If cos^210^∘cos20^∘cos40^∘cos50^∘cos70^∘=α +√(3)16cos10^∘, then 3α ^-1 is equal to

B. 64. Step 1: cosxcos(60^∘-x)cos(60^∘+x)=1/4cos3x; with x=10^∘: cos10^∘cos50^∘cos70^∘=√(3)8. Step 2: The left side =√(3)8cos10^∘cos20^∘cos40^∘, and cos20^∘cos40^∘=1/4+1/2cos20^∘. Step 3: =√(3)32cos10^∘+√(3)16cos10^∘cos20^∘, and cos10^∘cos20^∘=√(3)4+1/2cos10^∘. Step 4: =3/64+√(3)16cos10^∘, so α =3/64 and 3α ^-1=64. Check: Numerically the left side is 0.15348 and √(3)16cos10^∘=0.10661; the difference is 0.046875=3/64. Trap: 3α =9/64 is not 3α ^-1; the question asks for 3/α.

Question 47 of 47 · Trigonometric Functions & Identities

Questionhard

This question was previously asked in

NIMCET 2025

If $cos^{2} 10^{\circ} cos 20^{\circ} cos 40^{\circ} cos 50^{\circ} cos 70^{\circ} = α + \frac{3}{16} cos 10^{\circ}$ , then $3 α^{- 1}$ is equal to

Answer (Detailed Solution Below)View answer ▾

Answer: Option B — 64

Detailed Solution

$cos x cos (60^{\circ} - x) cos (60^{\circ} + x) = \frac{1}{4} cos 3 x$ ; with $x = 10^{\circ}$ : $cos 10^{\circ} cos 50^{\circ} cos 70^{\circ} = \frac{3}{8}$ .

The left side $= \frac{3}{8} cos 10^{\circ} cos 20^{\circ} cos 40^{\circ}$ , and $cos 20^{\circ} cos 40^{\circ} = \frac{1}{4} + \frac{1}{2} cos 20^{\circ}$ .

$= \frac{3}{32} cos 10^{\circ} + \frac{3}{16} cos 10^{\circ} cos 20^{\circ}$ , and $cos 10^{\circ} cos 20^{\circ} = \frac{3}{4} + \frac{1}{2} cos 10^{\circ}$ .

$= \frac{3}{64} + \frac{3}{16} cos 10^{\circ}$ , so $α = \frac{3}{64}$ and $3 α^{- 1} = 64$ .

✓ Check

Numerically the left side is

0.15348

and

\frac{3}{16} cos 10^{\circ} = 0.10661

; the difference is

0.046875 = \frac{3}{64}

⚠ Common trap

3 α = \frac{9}{64}

is not

3 α^{- 1}

; the question asks for

\frac{3}{α}

Quick tip

$cos x cos (6 0^{\circ} - x) cos (6 0^{\circ} + x) = \frac{1}{4} cos 3 x$

💬 Discussions on this question

Ask a doubt ›

Still stuck on this question? Ask our faculty and fellow students — answers appear here.

Ask a doubt

More Trigonometric Functions & Identities Questions

View all ›

Trigonometric Functions & Identities — Preparation

This question tests your understanding of Trigonometric Functions & Identities, a key topic in the Quantitative Aptitude section. The detailed solution above uses a step-by-step approach to show the reasoning, and where available, a shortcut that helps you solve similar questions in under 90 seconds during the actual exam.

Trigonometric functions (sine, cosine, tangent and their reciprocals) relate the angles of a right triangle to the ratios of its sides, and extend to any angle using the unit circle. This topic covers degree and radian measure, values of trigonometric ratios at standard angles, signs in the four quadrants, and allied angles such as (90 - theta) and (180 + theta). It also covers the main identities: sin^2 + cos^2 = 1 and its variations, compound angle formulas such as sin(A + B), multiple and sub-multiple angle formulas such as sin 2A and cos 2A, and the sum-to-product and product-to-sum formulas. Most questions are about choosing the right identity to simplify an expression or prove an equality, so learning these formulas thoroughly pays off. Practise the trigonometric functions and identities questions below, each with a step-by-step solution.

‹ All Trigonometric Functions & Identities questions Browse all topics ›

If cos210∘cos20∘cos40∘cos50∘cos70∘=α+163​​cos10∘, then 3α−1 is equal to

💬 Discussions on this question

More Trigonometric Functions & Identities Questions

More Quantitative Aptitude Questions

Trigonometric Functions & Identities — Preparation

If $cos^{2} 10^{\circ} cos 20^{\circ} cos 40^{\circ} cos 50^{\circ} cos 70^{\circ} = α + \frac{3}{16} cos 10^{\circ}$ , then $3 α^{- 1}$ is equal to