Math, asked by A1111, 1 year ago

sinx = 0 => x = nπ as I learnt,
but x = sin^-1(0) => x = 0 since the range of arcsinx is [-π/2, π/2]. So, sin^-1(0) has to be in between [-π/2, π/2], that's why 0.

Aren't these two statements contradictory ?

Where am I going wrong ? ​

Answers

Answered by mayank8913
1

Answer:

When we are talking about sinx, it is not a function, it is a type of relation so there are no conditions on it's domain and range. But arcsinx is a function, so for one value of x there must be one and only one value of arcsinx(by definition of function) and that is why we limit the range to -π/2,π/2 as it takes all values from -1 to 1

Answered by Anonymous
1

Answer:

At number line ,negative is less than zero

that's why 0 is lie between /2 and π/2.

range of sin X is /2 to π/2 .

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