Quick Summary

α = 3sin⁻¹(6/11) and β = 3cos⁻¹(4/9) are both triple angles, so the triple-angle formulas cos3θ = 4cos³θ−3cosθ and sin3θ = 3sinθ−4sin³θ do the work directly. Building cosθ = √85/11 from sinθ=6/11 gives cos(α) = −23√85/1331 ≈ −0.159, which is negative — Statement II is true. For β, cosφ=4/9 gives cos(3φ)≈−0.982 and sin(3φ)≈−0.188. Combining via cos(α+β)=cosα cosβ − sinα sinβ gives ≈ 0.342 > 0 — Statement I is also true. Both statements true → option (1).

Question Replay

Options

Correct Answer

Why This Answer

The trap in this question is intuition about magnitudes: sin⁻¹(6/11) is a modest first-quadrant angle (about 33°), so it is tempting to assume its triple α ≈ 99° "should" still behave like a first-quadrant angle. But tripling an angle past 30° pushes it past 90°, flipping the sign of its cosine — exactly what the exact computation shows: cos(α) = −23√85/1331, clearly negative. Statement II survives this check.

For Statement I, the two triple angles must be combined with the compound-angle formula, using the sign of each piece computed from the triple-angle formulas rather than approximated angles. The negative cosα and negative sinβ ((sinβ)=sin(3φ)≈−0.188, since 3φ is also past 90° given cos φ=4/9 gives φ≈63.6°, so 3φ≈190.7°, in the third quadrant) combine so that −sinα·sinβ contributes positively enough to outweigh the negative cosα·cosβ term, landing the sum just above zero. Both statements check out as true.

Core Concept

Triple-angle formulas. cos(3θ) = 4cos³θ − 3cosθ and sin(3θ) = 3sinθ − 4sin³θ convert a triple angle entirely into powers of the single-angle sine or cosine — no need to ever find θ itself in degrees. This is the standard tool whenever an inverse-trig expression is multiplied by an integer before combining with another.

Building the missing ratio. Given sinθ = 6/11 with θ = sin⁻¹(6/11) ∈ (0, π/2) (principal range), cosθ = +√(1−36/121) = √85/11 (positive, since θ is in the first quadrant). Similarly with cosφ=4/9, φ=cos⁻¹(4/9) ∈ (0,π/2) (principal range of cos⁻¹ restricted here since 4/9>0), sinφ = +√65/9.

Step-by-Step Solution

1. Find cosθ. sinθ=6/11 ⇒ cosθ = √(1−36/121) = √85/11 (positive, principal range).

2. Apply the triple-angle formula for cosα:

   cosα = 4cos³θ − 3cosθ = cosθ(4cos²θ−3)
   = (√85/11)(4·85/121 − 3) = (√85/11)(−23/121) = −23√85/1331 ≈ −0.159

   This is negative, so Statement II is true.

3. Find sinα for the compound-angle step: sinα = 3sinθ − 4sin³θ = 3(6/11) − 4(216/1331) = 1314/1331 ≈ 0.987.

4. Find sinφ and apply the triple-angle formulas for β:

   cosφ=4/9 ⇒ sinφ=√65/9
   cosβ = 4cos³φ−3cosφ = 256/729 − 12/9 = −716/729 ≈ −0.982
   sinβ = 3sinφ−4sin³φ = −17√65/729 ≈ −0.188

5. Combine via the compound-angle formula:

   cos(α+β) = cosα cosβ − sinα sinβ
   ≈ (−0.159)(−0.982) − (0.987)(−0.188)
   ≈ 0.1564 + 0.1856 = 0.342 > 0

   So Statement I is also true.

Final Answer

Both cos(α)<0 and cos(α+β)>0 hold exactly. The exam lesson: never eyeball the quadrant of a triple angle from the "size" of the base angle — always run the triple-angle formula and check the sign explicitly.

Key Facts to Remember

• cos(3θ) = 4cos³θ − 3cosθ; sin(3θ) = 3sinθ − 4sin³θ.
• Principal range of sin⁻¹ is [−π/2, π/2]; of cos⁻¹ is [0, π].
• Tripling an angle can push it into a different quadrant even when the base angle looks "safe."
• Always determine the sign of cosθ/sinθ from the principal-range quadrant before applying identities.

Common Mistakes

FAQs

Prerequisites

Before practising this question type, make sure the following are in place:

• Triple-angle formulas for sine and cosine.
• Principal ranges of sin⁻¹ and cos⁻¹, and how they fix the sign of the companion ratio.
• Compound-angle formula cos(A+B) = cosA cosB − sinA sinB.

Revise these from the Inverse Trigonometric Functions chapter, then attempt the related questions below.

Related Questions