Q.Integrate the following function: $\sin 4x \sin 8x$

Showing the 12 most recent of 42 on this concept.

1. 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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2. TG EAPCET 2021Set eng-2021-08-04-FN1 markMCQ
   Q.$\cot \dfrac{\pi}{16} \cdot \cot \dfrac{2\pi}{16} \cdot \cot \dfrac{3\pi}{16} \cdot \cot \dfrac{4\pi}{16} \cdot \cot \dfrac{5\pi}{16} \cdot \cot \dfrac{6\pi}{16} \cdot \cot \dfrac{7\pi}{16} =$ (A) $0$ (B) $1$ (C) $\dfrac{1}{2}$ (D) $2$

   ›Reveal solutionSolution

   The product of cotangents of complementary angles simplifies to 1; pairing each angle with its complement gives the result 1.

   The key insight is that cotangent has a beautiful symmetry: $\cot(\frac{\pi}{2} - \theta) = \tan \theta$, and $\cot \theta \cdot \tan \theta = 1$. So if we can pair each angle in the product with its complement (adding to $\frac{\pi}{2}$), the whole product collapses to 1 — except possibly for the middle term where the angle is exactly $\frac{\pi}{4}$, but $\cot \frac{\pi}{4} = 1$, so it doesn't break the pattern.

   Let's see this in action.

   1. Write the angles in radians:

      $\frac{\pi}{16}, \frac{2\pi}{16}, \frac{3\pi}{16}, \frac{4\pi}{16}, \frac{5\pi}{16}, \frac{6\pi}{16}, \frac{7\pi}{16}$.

   2. Notice that $\frac{\pi}{2} = \frac{8\pi}{16}$. So the complement of $\frac{k\pi}{16}$ is $\frac{8\pi}{16} - \frac{k\pi}{16} = \frac{(8-k)\pi}{16}$.

   3. Pair the terms:

      • $\frac{\pi}{16}$ pairs with $\frac{7\pi}{16}$ (since $1+7=8$)
      • $\frac{2\pi}{16}$ pairs with $\frac{6\pi}{16}$
      • $\frac{3\pi}{16}$ pairs with $\frac{5\pi}{16}$
      • $\frac{4\pi}{16}$ is its own complement (since $4+4=8$), i.e., $\frac{\pi}{4}$.

   4. For any pair $(\theta, \frac{\pi}{2} - \theta)$:

   $\cot \theta \cdot \cot\left(\frac{\pi}{2} - \theta\right) = \cot \theta \cdot \tan \theta = 1.$

   So each pair contributes a factor of 1.

   5. The middle term is $\cot \frac{4\pi}{16} = \cot \frac{\pi}{4} = 1$. …

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3. 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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4. 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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5. 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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6. TG EAPCET 2021Set eng-2021-08-04-FN1 markMCQ
   Q.

   [!FORMULA] $\frac{1 - \cos 2\theta + \sin 2\theta}{1 + \cos 2\theta + \sin 2\theta} =$

   (A) $\cot \theta$ (B) $\cos 2\theta$ (C) $\tan \theta$ (D) $\tan 2\theta$

   ›Reveal solutionSolution

   Use double-angle identities to rewrite the numerator and denominator, then simplify to a single trigonometric ratio. The expression simplifies to $\tan \theta$, so the correct option is (C).

   The core idea here is that when you see $\cos 2\theta$ and $\sin 2\theta$ together, your first instinct should be to replace them with their basic forms in terms of $\sin \theta$ and $\cos \theta$. That turns a messy fraction into something algebraic you can factor and cancel.

   Let’s walk through it.

   1. Rewrite $\cos 2\theta$ and $\sin 2\theta$ using standard double-angle formulas.

      We have:

      $\cos 2\theta = 1 - 2\sin^2 \theta$ (or $\cos^2 \theta - \sin^2 \theta$, but the $1 - 2\sin^2 \theta$ form is handy here because of the $1$ in the numerator).

