Answer to Question #50655 in Trigonometry for labu

Question #50655

Sin^(-1)⁡(cos⁡θ )= (π/2) - θ or (π/2) + θ I think both are correct but not sure.?
sin^(-1)⁡(cos⁡θ )= (π/2) - θ or (π/2) + θ
I think both are correct but not sure.

Here my thinking
sin^(-1)⁡(cos⁡θ )=sin^(-1)⁡(sin ((π/2) - θ))=(π/2) - θ

Or in another way
sin^(-1)⁡(cos⁡θ )=sin^(-1)⁡(sin (⁡(π/2)+ θ))= (π/2) + θ)

i want to differentiate sin^(-1)⁡(cos⁡θ ). so what should we differentiate (π/2) - θ or (π/2) +θ.please let me know

Expert's answer

Download Answer

Learn more about our help with Assignments: Trigonometry

Comments

Assignment Expert

17.02.15, 17:27

Dear tinn. Thank you for adding information. If the range of sin^(-1) function (arcsine function) does not fall into segment [-pi/2;pi/2] in an expression, then this expression is false. It is possible to transform expressions to get proper values of argument, for example, applying formulas like sin(θ)=sin(pi-θ). However, you have not specified values of θ (for example, θ takes on zero), therefore the expression is true if it is well defined. To differentiate sin^-1[(1-x^2)/(1+x^2)], you should use the chain rule and quotient rule for derivatives.

tinn

13.02.15, 18:45

if i use (π/2) + θ) then it breaks the range of sin^-1 x . because , its [-π/2,π/2]. in the problem., sin^-1(cos θ) . if u put θ= 30 degree then the value comes sin^-1 (sqrt 3 / 2)= 60 degree. which is in the range [-π/2,π/2]. but if u sin^-1 (sqrt 3 / 2)= 120 . then it violates the range. so that's why i think we can't use (π/2 + θ) in the above problem. i am not sure. please help. ant the problem i want to differentiate is sin^-1[(1-x^2)/(1+x^2)]. so here let x = cos θ then its sin^-1(cos θ)

Answer to Question #50655 in Trigonometry for labu

Comments

Leave a comment

Related Questions