Q.Prove that : $3\sin^{-1}x = \sin^{-1}(3x - 4x^3),\ x \in \left[-\dfrac{1}{2}, \dfrac{1}{2}\right]$

Showing the 12 most recent of 15 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.For $n \in Z$, the general solution of the equation $\cos 3x \cos^3 x + \sin 3x \sin^3 x = 0$ is (A) $(2n+1)\dfrac{\pi}{2}$ (B) $(2n+1)\dfrac{\pi}{3}$ (C) $(2n+1)\dfrac{\pi}{4}$ (D) $(2n+1)\dfrac{\pi}{6}$

   ›Reveal solutionSolution

   This tests reducing a mixed cube/triple-angle trig equation using standard multiple-angle identities down to a simple $\cos2x=0$. General solution: $x=(2n+1)\pi/4$.

   Concept and Intuition

   Expressions like $\cos^3x$ and $\sin^3x$ can always be rewritten in terms of $\cos x,\cos3x$ (or $\sin x,\sin3x$) using the standard triple-angle identities. Doing this systematically collapses seemingly complicated equations into simple ones in a single multiple angle.

   Step-by-Step Solution

   1. Recall the identities: $\cos3x = 4\cos^3x-3\cos x \implies \cos^3x=\dfrac{3\cos x+\cos3x}{4}$, and $\sin3x=3\sin x-4\sin^3x \implies \sin^3x=\dfrac{3\sin x-\sin3x}{4}$.
   2. Substitute into $\cos3x\cos^3x+\sin3x\sin^3x$:

   $\cos3x\cdot\dfrac{3\cos x+\cos3x}{4}+\sin3x\cdot\dfrac{3\sin x-\sin3x}{4}$

   $=\dfrac{1}{4}\Big[3(\cos3x\cos x+\sin3x\sin x)+(\cos^23x-\sin^23x)\Big]$

   3. Using $\cos3x\cos x+\sin3x\sin x=\cos(3x-x)=\cos2x$ and $\cos^23x-\sin^23x=\cos6x$, the equation becomes:

   $\dfrac{1}{4}\left[3\cos2x+\cos6x\right]=0 \implies 3\cos2x+\cos6x=0$

   4. Using $\cos6x=4\cos^32x-3\cos2x$ (triple-angle formula with $\theta=2x$): …

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2. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.$\cos^3\theta + \cos^3(\theta + 120°) + \cos^3(\theta - 120°) =$ (A) $\dfrac{\sqrt{3}}{2}\cos 3\theta$ (B) $\dfrac{3}{4}\sec^3\theta$ (C) $\dfrac{3}{2}\tan^3\theta$ (D) $\dfrac{3}{4}\cos 3\theta$

   ›Reveal solutionSolution

   Expanding each cube via the triple-angle identity and using that cosines $120^\circ$ apart sum to zero collapses the whole expression to $\dfrac34\cos3\theta$.

   Concept and Intuition

   Whenever a sum of trigonometric cubes at angles spaced $120^\circ$ apart appears, the triple-angle formula $4\cos^3\alpha=3\cos\alpha+\cos3\alpha$ is the standard tool: it converts every cubic term into a linear combination of a "first harmonic" (which cancels by symmetry over three equally spaced angles) and a "third harmonic" (which reinforces, since tripling a $120^\circ$ shift becomes a full $360^\circ$ shift).

   Step-by-Step Solution

   1. Recall the identity $\cos3\alpha=4\cos^3\alpha-3\cos\alpha \Rightarrow \cos^3\alpha=\dfrac{3\cos\alpha+\cos3\alpha}{4}$.
   2. Apply it to each of the three angles $\theta,\ \theta+120^\circ,\ \theta-120^\circ$ and add:

   $\sum\cos^3\alpha=\frac14\Big[3\big(\cos\theta+\cos(\theta+120^\circ)+\cos(\theta-120^\circ)\big)+\big(\cos3\theta+\cos(3\theta+360^\circ)+\cos(3\theta-360^\circ)\big)\Big].$

   3. The first bracket is the classic identity: three cosines of angles equally spaced by $120^\circ$ always sum to zero, $\cos\theta+\cos(\theta+120^\circ)+\cos(\theta-120^\circ)=0$. …