      Also, $\sin 2\theta = 2\sin \theta \cos \theta$.

   2. Substitute into the numerator.

      Numerator: $1 - \cos 2\theta + \sin 2\theta$

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

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

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

      $= 2\sin \theta (\sin \theta + \cos \theta)$.

   3. Substitute into the denominator.

      Denominator: $1 + \cos 2\theta + \sin 2\theta$

      Here, use $\cos 2\theta = 2\cos^2 \theta - 1$ (or $1 - 2\sin^2 \theta$, but the $2\cos^2 \theta - 1$ form pairs nicely with the $+1$).

      So: $1 + (2\cos^2 \theta - 1) + 2\sin \theta \cos \theta$

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

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

      $= 2\cos \theta (\cos \theta + \sin \theta)$.

   4. Form the fraction and simplify.

      The whole expression becomes:

      $\frac{2\sin \theta (\sin \theta + \cos \theta)}{2\cos \theta (\sin \theta + \cos \theta)}$ …

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7. TG EAPCET 2022Set eng-2022-07-20-FN1 markMCQ
   Q.

   [!FORMULA] $\frac{1}{\sin 250^\circ} + \frac{\sqrt{3}}{\cos 290^\circ} =$

   (A) $\frac{1}{\sqrt{3}}$ (B) $4$ (C) $\frac{4}{\sqrt{3}}$ (D) $1$

   ›Reveal solutionSolution

   We simplify the given expression by first reducing the angles to acute angles using quadrant rules, then combining the fractions, and finally applying trigonometric identities for sum/difference of angles and double angles. The expression simplifies to $\boxed{4}$.

   The problem asks us to evaluate a trigonometric expression involving angles outside the first quadrant. The core idea is to first simplify these angles to their acute equivalents using quadrant rules and then use standard trigonometric identities to simplify the resulting expression.

   Here's the breakdown of the approach:

   1. Reduce angles: Convert $250^\circ$ and $290^\circ$ to equivalent acute angles ($0^\circ$ to $90^\circ$) using the properties of trigonometric functions in different quadrants. This involves determining the sign of the function in that quadrant and using the appropriate reduction formula (e.g., $180^\circ \pm \theta$ or $360^\circ - \theta$).
   2. Combine fractions: Once the angles are simplified, the expression will be in terms of acute angles. Combine the two fractions into a single fraction.
   3. Apply identities: The numerator and denominator of the combined fraction will likely simplify further using standard trigonometric identities, such as the sum/difference formulas for sine or cosine, or double angle formulas.

   Let's work through the steps.

   1. Reduce $\sin 250^\circ$: The angle $250^\circ$ lies in the third quadrant ($180^\circ < 250^\circ < 270^\circ$). In the third quadrant, the sine function is negative. We can write $250^\circ$ as $180^\circ + 70^\circ$. Using the reduction formula $\sin(180^\circ + \theta) = -\sin \theta$:

   $\sin 250^\circ = \sin(180^\circ + 70^\circ) = -\sin 70^\circ$

   2. Reduce $\cos 290^\circ$: The angle $290^\circ$ lies in the fourth quadrant ($270^\circ < 290^\circ < 360^\circ$). In the fourth quadrant, the cosine function is positive. We can write $290^\circ$ as $360^\circ - 70^\circ$. Using the reduction formula $\cos(360^\circ - \theta) = \cos \theta$:

   $\cos 290^\circ = \cos(360^\circ - 70^\circ) = \cos 70^\circ$

   3. Substitute the reduced values into the expression: Now, substitute these simplified terms back into the original expression:

   $\frac{1}{\sin 250^\circ} + \frac{\sqrt{3}}{\cos 290^\circ} = \frac{1}{-\sin 70^\circ} + \frac{\sqrt{3}}{\cos 70^\circ}$

   This can be rewritten as:
   

