What is α=3sin⁻¹(6/11) in terms of quadrant?

sin⁻¹(6/11)∈(π/6,π/3) since 6/11 is between sin(π/6)=1/2 and sin(π/3)=√3/2. So α=3sin⁻¹(6/11)∈(π/2,π) — second quadrant, cos α<0.

Are both statements true for α=3sin⁻¹(6/11) and β=3cos⁻¹(4/9)?

Yes. cos(α) 0 since α+β≈1.61π∈(3π/2,2π) where cosine is positive. Both statements are true.

α=3sin⁻¹(6/11), β=3cos⁻¹(4/9) — check if cos(α+β)>0 and cos(α)<0

Q: Are both statements true for α=3sin⁻¹(6/11) and β=3cos⁻¹(4/9)?

Yes. cos(α) 0 since α+β≈1.61π∈(3π/2,2π) where cosine is positive. Both statements are true.

JEERANKERS

MathInverse Trig

QMCQInverse Trigonometry

Let $\alpha=3\sin^{-1}\!\dfrac{6}{11}$ and $\beta=3\cos^{-1}\!\dfrac{4}{9}$, where inverse trigonometric functions take only the principal values.

Statement I: $\cos(\alpha+\beta)>0$
Statement II: $\cos\alpha<0$

A) Both true B) Both false C) Only I true D) Only II true

✅ Correct Answer

A) Both Statements are true

✓

Solution

Locate α = 3sin⁻¹(6/11)

Since $\sin(\pi/6)=1/2=5.5/11<6/11$ and $\sin(\pi/3)=\sqrt{3}/2\approx9.526/11>6/11$:

$\sin^{-1}\!\dfrac{6}{11}\in\left(\dfrac{\pi}{6},\dfrac{\pi}{3}\right)\quad\Rightarrow\quad\alpha=3\sin^{-1}\!\dfrac{6}{11}\in\left(\dfrac{\pi}{2},\pi\right)$

$\alpha\in(\pi/2,\pi)$ means $\cos\alpha<0$. ✅ Statement II is TRUE.

Locate β = 3cos⁻¹(4/9)

$\cos(\pi/3)=1/2=4.5/9>4/9$ and $\cos(\pi/2)=0<4/9$, so:

$\cos^{-1}\!\dfrac{4}{9}\in\left(\dfrac{\pi}{3},\dfrac{\pi}{2}\right)\quad\Rightarrow\quad\beta=3\cos^{-1}\!\dfrac{4}{9}\in\left(\pi,\dfrac{3\pi}{2}\right)$

Determine the range of α+β

$\alpha\in\left(\dfrac{\pi}{2},\pi\right),\quad\beta\in\left(\pi,\dfrac{3\pi}{2}\right)$

$\alpha+\beta\in\left(\dfrac{3\pi}{2},\dfrac{5\pi}{2}\right)$

Numerically: $\sin^{-1}(6/11)\approx0.5769$, so $\alpha\approx1.731$ rad. $\cos^{-1}(4/9)\approx1.110$, so $\beta\approx3.331$ rad.

$\alpha+\beta\approx5.062$ rad $\approx 1.611\pi$

This is in $\left(\dfrac{3\pi}{2},2\pi\right)$, where $\cos>0$.

✅ Statement I: $\cos(\alpha+\beta)>0$ is TRUE.

Conclusion

Both Statement I and Statement II are TRUE → Answer: A

📘 Bounding Inverse Trig Values

To determine the quadrant of $n\cdot\sin^{-1}(x)$ or $n\cdot\cos^{-1}(x)$: bound the inverse trig between known special angles ($\pi/6, \pi/4, \pi/3, \pi/2$), multiply bounds by $n$, and identify the resulting quadrant. Numerical approximation can confirm when bounds are tight.

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Important Concepts

Bounding sin⁻¹(6/11)$\sin(\pi/6)=0.5<6/11\approx0.545<\sin(\pi/3)\approx0.866$. So $\sin^{-1}(6/11)\in(\pi/6,\pi/3)$, and $\alpha=3\sin^{-1}(6/11)\in(\pi/2,\pi)$. Cos is negative in second quadrant.

Bounding cos⁻¹(4/9)$\cos(\pi/3)=0.5>4/9\approx0.444>0=\cos(\pi/2)$. So $\cos^{-1}(4/9)\in(\pi/3,\pi/2)$, and $\beta=3\cos^{-1}(4/9)\in(\pi,3\pi/2)$.

Determining sign of cos(α+β)$\alpha+\beta\in(3\pi/2,5\pi/2)$. More precisely, numerically $\alpha+\beta\approx1.611\pi$, which lies in $(3\pi/2,2\pi)$ where cosine is positive.

Both Statements True$\cos\alpha<0$ (Statement II, $\alpha$ in 2nd quadrant) ✓ and $\cos(\alpha+\beta)>0$ (Statement I, $\alpha+\beta$ in 4th quadrant equivalent) ✓. Answer: A.

FAQs

How to bound sin⁻¹(6/11)?

⌄

Compare $6/11\approx0.545$ with $\sin(\pi/6)=0.5$ and $\sin(\pi/3)\approx0.866$. Since $0.5<0.545<0.866$, we get $\pi/6<\sin^{-1}(6/11)<\pi/3$.

Why multiply by 3?

⌄

The problem defines $\alpha=3\sin^{-1}(6/11)$. Multiplying the bounds $\pi/6<\cdot<\pi/3$ by 3 gives $\pi/2<\alpha<\pi$ — directly placing $\alpha$ in the second quadrant.

Is the numerical approximation needed?

⌄

For Statement I, the range of $\alpha+\beta$ spans $(3\pi/2,5\pi/2)$, which includes both positive and negative cosine regions. Numerical values confirm $\alpha+\beta\approx1.61\pi\in(3\pi/2,2\pi)$ where cosine is positive.

What if α+β were exactly 2π?

⌄

Then $\cos(2\pi)=1>0$. The conclusion would still be Statement I is true.

Is this from JEE Main 2026?

⌄

Yes, this question appeared in JEE Main 2026.

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