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3. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.All the values of $\alpha$ satisfying the equations $2\cos^2\alpha - 3\cos\alpha = 32\tan^8\theta$ and $3\cos 2\theta = 1$ are (A) $n\pi \pm \dfrac{\pi}{3}, n \in Z$ (B) $n\pi \pm \dfrac{2\pi}{3}, n \in Z$ (C) $2n\pi \pm \dfrac{\pi}{3}, n \in Z$ (D) $2n\pi \pm \dfrac{2\pi}{3}, n \in Z$

   ›Reveal solutionSolution

   Solve the $\theta$ equation first to pin down the numeric value $32\tan^8\theta=2$, then solve the resulting quadratic in $\cos\alpha$, rejecting the invalid root, to get $\cos\alpha=-\tfrac12$.

   Concept and Intuition

   This is really two separate, sequential trigonometric equations: the second equation ($3\cos2\theta=1$) is self-contained and lets us compute a definite numeric value for $\tan^2\theta$, which then substitutes into the first equation to convert it into an ordinary quadratic in $\cos\alpha$.

   Step-by-Step Solution

   1. From $3\cos2\theta=1$: $\cos2\theta=\tfrac13$. Using $\cos2\theta=\dfrac{1-\tan^2\theta}{1+\tan^2\theta}$, let $t=\tan^2\theta$: $\dfrac{1-t}{1+t}=\dfrac13 \Rightarrow 3(1-t)=1+t \Rightarrow 3-3t=1+t \Rightarrow 2=4t \Rightarrow t=\tfrac12$.
   2. So $\tan^2\theta=\tfrac12$, and $\tan^8\theta = (\tan^2\theta)^4 = \left(\tfrac12\right)^4=\tfrac1{16}$.
   3. Then $32\tan^8\theta = 32\times\tfrac1{16}=2$.
   4. Substitute into the first equation: $2\cos^2\alpha-3\cos\alpha = 2 \Rightarrow 2\cos^2\alpha-3\cos\alpha-2=0$. …

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4. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.$\cos^3\dfrac{\pi}{8}\cos\dfrac{3\pi}{8}+\sin^3\dfrac{\pi}{8}\sin\dfrac{3\pi}{8}=$ (A) $\dfrac{1}{2\sqrt{2}}$ (B) $\dfrac{1}{2}$ (C) $\dfrac{1}{\sqrt{2}}$ (D) $\dfrac{1}{4}$

   ›Reveal solutionSolution

   This tests reducing a triple-angle trig expression to a clean function of the double angle, then evaluating. Answer: $\dfrac{1}{2\sqrt2}$.

   Concept and Intuition

   Expressions of the form $\cos^3a\cos3a+\sin^3a\sin3a$ collapse nicely when you substitute the triple-angle formulas for $\cos3a,\sin3a$ in terms of $\cos a,\sin a$: everything becomes a polynomial in $\cos a,\sin a$, which can then be rewritten purely in terms of $\cos2a$ using $\cos^4a-\sin^4a=\cos2a$ and a similar identity for the degree-6 difference. This turns a seemingly complicated expression into $\cos^3(2a)$ — clean enough to evaluate directly.

   Step-by-Step Solution

   1. Let $a=\pi/8$. Use $\cos3a=4\cos^3a-3\cos a$, $\sin3a=3\sin a-4\sin^3a$.
   2. $\cos^3a\cos3a=\cos^3a(4\cos^3a-3\cos a)=4\cos^6a-3\cos^4a$.
   3. $\sin^3a\sin3a=\sin^3a(3\sin a-4\sin^3a)=3\sin^4a-4\sin^6a$.
   4. Sum $=4\cos^6a-3\cos^4a+3\sin^4a-4\sin^6a=4(\cos^6a-\sin^6a)-3(\cos^4a-\sin^4a)$.
   5. $\cos^4a-\sin^4a=(\cos^2a-\sin^2a)(\cos^2a+\sin^2a)=\cos2a$.
   6. $\cos^6a-\sin^6a=(\cos^2a-\sin^2a)(\cos^4a+\cos^2a\sin^2a+\sin^4a)=\cos2a\left[(\cos^2a+\sin^2a)^2-\cos^2a\sin^2a\right]=\cos2a\left(1-\tfrac14\sin^22a\right)$. …

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5. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.If $\left(\dfrac{\sin 3\theta}{\sin\theta}\right)^2 - \left(\dfrac{\cos 3\theta}{\cos\theta}\right)^2 = a\cos b\theta$, then $a:b =$ (A) $4:1$ (B) $8:1$ (C) $3:2$ (D) $2:1$

   ›Reveal solutionSolution

   Simplify $\sin3\theta/\sin\theta$ and $\cos3\theta/\cos\theta$ to algebraic expressions in $\sin\theta,\cos\theta$, then use a difference-of-squares factoring to collapse everything to a single cosine term. Answer: $a:b=4:1$.

