(2sinα+3cosα)^2+(3sinα−2cosα)^2=13

Annette Arroyo

Annette Arroyo

Answered question

2020-11-05

(2sinα+3cosα)2+(3sinα2cosα)2=13

Answer & Explanation

svartmaleJ

svartmaleJ

Skilled2020-11-06Added 92 answers

Work on the left side. Expand as: (2sinα+3cosα)2+(3sinα2cosα)=(4sin2+12sinαcosα+9cos2α)+(9sin2α12sinαcosα+4cos2α)
12sinacosa cancels out so:
(2sina+3cosa)2+(3sina2cosa)2=(4sin2a+9cos2a)+(9sin2a+4cos2a) Regroup as:
(2sina+3c0sa)2+(3sina2cosa)2=4sin2a+4cos2a+9cos2a+9sin2a Factor as:
(2sina+3cosa)2+(3sina2cosa)2=4(sin2a+4cos2a)+9(cos2a+9sin2a)
Use the Pythagorean identity: sin2x+cos2x=1 (2sina+3cosa)2+(3sina2cosa)2=4(1)+9(1)
(2sina+3cosa)2+(3sina2cosa)2=4+9
(sina+3cosa)2+(3sina2cosa)2=13

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