)Suppose is a real number with 0 < < 4 . Set = sin( − ), = cos( + ), = sin( − ), and = cos(2 − ). Arrange the values , , , and in increasing order, and explain how you determined their order.

Answer:

b < c < a < d

Step-by-step explanation:

P.S - The exact question is -

Given - Suppose x is a real number with 0 < x < [tex]\frac{\pi }{4}[/tex] . Set a = sin([tex]\pi[/tex] - x) ,                                    

             b = cos([tex]\pi[/tex] + x) , c = sin( x - [tex]\pi[/tex]) , d = cos(2[tex]\pi[/tex] - x)

To find - Arrange the values of a, b, c, d in increasing order , and explain

              how you determined their order.

Proof -

Firstly draw a unit circle where you indicate the following

A = ( cos(x), sin(x) )

B = ( cos( [tex]\pi[/tex] - x) , sin ([tex]\pi[/tex] - x) )

C = ( cos([tex]\pi[/tex] + x), sin ([tex]\pi[/tex] + x) )

D = ( cos (2[tex]\pi[/tex] - x) , sin (2[tex]\pi[/tex] - x) )

As we have given that 0 < x < [tex]\frac{\pi }{4}[/tex]

⇒sin (x) and cos(x) are positive and sin(x) < cos(x)

The figure is as follows :

Now

By seeing point B , we can say that

sin ([tex]\pi[/tex] - x) = sin (x) = a

By seeing point C , we can say that

cos([tex]\pi[/tex] + x) = -cos (x) = b

By seeing point C , we can say that

sin( x - [tex]\pi[/tex]) = -sin(x) = c

By seeing point D , we can say that

cos(2[tex]\pi[/tex] - x) = cos(x) = d

As sin(x) < cos(x) ⇒ a < d

   -sin(x) > -cos(x) ⇒ c > b ⇒ b< c

and as we know that -sin(x) < sin(x) ⇒ c < a

∴ we get

b < c < a < d

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