2015年10月21日 · #arcsin(x) = sin^-1(x)# is the inverse function of the function #sin(x)# That is: If #x in (-pi/2, pi/2)#, then #arcsin(sin(x)) = x#

2016年1月1日 · How do you use inverse trigonometric functions to find the solutions of the equation that are in

2017年7月13日 · Given: f(x)=cos(sin^-1(x)) The domain for the inverse sine function is -1<=x<=1 because this is the range for the sine function. The range for the function is the same as the range for the cosine function, -1<=f(x)<=1 Use the identity cos(x) = +-sqrt(1-sin^2(x)) f(x) = +-sqrt(1-sin^2(sin^-1(x))) The sine and the inverse sine function cancel; leaving us with x^2: f(x) = +-sqrt(1-x^2); -1<=x<=1 ...

What does #arcsin(cos ((5pi)/6)) # equal? Trigonometry Inverse Trigonometric Functions Basic Inverse ...

2018年7月25日 · We know that , #color(red)((1)arc sin (sintheta)=theta)#, #color(red)(theta in[-pi/2,pi/2])# Let , #X=arc sin(sin(-pi/6))....tocolor(red)(-pi/6 in [-pi/2,pi/2])#

2016年3月12日 · A general convention is to use the small letter #arccos(x)# and #arcsin(x)# as multivalued expressions, each respectively indicating all the values whose cosine or sine has a given value #x#. The meaning of the sum of those is really every possible combination, and those wouldn't always give #pi/2.#

2016年1月20日 · -pi/2 If you want to know how please follow. sin(-pi/2) = -sin(pi/2) since sin(x) is a odd function. sin(-pi/2) = -1 since sin(pi/2) = 1 arcsin(sin(-pi/2)) = arcsin(-1) Now comes the range of arcsin(x) The range of arcsin(x) is [-pi/2,pi/2] So arcsin(-1) would give -pi/2 Therefore, arcsin(sin(-pi/2)) = -pi/2

2016年2月26日 · -pi/6 cos (-2pi/3 ) = sin ( pi/2 + (-2pi/3)) = sin (-pi#/6) arc sin (sin) is unit operator 1.

2016年2月11日 · cos(-pi/3) = cos(pi/3) because cosinus is even watch your trigonometric circle you get cos(pi/3) = 1/2 you search a number x such as arcsin(1/2) = x 1/2 = sin(x) watch your circle again you get x_1 = pi/6 + 2kpi where k in ZZ x_2 = pi-pi/6 + 2kpi where k in ZZ

2017年8月17日 · arcsinx-=sin^-1(x) arcsinx" is used in place of "sin^-1x" to avoid confusion" " misinterpretation with "sin^-1x-==1/sinx "of course "sin^-1x!=1/sinx (sinx)^-1=1/sinx