   $\frac{\sqrt{3}}{\cos 70^\circ} - \frac{1}{\sin 70^\circ}$

   4. Combine the fractions: To combine these, we find a common denominator, which is $\sin 70^\circ \cos 70^\circ$:

   $\frac{\sqrt{3} \sin 70^\circ - \cos 70^\circ}{\sin 70^\circ \cos 70^\circ}$

   5. Simplify the numerator: The numerator is in the form $A \sin \theta - B \cos \theta$. We can express this in the form $R \sin(\theta - \alpha)$. Factor out $2$ from the numerator:

   $2 \left( \frac{\sqrt{3}}{2} \sin 70^\circ - \frac{1}{2} \cos 70^\circ \right)$

   We know that $\cos 30^\circ = \frac{\sqrt{3}}{2}$ and $\sin 30^\circ = \frac{1}{2}$. Substitute these values:
   $$ 2 (\cos 30^\circ \sin 70^\circ - \sin 30^\circ \cos 70^\circ) $$ …
   

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8. TG EAPCET 2022Set eng-2022-07-20-AN1 markMCQ
   Q.$\lim_{x \to 0} \frac{4 \left[ \sin(2022x) - \sin(2020x) \right]}{x \left[ \cos(2022x) + 2\cos(2021x) + \cos(2020x) \right]} =$ (A) $1$ (B) $2$ (C) $2020$ (D) $2021$

   ›Reveal solutionSolution

   Use sum-to-product identities to simplify the numerator and denominator, then apply the standard limit $\lim_{u\to 0} \frac{\sin u}{u}=1$ to get the answer 2.

   The problem asks for a limit that looks messy at first glance — sines and cosines with large coefficients, all divided by $x$. The natural instinct is to try to factor something out so that each term becomes a $\frac{\sin(\text{something})}{\text{something}}$ form. That’s exactly what sum-to-product formulas are for: they turn sums of sines or cosines into products, which often reveal a common factor of $x$ in both numerator and denominator.

   Let’s work through it step by step.

   1. Simplify the numerator Use the identity $\sin A - \sin B = 2\cos\frac{A+B}{2}\sin\frac{A-B}{2}$. Here $A = 2022x$, $B = 2020x$, so

   $\sin(2022x) - \sin(2020x) = 2\cos\left(\frac{2022x+2020x}{2}\right) \sin\left(\frac{2022x-2020x}{2}\right) = 2\cos(2021x)\,\sin(x).$

   Therefore the numerator becomes

   $4\left[\sin(2022x) - \sin(2020x)\right] = 4 \cdot 2\cos(2021x)\,\sin(x) = 8\cos(2021x)\,\sin(x).$

   2. Simplify the denominator The denominator is $x\left[\cos(2022x) + 2\cos(2021x) + \cos(2020x)\right]$. Group the two outer cosines: $\cos(2022x) + \cos(2020x)$. Use the identity $\cos P + \cos Q = 2\cos\frac{P+Q}{2}\cos\frac{P-Q}{2}$. With $P=2022x$, $Q=2020x$,

   $\cos(2022x) + \cos(2020x) = 2\cos(2021x)\,\cos(x).$

   So the bracket becomes

   $2\cos(2021x)\,\cos(x) + 2\cos(2021x) = 2\cos(2021x)\left[\cos(x) + 1\right].$

   Hence the denominator is

   $x \cdot 2\cos(2021x)\left[\cos(x) + 1\right] = 2x\cos(2021x)\left(\cos x + 1\right).$

   3. Cancel the common factor The whole limit now reads

   $\lim_{x\to 0} \frac{8\cos(2021x)\,\sin(x)}{2x\cos(2021x)\left(\cos x + 1\right)}.$

   Provided $\cos(2021x) \neq 0$ near $x=0$ (it’s $1$ at $x=0$, so fine), we cancel $\cos(2021x)$:

   $$$$

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9. 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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10. 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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11. 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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12. 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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