   Concept and Intuition

   The triple-angle formulas $\sin3\theta=3\sin\theta-4\sin^3\theta$ and $\cos3\theta=4\cos^3\theta-3\cos\theta$ let us write $\dfrac{\sin3\theta}{\sin\theta}$ and $\dfrac{\cos3\theta}{\cos\theta}$ as clean polynomial expressions, after which a difference-of-squares factoring turns the whole thing into a multiple of $\cos2\theta$.

   Step-by-Step Solution

   1. $\sin3\theta=3\sin\theta-4\sin^3\theta=\sin\theta(3-4\sin^2\theta)\Rightarrow\dfrac{\sin3\theta}{\sin\theta}=3-4\sin^2\theta$.
   2. $\cos3\theta=4\cos^3\theta-3\cos\theta=\cos\theta(4\cos^2\theta-3)\Rightarrow\dfrac{\cos3\theta}{\cos\theta}=4\cos^2\theta-3$.
   3. The expression is $(3-4\sin^2\theta)^2-(4\cos^2\theta-3)^2$, a difference of squares $=(A-B)(A+B)$ with $A=3-4\sin^2\theta$, $B=4\cos^2\theta-3$.
   4. $A-B=3-4\sin^2\theta-4\cos^2\theta+3=6-4(\sin^2\theta+\cos^2\theta)=6-4=2$. …

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6. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
   Q.$\left(4\cos^2\dfrac{\pi}{20} - 1\right)\left(4\cos^2\dfrac{3\pi}{20} - 1\right)\left(4\cos^2\dfrac{5\pi}{20} + 1\right)\left(4\cos^2\dfrac{7\pi}{20} - 1\right)\left(4\cos^2\dfrac{9\pi}{20} - 1\right) =$ (A) $1$ (B) $\dfrac{1}{2}$ (C) $2$ (D) $3$

   ›Reveal solutionSolution

   Four of the five factors telescope to $1$ via the identity $4\cos^2\theta-1=\sin3\theta/\sin\theta$; the odd one out (at $\pi/4$, which has a "$+1$" instead of "$-1$") simply evaluates to $3$ directly. The full product is $3$.

   Concept and Intuition

   The key identity here is $\sin3\theta=\sin\theta(3-4\sin^2\theta)=\sin\theta(4\cos^2\theta-1)$, i.e. $4\cos^2\theta-1=\dfrac{\sin3\theta}{\sin\theta}$. When several factors of this form are multiplied and the angles are related by tripling (mod adjustments of $\pi$), the numerators and denominators cancel in a telescoping pattern. The fifth angle here, $\pi/4$, is a special value where $\cos^2(\pi/4)=1/2$ is exactly computable, and its factor carries a '+1' rather than '-1,' so it doesn't participate in the telescoping — it's just evaluated directly.

   Step-by-Step Solution

   1. Write each "$-1$" factor using $4\cos^2\theta-1=\dfrac{\sin3\theta}{\sin\theta}$:

   • $\theta=\pi/20$: $\dfrac{\sin(3\pi/20)}{\sin(\pi/20)}$
   • $\theta=3\pi/20$: $\dfrac{\sin(9\pi/20)}{\sin(3\pi/20)}$
   • $\theta=7\pi/20$: $\dfrac{\sin(21\pi/20)}{\sin(7\pi/20)}$
   • $\theta=9\pi/20$: $\dfrac{\sin(27\pi/20)}{\sin(9\pi/20)}$

   2. Multiply the first two: $\dfrac{\sin(3\pi/20)}{\sin(\pi/20)}\cdot\dfrac{\sin(9\pi/20)}{\sin(3\pi/20)}=\dfrac{\sin(9\pi/20)}{\sin(\pi/20)}$.
   3. For the last two, use $\sin(21\pi/20)=\sin(\pi+\pi/20)=-\sin(\pi/20)$ and $\sin(27\pi/20)=\sin(\pi+7\pi/20)=-\sin(7\pi/20)$: …

