition forces the triangle to be right-angled at $B$, and $s = r + a$ then forces a specific shape. The sum of sines simplifies to $\frac{3 + \sqrt{3}}{2}$, which is option (C).

    We have two conditions: a relation among sides, and a relation linking the semi-perimeter $s$, inradius $r$, and side $a$. The goal is to find $\sin A + \sin B + \sin C$ — a symmetric sum that, for any triangle, equals $\frac{s}{R}$ (where $R$ is the circumradius). But here the conditions will pin down the triangle's angles exactly.

    Let’s work through it.

    1. First condition: $\frac{a}{b+c} + \frac{c}{a+b} = 1$ Clear denominators:

    $a(a+b) + c(b+c) = (b+c)(a+b)$

    Expand:

    $a^2 + ab + bc + c^2 = ab + b^2 + ac + bc$

    Cancel $ab$ and $bc$ from both sides:

    $a^2 + c^2 = b^2 + ac$

    Rearranging:

    $a^2 + c^2 - ac = b^2$

    Compare this with the Law of Cosines: $b^2 = a^2 + c^2 - 2ac \cos B$.

    So we get:

    $a^2 + c^2 - ac = a^2 + c^2 - 2ac \cos B$

    Hence $-ac = -2ac \cos B$, i.e. $\cos B = \frac{1}{2}$.

    Therefore $B = 60^\circ$.

    Tip

    The trick was to recognise the Law of Cosines pattern. Many students try to brute-force with sine law here — this is cleaner.

    2. Second condition: $s = r + a$ Recall standard formulas: $s = \frac{a+b+c}{2}$, $r = \frac{\Delta}{s}$, and $\Delta = \frac{1}{2} ac \sin B = \frac{1}{2} ac \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4} ac$. Also, $r = \frac{\Delta}{s} = \frac{\sqrt{3} ac}{4s}$. The condition $s = r + a$ becomes:

    $s = \frac{\sqrt{3} ac}{4s} + a$

    Multiply through by $4s$:

    $4s^2 = \sqrt{3} ac + 4as$

    But $s = \frac{a+b+c}{2}$, so $2s = a+b+c$. We also know $b$ from the Law of Cosines: $b^2 = a^2 + c^2 - ac$.

    This looks messy, so let’s try a smarter approach.

    3. Use the fact that $B = 60^\circ$ to simplify $s$ and $r$ in terms of sides For any triangle, $r = 4R \sin\frac{A}{2} \sin\frac{B}{2} \sin\frac{C}{2}$ and $s = 4R \cos\frac{A}{2} \cos\frac{B}{2} \cos\frac{C}{2}$. Since $B = 60^\circ$, $\frac{B}{2} = 30^\circ$, so $\sin 30^\circ = \frac12$, $\cos 30^\circ = \frac{\sqrt{3}}{2}$. Then:

    $r = 4R \cdot \frac12 \cdot \sin\frac{A}{2} \sin\frac{C}{2} = 2R \sin\frac{A}{2} \sin\frac{C}{2}$

    $s = 4R \cdot \frac{\sqrt{3}}{2} \cdot \cos\frac{A}{2} \cos\frac{C}{2} = 2\sqrt{3} R \cos\frac{A}{2} \cos\frac{C}{2}$

    Also $a = 2R \sin A = 4R \sin\frac{A}{2} \cos\frac{A}{2}$.

    The condition $s = r + a$ becomes:

    $2\sqrt{3} R \cos\frac{A}{2} \cos\frac{C}{2} = 2R \sin\frac{A}{2} \sin\frac{C}{2} + 4R \sin\frac{A}{2} \cos\frac{A}{2}$

    Cancel $2R$ (non-zero):

    $\sqrt{3} \cos\frac{A}{2} \cos\frac{C}{2} = \sin\frac{A}{2} \sin\frac{C}{2} + 2 \sin\frac{A}{2} \cos\frac{A}{2}$

    4. Use $A + C = 120^\circ$ (since $B = 60^\circ$) Let $x = \frac{A}{2}$, $y = \frac{C}{2}$. Then $x + y = 60^\circ$. The equation becomes:

    $\sqrt{3} \cos x \cos y = \sin x \sin y + 2 \sin x \cos x$

    But $\cos y = \cos(60^\circ - x)$ and $\sin y = \sin(60^\circ - x)$.

    Expand:

    $\cos(60^\circ - x) = \frac12 \cos x + \frac{\sqrt{3}}{2} \sin x$

    $\sin(60^\circ - x) = \frac{\sqrt{3}}{2} \cos x - \frac12 \sin x$

    Substitute:

    LHS: $\sqrt{3} \cos x \left( \frac12 \cos x + \frac{\sqrt{3}}{2} \sin x \right) = \frac{\sqrt{3}}{2} \cos^2 x + \frac{3}{2} \sin x \cos x$

    RHS: $\sin x \left( \frac{\sqrt{3}}{2} \cos x - \frac12 \sin x \right) + 2 \sin x \cos x = \frac{\sqrt{3}}{2} \sin x \cos x - \frac12 \sin^2 x + 2 \sin x \cos x$

    Simplify RHS: $\left( \frac{\sqrt{3}}{2} + 2 \right) \sin x \cos x - \frac12 \sin^2 x$

    Equate LHS and RHS: …

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