Question

simplify
(cos3A+isin3A)^5(cosA-isinA)^3/(cos5A+isin5A)^7(cos2A-isin2A)^5
​

Answer 1

Answer:

cos13A-isin13A

Step-by-step explanation:

e^(iA) = cosA + isinA

so as per above condition

(cos3A+isin3A)^5= {e^(3iA)}^5 = e^(15iA)

cosA-isinA = cos(-A)+isin(-A) because cos(-x) = cosx and sin(-x) = -sinx

(cosA- isinA)^3 = {e^(iA)}^3= e^(-3iA)

(cos5A+isin5A)^7 = e^(35iA)

(cos2A-isin2A)^5 = e^(-10iA)

so the given equitation will be written as

e^(15iA) × e^(-3iA)/e^(35iA) × e^(-10iA)

= e^(15iA-3iA-35iA+10iA)

= e^(-13iA)

= cos13A - isin13A