tan-1(tan(3pi/5))
To find the value of tan^(-1)(tan(3π/5)), we can use the identity:
tan^(-1)(tan(x)) = x + nπ, where n is an integer.
In this case, x = 3π/5.
So, tan^(-1)(tan(3π/5)) = 3π/5 + nπ.
Since n can be any integer, we need to simplify the expression to the principal value, which is within the range of -π/2 to π/2.
To do this, we need to convert 3π/5 into an angle within this range.
Dividing 3π/5 by π, we get:
3π/5 ÷ π = 3/5.
This means 3π/5 represents an angle that is greater than π/2, since the numerator is greater than the denominator.
To find an equivalent angle within the range of -π/2 to π/2, we can subtract 2π until we get a value in this range.
Let's reduce the angle step by step:
3/5 - 2 = -7/5.
So, -7/5 represents an angle within the range of -π/2 to π/2.
Finally, we can express tan^(-1)(tan(3π/5)) as:
tan^(-1)(tan(3π/5)) = -7/5 + nπ, where n is an integer.
To simplify the expression tan-1(tan(3π/5)), we need to understand the properties of inverse trigonometric functions.
The inverse trigonometric function tan^(-1)(x) or arctan(x) gives the angle whose tangent is x.
In this case, we have tan-1(tan(3π/5)). Since the argument of the tangent function is 3π/5, we can simplify the equation as follows:
Since 3π/5 lies in the range of -π/2 to π/2, the tangent function is positive. Hence, we have:
tan-1(tan(3π/5)) = 3π/5.
So, the simplified expression for tan-1(tan(3π/5)) is 3π/5.
by definition, arctan(tan(x)) = x
But if you want the principal value for arctan, that is between -pi/2 and pi/2