Skip to content
Exercise 2.9 · Q17

Q.The principal argument of (sin⁡40∘+icos⁡40∘)5(\sin40^\circ+i\cos40^\circ)^5 is

(1) −110∘-110^\circ
(2) −70∘-70^\circ
(3) 70∘70^\circ
(4) 110∘110^\circ
Tamil Nadu DgeTextbookSubjectiveImportance★★★★★
54% · 66/122 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

de Moivre's theorem needs the base written as cos⁡θ+isin⁡θ\cos\theta+i\sin\theta; a co-function swap converts sin⁡40∘+icos⁡40∘\sin40^\circ+i\cos40^\circ into that form before raising to the 5th power.

Step 1. Convert to cosine-sine form using co-function identities.

sin⁡40∘=cos⁡(90∘−40∘)=cos⁡50∘,cos⁡40∘=sin⁡(90∘−40∘)=sin⁡50∘.\sin40^\circ=\cos(90^\circ-40^\circ)=\cos50^\circ,\qquad \cos40^\circ=\sin(90^\circ-40^\circ)=\sin50^\circ.

So sin⁡40∘+icos⁡40∘=cos⁡50∘+isin⁡50∘=cis⁡50∘\sin40^\circ+i\cos40^\circ=\cos50^\circ+i\sin50^\circ=\operatorname{cis}50^\circ.

Step 2. Apply de Moivre's theorem with n=5n=5.

(cis⁡50∘)5=cis⁡(5×50∘)=cis⁡250∘.(\operatorname{cis}50^\circ)^5=\operatorname{cis}(5\times50^\circ)=\operatorname{cis}250^\circ. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.