Let alpha = 3 sin^-1(6/11) and beta = 3 cos^-1(4/9), where inverse trigonometric functions take only the principal values. Given below are two statements : Statement I: cos(alpha + beta) > 0. Statement II: cos(alpha) < 0. In the light of the above statements, choose the correct answer from the options given below :

The correct answer is Option 1. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q12 – JEE-MAIN 2026

← Back

Home Maths Inverse TrigonometryQuestion #231

Let

α = 3 \sin^{- 1} (\frac{6}{11})

and

β = 3 \cos^{- 1} (\frac{4}{9})

, where inverse trigonometric functions take only the principal values. Given below are two statements : Statement I:

\cos (α + β) > 0

. Statement II:

\cos (α) < 0

. In the light of the above statements, choose the correct answer from the options given below :

Clue 1 of 3: Determine the Quadrants

First, find the principal value ranges for $\sin^{- 1} (x)$ and $\cos^{- 1} (x)$ , then use the given expressions for $α$ and $β$ to determine the quadrants in which $α$ and $β$ lie.

🧠 Full Solution Path

Step 1: Determine the Quadrants of

α

and

β

✦ Active

Let $x = \sin^{- 1} (\frac{6}{11})$ . Since $0 < \frac{6}{11} < 1$ , $x \in (0, \frac{π}{2})$ . We know $\sin (\frac{π}{6}) = 0.5$ and $\sin (\frac{π}{4}) = \frac{\sqrt{2}}{2} \approx 0.707$ . Since $0.5 < \frac{6}{11} \approx 0.545 < 0.707$ , we have $\frac{π}{6} < x < \frac{π}{4}$ . Thus, $α = 3 x \in (\frac{3 π}{6}, \frac{3 π}{4}) = (\frac{π}{2}, \frac{3 π}{4})$ . This means $α$ lies in the second quadrant.

Let $y = \cos^{- 1} (\frac{4}{9})$ . Since $0 < \frac{4}{9} < 1$ , $y \in (0, \frac{π}{2})$ . We know $\cos (\frac{π}{3}) = 0.5$ and $\cos (\frac{π}{2}) = 0$ . Since $0 < \frac{4}{9} \approx 0.444 < 0.5$ , we have $\frac{π}{3} < y < \frac{π}{2}$ . Thus, $β = 3 y \in (\frac{3 π}{3}, \frac{3 π}{2}) = (π, \frac{3 π}{2})$ . This means $β$ lies in the third quadrant.

💡 Teacher's Secret Hint

Remember the principal value ranges for inverse trigonometric functions: $\sin^{- 1} (x) \in [- π / 2, π / 2]$ and $\cos^{- 1} (x) \in [0, π]$ .

Step 2: Evaluate Statement II:

\cos (α) < 0

○ Expand

From Step 1, we found that $α \in (\frac{π}{2}, \frac{3 π}{4})$ , which is in the second quadrant. In the second quadrant, the cosine function is negative. Therefore, $\cos (α) < 0$ . Statement II is true.

💡 Teacher's Secret Hint

Recall the signs of trigonometric functions in different quadrants.

Step 3: Evaluate Statement I:

\cos (α + β) > 0

○ Expand

From Step 1, we have $\frac{π}{2} < α < \frac{3 π}{4}$ and $π < β < \frac{3 π}{2}$ . Adding these inequalities, we get:

\frac{π}{2} + π < α + β < \frac{3 π}{4} + \frac{3 π}{2}

\frac{3 π}{2} < α + β < \frac{3 π}{4} + \frac{6 π}{4}

\frac{3 π}{2} < α + β < \frac{9 π}{4}

The interval $(\frac{3 π}{2}, \frac{9 π}{4})$ can be rewritten as $(\frac{3 π}{2}, 2 π + \frac{π}{4})$ . This means $α + β$ lies in the fourth quadrant (where $\cos$ is positive) or the first quadrant (where $\cos$ is positive) after one full rotation. Specifically, if $α + β \in (\frac{3 π}{2}, 2 π)$ , $\cos (α + β) > 0$ . If $α + β \in (2 π, \frac{9 π}{4})$ , then $\cos (α + β) = \cos (α + β - 2 π)$ , and $α + β - 2 π \in (0, \frac{π}{4})$ , so $\cos (α + β) > 0$ . In either case, $\cos (α + β) > 0$ . Statement I is true.

To be more rigorous, let's check if $α + β$ crosses $2 π$ . We need to compare $x + y$ with $2 π / 3$ . We have $\sin x = 6 / 11$ and $\cos y = 4 / 9$ . Then $\cos x = \sqrt{1 - (6 / 11)^{2}} = \frac{\sqrt{85}}{11}$ and $\sin y = \sqrt{1 - (4 / 9)^{2}} = \frac{\sqrt{65}}{9}$ . $\cos (x + y) = \cos x \cos y - \sin x \sin y = \frac{\sqrt{85}}{11} \cdot \frac{4}{9} - \frac{6}{11} \cdot \frac{\sqrt{65}}{9} = \frac{4 \sqrt{85} - 6 \sqrt{65}}{99}$ . We compare this with $\cos (2 π / 3) = - 1 / 2$ . Numerically, $4 \sqrt{85} \approx 4 \times 9.219 = 36.876$ and $6 \sqrt{65} \approx 6 \times 8.062 = 48.372$ . So, $\cos (x + y) \approx \frac{36.876 - 48.372}{99} = \frac{- 11.496}{99} \approx - 0.116$ . Since $- 0.116 > - 0.5$ , we have $\cos (x + y) > \cos (2 π / 3)$ . As $x + y \in (\frac{π}{2}, \frac{5 π}{6})$ and $\cos$ is decreasing in this interval, $x + y < \frac{2 π}{3}$ . Multiplying by 3, we get $3 (x + y) < 2 π$ , so $α + β < 2 π$ . Combined with $α + β > \frac{3 π}{2}$ , we have $\frac{3 π}{2} < α + β < 2 π$ . This interval is strictly in the fourth quadrant, where $\cos (α + β) > 0$ . Thus, Statement I is true.

💡 Teacher's Secret Hint

When comparing values like $\cos (x + y)$ with a known value, it's often easier to compare the numerical values or use the monotonicity of the cosine function in the relevant interval.

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Maths Question 12 – JEE-MAIN 2026

Clue 1 of 3: Determine the Quadrants

🧠 Full Solution Path