Question

$\display \text{Show that}\\\\\displaystyle{\{(\cos\theta+\sin\phi)+i(\sin\theta-\cos\phi)\}^n+\{(\cos\theta+\sin\phi)-i(\sin\theta-\cos\phi)\}^n}\\\\\displaystyle{=2^{n+1}\cos^n\left(\dfrac{\pi}{4}+\dfrac{\theta-\phi}{2}\right)\cos\left[n\left(\dfrac{\pi}{4}-\dfrac{\theta+\phi}{2}\right)\right]}$

Answer 1

Answer:

Trigonometric formula used:

cosA-cosB=2•sin{(A+B)/2}•sin{(B-A)/2}

= -2•sin{(A+B)/2}•sin{(A-B)/2}

sinA-sinB=2•cos{(A+B)/2}•sin{(A-B)/2}

cos(π/2 + A) = - sinA

sin(π/2 + A) = cosA

De-Moivre's Formula;

(cosA + ísinA)^n= cos(nA) + ísin(nA)

Answer 2

Answer:

see solution in attachements ....