If 4 i 3 - i 2 i + 1 2 = r cos ⁡ θ + i sin ⁡ θ , then cos ⁡ θ + sin ⁡ θ is equal to (where, i 2 = - 1 )

Question

If 4 i 3 - i 2 i + 1 2 = r cos ⁡ θ + i sin ⁡ θ , then cos ⁡ θ + sin ⁡ θ is equal to (where, i 2 = - 1 ),

MARKS App · Accepted Answer

4 i 3 - i 2 i + 1 2 = - 4 i - i 2 i + 1 2 ⇒ - 5 i 2 4 i 2 + 1 + 4 i = 25 - 1 - 3 + 4 i = 25 3 - 4 i × 3 + 4 i 3 + 4 i = 25 3 2 + 4 2 3 + 4 i = 5 3 5 + 4 5 i = r cos ⁡ θ + i sin ⁡ θ ⇒ r = 5 , cos ⁡ θ = 3 5 , sin ⁡ θ = 4 5 ⇒ cos ⁡ θ + sin ⁡ θ = 7 5 = 1 . 4

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