What am I allowed to do in a deduce question? (1 Viewer)

[Tsylana]

My question states


(1+sinx+icosx)/(1+sinx-icosx) = sinx + icosx

DEDUCE that

(1+sin pi/5 + icos pi/5)^5 + i(1 + sin pi/5 - i cos pi/5) = 0

I have to prove that LHS = RHS right, but am i allowed to divide the whole equation by (1 + sin pi/5 - i cos pi/5)

so i get (sin pi/5 + icos pi/5)^5 + i = 0?

 

[tommykins]

Using the first part of the question, lead it onto the second part.

 

[Tsylana]

just to make it more readable x).




DEDUCE that





I have to prove that LHS = RHS right, but am i allowed to divide the whole equation by


so i get
= 0?

 

[tommykins]

Post your solution and I'll tell you if it's valid

 

[Tsylana]

then just de moivres and im done?

 

[tommykins]

then just de moivres and im done?

I don't think so because you assumed that
is true.

You need to use the fact that and manipulate it so it obtains

 

[jchoi]

Using what's given, you need to be able to answer the question.
This means

Using: (1+sinx+icosx)/(1+sinx-icosx) = sinx + icosx
(1+sin pi/5 + icos pi/5)^5 + i(1 + sin pi/5 - i cos pi/5) = 0

But, I can't get the answer.... lol
sad.
I'll ask my teacher tomoz.
I'm up til 4am doing maths..... sadder.....

 

[Drongoski]

My question states


(1+sinx+icosx)/(1+sinx-icosx) = sinx + icosx

DEDUCE that

(1+sin pi/5 + icos pi/5)^5 + i(1 + sin pi/5 - i cos pi/5) = 0

I have to prove that LHS = RHS right, but am i allowed to divide the whole equation by (1 + sin pi/5 - i cos pi/5)

so i get (sin pi/5 + icos pi/5)^5 + i = 0?

can you please check whether or not the 2nd term LHS has to be, like the 1st, raised to the power of 5. I remember proving it with this amendment.

 

[Drongoski]

can you please check whether or not the 2nd term LHS has to be, like the 1st, raised to the power of 5. I remember proving it with this amendment.

You subst x = pi/5 and therefore [(1 + sinx + icos x)/(1 + sin x - icosx)]^5 =
(sin x + i cos x)^5; Therefore [(1 + sin pi/5 + i cos pi/5) / (1 + sin pi/5 - i cos pi/5)]
= [i cis(-x)]^5 = [i ^5 cis( -pi/5 x 5)] = i cis(-pi) = -i
therefore [1 + sin pi/5 + i cos pi/5)]^5 = - i[1 + sin pi/5 - i cos pi/5)]^5
hence bringing all to the left we get the result.

 

[thenuker6]

Note: the first part of the question does say PROVE that (1+ sin x + icosx)/(1+sinx-icosx)=sinx + i cosx
then divide the whole equation by [1+sin pi/5 -icos pi/5]^5
now we get [sin pi/5+icos pi/5]^5 = -i
-i = -i which proves that equation