I really hope If you can solve this for me
asked Oct 30, 2016 in Trigonometry Answers by cautionc

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1 Answer

1.

Put A in place of ?.

Multiply top and bottom by sinA-icosA:

(1+a)/(1-a)=(1+cosA+isinA)/(1-cosA-isinA)=

(1+cosA+isinA)(1-cosA+isinA)/((1-cosA)^2+sin^2(A)).

Numerator: 1-cos^2(A)+isinA(1+cosA+1-cosA)-sin^2(A)=

sin^2(A)-sin^2(A)+2isinA=2isinA.

Denominator: 1-2cosA+cos^2(A)+sin^2(A)=2-2cosA.

Numerator/denominator=isinA/(1-cosA).

But sinA=2sin(A/2)cos(A/2) and cosA=1-2sin^2(A/2),

so isinA/(1-cosA)=2isin(A/2)cos(A/2)/2sin^2(A/2)=icot(A/2).

Therefore, (1+a)/(1-a)=icot(A/2) QED

2.

v has been assumed to represent square root.

(x+iy)(x-iy)=x^2+y^2.

(x+iy)/(x-iy)=(x+iy)^2/(x^2+y^2)=

(ac+bd+ibc-iad)/((c^2+d^2)(x^2+y^2))

(x+iy)^2=x^2-y^2+2ixy=

(a+ib)/(c+id)=(a+ib)(c-id)/(c^2+d^2)=(ac+bd+ibc-iad)/(c^2+d^2).

So x^2-y^2=(ac+bd)/(c^2+d^2) and 2xy=(bc-ad)/(c^2+d^2).

Also, 1/(x+iy)=(x-iy)/(x^2+y^2)=

√((c+id)/(a+ib))=√((c+id)(a-ib)/(a^2+b^2))=

√((ac+bd+iad-ibc)/(a^2+b^2)).

 

More to follow...

answered Nov 1, 2016 by Rod Top Rated User (424,820 points)
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