 Hi and welcome to the session. Let us discuss the following question which says evaluate sin of pi by 3 minus sin inverse minus 1 by 2. Now before moving on to the solution let's recall that sin inverse of minus x is equal to minus of sin inverse x where x belongs to closed interval minus 1 1. This is the key idea for this question. Now let's see its solution. We need to evaluate sin of pi by 3 minus sin inverse minus 1 by 2. Now we know that sin inverse of minus x is equal to minus of sin inverse x where x belongs to the closed interval minus 1 2 1. So here minus 1 by 2 belongs to this closed interval. That means sin inverse of minus 1 by 2 will be minus of sin inverse 1 by 2. So this will be equal to sin pi by 3 plus sin inverse 1 by 2. Now let's find out the value of sin inverse 1 by 2. We know that the range of principal values of sin inverse is closed interval from minus pi by 2 to pi by 2. So let us assume that sin inverse 1 by 2 is equal to y. So this implies that sin y is equal to 1 by 2 which is equal to sin pi by 6. So this means that y will be equal to pi by 6 which belongs to the closed interval minus pi by 2 to pi by 2 and y is equal to sin inverse 1 by 2. So that means sin inverse 1 by 2 is equal to pi by 6. So let us substitute the value of sin inverse 1 by 2 over here. So this will be equal to sin of pi by 3 plus pi by 6 which will be equal to sin pi by 2 which is equal to 1. Thus 1 is the required answer to this question. With this we finish this session. Hope you must have enjoyed it. Goodbye, take care and have a nice day.