Find the maximum and minimum value of 9 CosA + 40 SinA + 40
1). 89, -4
2). 85, -7
3). 87, -5
4). 81, -1
⇒ Always look out for Pythagoras theorem, we know that (9, 41, 40) is one
⇒ ∴ 41 (9/41 CosA + 40/41 SinA)
⇒ Let there be an angle B for which SinB = 9/41, CosB = 40/41
⇒ 41 (SinB CosA + CosB SinA) + 40
⇒ 41 (Sin(A + B)) + 40
⇒ We know that Sin(A + B)max = 1
⇒ Max value = 41 × 1 + 40 = 81
⇒ Min value = 41 × (-1) + 40 = -1
∴ Answer is -11. If $0^{0} < 0 < 90^{0}$ and $2sin^{2}\theta + 3cos\theta$ = 3, then the value of $\theta$ is
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