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7. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
   Q.If $\cos^3 80^\circ + \cos^3 40^\circ - \cos^3 20^\circ = k$, then $\dfrac{4k}{3} =$ (A) $\sin\left(\dfrac{4\pi}{3}\right)$ (B) $\cos\left(\dfrac{2\pi}{3}\right)$ (C) $\tan\left(\dfrac{\pi}{3}\right)$ (D) $\sec\left(\dfrac{2\pi}{3}\right)$

   ›Reveal solutionSolution

   Rewriting each cube using the triple-angle identity $\cos^3\theta=\tfrac14(\cos3\theta+3\cos\theta)$ makes the linear-cosine terms cancel via a sum-to-product identity, leaving a simple numeric value: $4k/3=-\tfrac12=\cos(2\pi/3)$.

   Concept and Intuition

   Sums of cubes of cosines at angles that are multiples of $20°$ (which relate nicely to $60°,120°,240°$ under tripling) are a classic setup for the identity $4\cos^3\theta-3\cos\theta=\cos3\theta$, i.e. $\cos^3\theta=\dfrac{\cos3\theta+3\cos\theta}{4}$. Applying it to each term converts the sum of cubes into a sum of cosines at "nice" tripled angles ($240°,120°,60°$) plus a residual linear-cosine sum that often collapses via sum-to-product.

   Step-by-Step Solution

   1. $\cos^3\theta=\dfrac{\cos3\theta+3\cos\theta}{4}$. Apply to each term: $\cos^380^\circ=\dfrac{\cos240^\circ+3\cos80^\circ}{4}$, $\cos^340^\circ=\dfrac{\cos120^\circ+3\cos40^\circ}{4}$, $\cos^320^\circ=\dfrac{\cos60^\circ+3\cos20^\circ}{4}$.
   2. $k=\cos^380^\circ+\cos^340^\circ-\cos^320^\circ=\dfrac{(\cos240^\circ+\cos120^\circ-\cos60^\circ)+3(\cos80^\circ+\cos40^\circ-\cos20^\circ)}{4}$.
   3. $\cos80^\circ+\cos40^\circ=2\cos\left(\frac{80+40}{2}\right)\cos\left(\frac{80-40}{2}\right)=2\cos60^\circ\cos20^\circ=2\cdot\tfrac12\cdot\cos20^\circ=\cos20^\circ$. So $\cos80^\circ+\cos40^\circ-\cos20^\circ=0$ — this whole bracket vanishes.
   4. $\cos240^\circ=-\tfrac12$, $\cos120^\circ=-\tfrac12$, $\cos60^\circ=\tfrac12$, so $\cos240^\circ+\cos120^\circ-\cos60^\circ=-\tfrac12-\tfrac12-\tfrac12=-\tfrac32$. …

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8. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
   Q.If $\cos\alpha + \cos\beta + \cos\gamma = \sin\alpha + \sin\beta + \sin\gamma = 0$ then $(\cos^3\alpha + \cos^3\beta + \cos^3\gamma)^2 + (\sin^3\alpha + \sin^3\beta + \sin^3\gamma)^2 =$ (A) $1$ (B) $\dfrac{3}{4}$ (C) $\dfrac{9}{16}$ (D) $\dfrac{9}{8}$

   ›Reveal solutionSolution

   Treating $e^{i\alpha},e^{i\beta},e^{i\gamma}$ as three complex numbers summing to zero lets the identity $x^3+y^3+z^3=3xyz$ collapse the whole expression to exactly $\dfrac{9}{16}$.

   Concept and Intuition

   Whenever both $\sum\cos\theta_i=0$ and $\sum\sin\theta_i=0$ are given together, it's a strong hint to combine them into one complex condition $\sum e^{i\theta_i}=0$, since that's a much more powerful single fact than the two separate real equations, and standard algebraic identities for sums of cubes can then be applied directly to the complex exponentials.

