Verified answer

You need to remember the following trigonometrical identity:

sin²x + cos²x ≡ 1

=> cos²x ≡ 1 - sin²x

=> cos²x = 1 - ((√3)/4)²

=> cos²x = 1 - (3/16)

i.e. cos²x = 13/16

so, cosx = ± √13/4

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Just draw a right angled triangle, observe that if sin theta = (√3)/4, then the hypotenuse of that triangle is 4, while the other two sides are (√3)/4 and √13 using pythagora's theorem

So cos theta adj/hyp = (√13)/4 which is equal to 0.901388....

cos^2 theta = 1 - sin^2 theta

= 1 - 3/16

= 13/16

cos theta = sqrt(13)/4

Sin A=sqrt(3)/4

=opp/hyp

adj=sqrt(hyp^2-adj^2)

=sqrt(16-3)

=sqrt(13)

cos A=adj/hyp

=sqrt(13)/4

a.) in "a+bi" kind, the "r" is sqrt(a^2+b^2) so (-5)^2 + b^2 = 80 one b= sqrt56 = 2sqrt14 b.) i.) cos 315 = sqrt(2)/2 sin 315 = -sqrt(2)/2 18cis315 = 9sqrt(2) + 9isqrt(2) ii.) cos a hundred and fifty = -sqrt(3)/2 sin a hundred and fifty = one million/2 (4sqrt(3))cis150 = -6 + 2sqrt(3) c.) i.) -3 + 4i ii.) 5 cis 127 stages