Step by step solution to "(2sin⁡α+3cos⁡α)2+(3sin⁡α−2cos⁡α)2=13"

Annette Arroyo

Answered question

2020-11-05

$(2 \sin α + 3 \cos α)^{2} + (3 \sin α - 2 \cos α)^{2} = 13$

Answer & Explanation

svartmaleJ

Skilled2020-11-06Added 92 answers

Work on the left side. Expand as: $(2 \sin α + 3 \cos α)^{2} + (3 \sin α — 2 \cos α) ” = (4 \sin^{2} + 12 \sin α \cos α + 9 \cos^{2} α) + (9 \sin^{2} α — 12 \sin α \cos α + 4 \cos^{2} α)$
$12 \sin a \cos a$ cancels out so:
${(2 \sin a + 3 \cos a)}^{2} + {(3 \sin a — 2 \cos a)}^{2} = (4 \sin^{2} a + 9 \cos^{2} a) + (9 \sin^{2} a + 4 \cos^{2} a)$ Regroup as:
${(2 \sin a + 3 c 0 s a)}^{2} + {(3 \sin a — 2 \cos a)}^{2} = 4 \sin^{2} a + 4 \cos^{2} a + 9 \cos^{2} a + 9 \sin^{2} a$ Factor as:
${(2 \sin a + 3 \cos a)}^{2} + {(3 \sin a — 2 \cos a)}^{2} = 4 (\sin^{2} a + 4 \cos^{2} a) + 9 (\cos^{2} a + 9 \sin^{2} a)$
Use the Pythagorean identity: $\sin^{2} x + \cos^{2} x = 1$ ${(2 \sin a + 3 \cos a)}^{2} + {(3 \sin a — 2 \cos a)}^{2} = 4 (1) + 9 (1)$
${(2 \sin a + 3 \cos a)}^{2} + {(3 \sin a — 2 \cos a)}^{2} = 4 + 9$
${(\sin a + 3 \cos a)}^{2} + {(3 \sin a — 2 \cos a)}^{2} = 13$

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