Question

Brainly Stars , Brainly teachers or moderators please help ​

Answer 1

$1+sin(\dfrac{\pi}{8})=1+cos(\dfrac{\pi}{2}-\dfrac{\pi}{8})\\\\\implies 1+cos(\dfrac{3\pi}{8})$

$icos(\dfrac{\pi}{8})=isin(\dfrac{\pi}{2}-\dfrac{\pi}{8})\\\\\implies isin(\dfrac{3\pi}{8})$

$[\dfrac{1+sin(\pi/8)+icos(\pi/8)}{1+sin(\pi/8)-icos(\pi/8)}]^{8/3}\\\\\implies [\dfrac{1+cos(3\pi/8)+isin(3 \pi/8)}{1+cos(3\pi/8)-isin(3\pi/8)}]^{8/3}\\\\\textbf{Use the formulas : 1+cosA=2cos^2(\dfrac{A}{2}) }\\sin2A=2sinAcosA\\\\\implies [\dfrac{2cos^2(3\pi/16)+2isin(3\pi/16)cos(3\pi/16)}{2cos^2(3\pi/16)-2isin(3\pi/16)cos(2\pi/16)}]^{8/3}$

$[\dfrac{cos(3\pi/16)+isin(3\pi/16)}{cos(3\pi/16)-isin(3\pi/16)}]^{8/3}\\\\\implies [\dfrac{e^{3i\pi/16}}{e^{-3i\pi/16}}]^{8/3}[\textbf{Use e^{iA}=cosA+isinA ]}\\\\\implies [e^{3i\pi/16+3i\pi/16}]^{8/3}\\\\\implies [e^{3i\pi/8}]^{8/3}\\\\\implies e^{i\pi}\\\\\implies cos\pi+isin\pi\\\\\implies -1+i\times 0\\\\\implies \boxed{-1}$

Answer 2

Answer:

Step-by-step explanation: