 Hello and welcome to this session. In this session we are going to discuss the following question which says that simplify sine of 3 cos inverse of x and hence find the value of sine of 3 cos inverse of vanapal square root of 5, 3 cos inverse of x is equal to cos inverse of 4x cubed minus 3x. With this key idea let us proceed with the solution. We are given the expression sine of 3 cos inverse of x and we know that 3 cos inverse of x is given by cos inverse of 4x cubed minus 3x. Therefore we can replace 3 cos inverse of x with cos inverse of 4x cubed minus 3x in the given expression and we get sine of cos inverse of 4x cubed minus 3x let cos inverse of 4x cubed minus 3x be theta which implies that cos of theta is equal to 4x cubed minus 3x in a triangle p through r if theta is the angle between the lines p r and r cubed then cos of angle theta is given by base of the hypotenuse that is r cubed upon rp which is equal to 4x cubed minus 3x by 1 if we are given base r cubed as 4x cubed minus 3x and hypotenuse rp as 1 then we can find the value of the prependicular pq by using Pythagoras theorem. By Pythagoras theorem we have prependicular pq is equal to square root of hypotenuse rp square minus base rq square which is equal to square root of 1 square minus 4x cubed minus 3x the wave square which is equal to the square root of 1 minus rp 16 x raised to the power 6 plus 9x square minus 24 x raised to the power 4 which is equal to square root of 1 minus 16 x raised to the power 6 minus 9x square plus 24 x raised to the power 4 that is prependicular pq is given by square root of 1 minus 16 x raised to the power 6 minus 9x square plus 24 x raised to the power 4 and we know that sin of angle theta is equal to prependicular upon hypotenuse that is pq upon rp which is equal to square root of 1 minus 16 x raised to the power 6 minus 9x square plus 24 x raised to the power 4 pole upon 1 and we have assumed the value of theta as cos inverse of 4x cubed minus 3x therefore we can write sin of cos inverse of 4x cubed minus 3x is equal to square root of 1 minus 16 x raised to the power 6 minus 9x square plus 24 x raised to the power 4 and we know that cos inverse of 4x cubed minus 3x is equal to 3 cos inverse of x therefore the value of the expression sin of 3 cos inverse of x is equal to square root of 1 minus 16 x raised to the power 6 minus 9x square plus 24 x raised to the power 4 now we shall find the value of the expression sin of 3 cos inverse of 1 upon square root of 5 in the above equation we get sin of 3 cos inverse of 1 upon square root of 5 is equal to square root of 1 minus 16 into 1 upon square root of 5 raised to the power 6 minus 9 into 1 upon square root of 5 square plus 24 into 1 upon square root of 5 raised to the power 4 which is equal to square root of 1 minus 16 upon 125 minus 9 upon 5 plus 24 upon 25 taking the a sin of square root we get square root of 125 minus 16 minus 225 plus 120 upon 125 which is equal to square root of 4 upon 125 which is equal to 2 upon 5 into square root of 5 therefore the value of the expression sin of 3 cos inverse of 1 upon square root of 5 is equal to 2 upon 5 into square root of 5 which is the required answer this completes our session hope you enjoyed this session.