Find the principal values of the following:

    2. \cos^{-1}\left(\frac{\sqrt3}{2} \right )

Answers (1)

So, let us assume that \cos^{-1}\left(\frac{\sqrt3}{2} \right ) = x  then,

Taking inverse both sides we get;

cos\ x = (\frac{\sqrt{3}}{2}),  or   cos (\frac{\pi}{6}) = (\frac{\sqrt{3}}{2})         

and as we know that the principal values of cos^{-1} is from [0,\pi],

Hence cos\ x = (\frac{\sqrt{3}}{2})     when x = \frac{\pi}{6} . 

Therefore, the principal value for \cos^{-1}\left(\frac{\sqrt3}{2} \right )is   \frac{\pi}{6}.

 

Most Viewed Questions

Preparation Products

Knockout NEET May 2024

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 40000/-
Buy Now
Knockout NEET May 2025

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 45000/-
Buy Now
NEET Foundation + Knockout NEET 2024

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 54999/- ₹ 42499/-
Buy Now
NEET Foundation + Knockout NEET 2024 (Easy Installment)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 3999/-
Buy Now
NEET Foundation + Knockout NEET 2025 (Easy Installment)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 3999/-
Buy Now
Boost your Preparation for JEE Main with our Foundation Course
 
Exams
Articles
Questions