   Step-by-Step Solution

   1. Let $x=e^{i\alpha},\ y=e^{i\beta},\ z=e^{i\gamma}$ — each has modulus $1$. The given conditions combine to $x+y+z = (\cos\alpha+\cos\beta+\cos\gamma) + i(\sin\alpha+\sin\beta+\sin\gamma) = 0$.
   2. Standard algebraic identity: if $x+y+z=0$ then $x^3+y^3+z^3=3xyz$ (since $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$, and the first factor is zero).
   3. So $e^{i3\alpha}+e^{i3\beta}+e^{i3\gamma} = 3\,e^{i\alpha}e^{i\beta}e^{i\gamma} = 3e^{i(\alpha+\beta+\gamma)}$.
   4. Taking real and imaginary parts: $\cos3\alpha+\cos3\beta+\cos3\gamma = 3\cos(\alpha+\beta+\gamma)$ and $\sin3\alpha+\sin3\beta+\sin3\gamma=3\sin(\alpha+\beta+\gamma)$.
   5. Now use the triple-angle identities in reverse: $\cos^3\theta = \dfrac{3\cos\theta+\cos3\theta}{4}$ and $\sin^3\theta=\dfrac{3\sin\theta-\sin3\theta}{4}$.
   6. Sum of cubes of cosines: $\displaystyle\sum\cos^3\theta_i = \frac{3\sum\cos\theta_i + \sum\cos3\theta_i}{4} = \frac{3(0) + 3\cos(\alpha+\beta+\gamma)}{4} = \frac{3\cos(\alpha+\beta+\gamma)}{4}$. …

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9. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
   Q.The general solution of the equation $\tan x + \tan 2x - \tan 3x = 0$ is (A) $\{x \mid x = n\pi \pm \dfrac{\pi}{3} \text{ or } \dfrac{n\pi}{2}, n \in Z\}$ (B) $\{x \mid x = n\pi \pm \dfrac{\pi}{3} \text{ or } n\pi, n \in Z\}$ (C) $\{x \mid x = n\pi \pm \dfrac{\pi}{3} \text{ or } \dfrac{n\pi}{2} \text{ or } n\pi, n \in Z\}$ (D) $\{x \mid x = n\pi \pm \dfrac{\pi}{6} \text{ or } \dfrac{n\pi}{2}, n \in Z\}$

   ›Reveal solutionSolution

   This tests solving a trigonometric equation by combining tangent differences into sines/cosines of multiple angles, then discarding spurious roots where the original expression is undefined. Answer: $x=n\pi\pm\pi/3$ or $n\pi$.

   Concept and Intuition

   $\tan x+\tan2x-\tan3x=0$ mixes three different multiples of $x$. The trick is to regroup as $\tan3x-\tan x=\tan2x$ and use the tangent-difference-as-sine identity $\tan A-\tan B=\dfrac{\sin(A-B)}{\cos A\cos B}$, which converts everything to a common $\sin2x$ factor — turning a messy tangent equation into a clean product-equals-zero form.

   Step-by-Step Solution

   1. Rearrange: $\tan3x-\tan x=\tan2x$.
   2. LHS $=\dfrac{\sin(3x-x)}{\cos3x\cos x}=\dfrac{\sin2x}{\cos3x\cos x}$. RHS $=\dfrac{\sin2x}{\cos2x}$.
   3. So $\sin2x\left[\dfrac{1}{\cos3x\cos x}-\dfrac{1}{\cos2x}\right]=0$.
   4. Case 1: $\sin2x=0 \Rightarrow x=\dfrac{n\pi}{2}$.
   5. Case 2: $\cos2x=\cos3x\cos x$. Using $\cos3x\cos x=\tfrac12(\cos4x+\cos2x)$: $\cos2x=\tfrac12\cos4x+\tfrac12\cos2x \Rightarrow \cos2x=\cos4x$.
   6. $\cos4x=\cos2x \Rightarrow 4x=2n\pi\pm2x \Rightarrow x=n\pi \text{ or } x=\dfrac{n\pi}{3}$.
   7. Combine all candidates: $x=\dfrac{n\pi}{2}$, $x=n\pi$, $x=\dfrac{n\pi}{3}$. …

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10. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
    Q.If $\theta = \dfrac{\pi}{9}$, then $1+27\tan^2\theta - 33\tan^4\theta + \tan^6\theta =$ (A) $3$ (B) $4$ (C) $-3$ (D) $-11$

    ›Reveal solutionSolution

    Since $\tan(3\times20^\circ)=\tan60^\circ=\sqrt3$, $t=\tan20^\circ$ satisfies a specific cubic, and the given sextic expression in $t$ evaluates to exactly $4$.

