Question

If sinA + cos2A = 1/2 and cosA+sin2A = 1/3, then find the value of sin3A.​

Answer 1

Answer:

it's your answer. ..........

Answer 2

$\huge\underline\blue{\sf Given:}$

1) sinA + cos2A = 1/2

2) cosA+sin2A = 1/3

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sin3A = ???

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sinA + cos2A = 1/2 ---------------- (Equation 1)

cosA+ sin2A = 1/3 ----------------( Equation 2)

Squaring both sides and adding both equations we get,

(sin²A+cos²2A+2sinA*cos2A)+(cos²A+sin²2A+2cosA*sin2A) = (1/4) + (1/9)

Re-arranging the equations :-

(sin²A+cos²A)+(sin²2A+cos²2A)+2(sinA*cos2A+cosA*sin2A) = $\huge{\frac{</strong><strong>1</strong><strong>3</strong><strong>}{</strong><strong>3</strong><strong>6</strong><strong>}}$

we know that,

sin²A+cos²A = 1

sinA*cosB+cosA*sinB = sin(A+B)

using both we get,

1 + 1 + 2sin(A+2A) = $\huge{\frac{13}{36}}$

2 + 2sin3A = $\huge{\frac{13}{36}}$

2sin3A = $\huge{\frac{13}{36}}$ - 2

2sin3A = $\huge{\frac{</strong><strong>(</strong><strong>-</strong><strong>5</strong><strong>9</strong><strong>)</strong><strong>}{36}}$

sin3A = $\huge{\frac{</strong><strong>(</strong><strong>-59</strong><strong>)</strong><strong>}{</strong><strong>7</strong><strong>2</strong><strong>}}$

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