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ngkamsengpeter
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Solve the following question. Give your answer in cert form .Thank you .
sin-1 x + sin-1 (2x) = pie/3
sin-1 x + sin-1 (2x) = pie/3
finchie_88 said:You could use numerical methods. Example:
f(x) = arcsinx+arcsin2x = pi/3 =>let x(n+1) = sin( pi/3 - arcsinxn), that should work...
finchie_88 said:You could use numerical methods. Example:
f(x) = arcsinx+arcsin2x = pi/3 =>let x(n+1) = sin( pi/3 - arcsinxn), that should work...
Tide said:If you take the sine of both sides and use the basic properties of inverse trig functions you will find:
[tex]x \left( \sqrt{1-4x^2} +2 \sqrt {1 - x^2} \right) = \frac {\sqrt 3}{2}[/tex]
I believe you can solve this equation without resorting to numerical approximation.
How much do you understand? Have you tried working through the suggested step yourself? Where did you get stuck?I still not understand .Can you explain more detail ? Thank You .
I do not know what I need to do with the sin-1(x) +sin-1 (2x) .Is it these two have relationship ? If yes please tell me . The suggested step is to short , I cannot understand it . So counld you give me more details on that .Hurkyl said:How much do you understand? Have you tried working through the suggested step yourself? Where did you get stuck?
Can you prove for me why [tex]\sin \cos^{-1} x = \sqrt {1-x^2}[/tex] . Thank You .Tide said:ng,
When I suggested taking the sine of both sides I assumed you would recognize you need to use a basic trig identity for the sum of angles:
[tex]\sin a + b = \sin a \cos b + \cos a \sin b[/tex]
Also, you will need basic relations for the inverse trig functions:
[tex]\sin \sin^{-1} x = x[/tex]
[tex]\sin \cos^{-1} x = \sqrt {1-x^2}[/tex]
and similarly for the cosines. Does that help?
ngkamsengpeter said:Can you prove for me why [tex]\sin \cos^{-1} x = \sqrt {1-x^2}[/tex] . Thank You .
My book does not teach me that . It only have sin-1 A + cos-1 B = pie/2 . So can you please prove for me because I have no idea about that ?Hurkyl said:Have you tried to do it yourself? How far did you get? Have you consulted your book at all?
Oh yes . I am sorry about the typing error . I have seen your answer but how can we know [tex]\sin{\left(\cos^{-1}{x}\right)}=\sqrt{1-\cos^2{\left(\cos^{-1}{x}}\right)}[/tex]apmcavoy said:[tex]\sin{\left(\cos^{-1}{x}\right)}=\sqrt{1-\cos^2{\left(\cos^{-1}{x}}\right)}[/tex]
You know that [itex]\cos{\left(\cos^{-1}{x}\right)}=x[/itex], so that above equation becomes:
[tex]\sqrt{1-x^2}[/tex]
You could also do this geometrically by drawing a right triangle and using the Pythagorean Theorem.
In your last post, did you mean?:
[tex]\sin^{-1}{x}+\cos^{-1}{x}=\frac{\pi}{2}[/tex]
Look at post #16.ngkamsengpeter said:Oh yes . I am sorry about the typing error . I have seen your answer but how can we know [tex]\sin{\left(\cos^{-1}{x}\right)}=\sqrt{1-\cos^2{\left(\cos^{-1}{x}}\right)}[/tex]
Based on my knowledge, [tex]\cos^2{\left(\cos^{-1}{x}\right)} [/tex] should become [tex]cos x[/tex] right ? How can [tex]cos^2{\left(\cos^{-1}{x}}\right)} [/tex] become [tex]{x}^2[/tex] ?ngkamsengpeter said:Oh yes . I am sorry about the typing error . I have seen your answer but how can we know [tex]\sin{\left(\cos^{-1}{x}\right)}=\sqrt{1-\cos^2{\left(\cos^{-1}{x}\right)}[/tex]
Someone please answer my question . Thank you .ngkamsengpeter said:Based on my knowledge, [tex]\cos^2{\left(\cos^{-1}{x}\right)} [/tex] should become [tex]cos x[/tex] right ? How can [tex]cos^2{\left(\cos^{-1}{x}}\right)} [/tex] become [tex]{x}^2[/tex] ?
The inverse function of sin is called arcsin or sin-1. It is the function that undoes the output of the sin function, and is defined as the angle whose sine is equal to a given value.
The inverse function of cos is called arccos or cos-1. It is the function that undoes the output of the cos function, and is defined as the angle whose cosine is equal to a given value.
The inverse function of tan is called arctan or tan-1. It is the function that undoes the output of the tan function, and is defined as the angle whose tangent is equal to a given value.
The domain of inverse trigonometric functions is the set of values that can be plugged into the function to produce a valid output. The range is the set of all possible outputs of the function. For inverse trigonometric functions, the domain is usually restricted to a specific interval in order to ensure a unique output.
The inverse trigonometric functions are essentially the inverse operations of their corresponding trigonometric functions. For example, arcsin undoes the output of sin, arccos undoes the output of cos, and arctan undoes the output of tan. They can also be thought of as the "undo" buttons for trigonometric functions.