    Concept and Intuition

    Whenever $\theta$ is such that $3\theta$ is a "nice" angle (here $60^\circ$), the triple-angle tangent formula $\tan3\theta=\dfrac{3t-t^3}{1-3t^2}$ ties $t=\tan\theta$ to an algebraic equation. Expressions like $1+27t^2-33t^4+t^6$ that mix even powers of $t$ are frequently disguised versions of quantities derivable from that same triple-angle relation (or can simply be checked numerically once the angle is fixed).

    Step-by-Step Solution

    1. $\theta=\dfrac{\pi}{9}=20^\circ\Rightarrow3\theta=60^\circ\Rightarrow\tan3\theta=\sqrt3$.
    2. From $\tan3\theta=\dfrac{3t-t^3}{1-3t^2}=\sqrt3$: $3t-t^3=\sqrt3(1-3t^2)\Rightarrow t^3-3\sqrt3t^2-3t+\sqrt3=0$, the governing equation for $t=\tan20^\circ$. …

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11. AP EAPCET 2022Set eng-2022-07-05-AN1 markMCQ
    Q.A true statement among the following identities is (A) $\cos5\theta = 16\cos^5\theta - 20\cos^3\theta - 5\cos\theta$ (B) $\cos5\theta = 20\cos^3\theta - 16\cos^5\theta + 5\cos\theta$ (C) $\cos5\theta = 16\cos^5\theta + 20\cos^3\theta - 5\cos\theta$ (D) $\cos5\theta = 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta$

    ›Reveal solutionSolution

    This is the standard $\cos5\theta$ multiple-angle expansion in terms of $\cos\theta$ alone; the correct signs are $+,-,+$: $16\cos^5\theta-20\cos^3\theta+5\cos\theta$.

    Concept and Intuition

    Using De Moivre's theorem, $\cos n\theta$ can always be written as a polynomial in $\cos\theta$ by taking the real part of $(\cos\theta+i\sin\theta)^n$ and replacing every even power of $\sin\theta$ with $1-\cos^2\theta$.

    Step-by-Step Solution

    1. $(\cos\theta+i\sin\theta)^5 = \cos5\theta + i\sin5\theta$ by De Moivre.
    2. Expanding via the binomial theorem and taking the real part (terms with even powers of $i\sin\theta$, i.e. $i^0,i^2,i^4$): $\cos5\theta = \cos^5\theta - 10\cos^3\theta\sin^2\theta + 5\cos\theta\sin^4\theta$.
    3. Substitute $\sin^2\theta = 1-\cos^2\theta$ throughout and simplify (a standard, well-known reduction) to get: $\cos5\theta = 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta$. …

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12. AP EAPCET 2022Set eng-2022-07-06-AN1 markMCQ
    Q.$(4\cos^2 9^0 - 3)(4\cos^2 27^0 - 3) =$ (A) $\sin 9^0$ (B) $\cos 9^0$ (C) $\tan 9^0$ (D) $\cot 9^0$

    ›Reveal solutionSolution

    This tests recognizing the disguised triple-angle identity $4\cos^2\theta-3=\cos3\theta/\cos\theta$, applied twice in a telescoping product. Answer: $\tan9^\circ$.

    Concept and Intuition

    The expression $4\cos^2\theta-3$ looks unfamiliar until you recall the triple angle formula $\cos3\theta = 4\cos^3\theta-3\cos\theta = \cos\theta(4\cos^2\theta-3)$. Dividing both sides by $\cos\theta$ reveals $4\cos^2\theta-3=\cos3\theta/\cos\theta$. Once each bracket is rewritten this way with $\theta=9^\circ$ and $\theta=27^\circ$ respectively, the product telescopes because $3\times9^\circ=27^\circ$ links the two factors, leaving a simple ratio of cosines that reduces via the complementary-angle identity.

    Step-by-Step Solution

    1. Recall $\cos3\theta = 4\cos^3\theta - 3\cos\theta = \cos\theta\,(4\cos^2\theta-3)$, so $4\cos^2\theta-3=\dfrac{\cos3\theta}{\cos\theta}$.
    2. Apply with $\theta=9^\circ$: $4\cos^2 9^\circ - 3 = \dfrac{\cos27^\circ}{\cos9^\circ}$.
    3. Apply with $\theta=27^\circ$: $4\cos^2 27^\circ - 3 = \dfrac{\cos81^\circ}{\cos27^\circ}$. …

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