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Name: Explain solution for RD Sharma class class 12 chapter 3 Inverse Trigonometric Functions exercise Very short answer question 20 math
Uploaded: Thu, 2021-08-05 23:50
Description: Explain solution for RD Sharma class class 12 chapter 3 Inverse Trigonometric Functions exercise Very short answer question 20 math

#Inverse Trigonometric Functions
#Maths
#School
#RD Sharma Vol. 1 Class 12
#R.D. Sharma
#CBSE 12 Class

#1.3
#3.1
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#3.9
#3.11
#3.12
#9.12
#3.10
#Fill in the blanks
#Very Short Answer type
#Multiple Choice Questions (MCQs)
#18.1
#3.14
#3.13

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Explain solution for RD Sharma class class 12 chapter 3 Inverse Trigonometric Functions exercise Very short answer question 20 math

Answers (1)

Answer:

$\frac{1}{2}$

Given:

$\cos \left(2 \sin ^{-1} \frac{1}{2}\right)$

Hint: Try to solve $\sin$ function.

Solution:

$\begin{aligned} \cos \left(2 \sin ^{-1} \frac{1}{2}\right) &=\cos \left(2 \times \frac{\pi}{6}\right)=\cos \left(\frac{\pi}{3}\right) \\ &=\frac{1}{2} \end{aligned}$

Posted by

infoexpert22

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Explain solution for RD Sharma class class 12 chapter 3 Inverse Trigonometric Functions exercise Very short answer question 20 math

Answers (1)

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