If sinθ + cosθ = m and sinθ·cosθ = n, then m² equals:

1 + 2n. m² = (sinθ + cosθ)² = sin²θ + cos²θ + 2sinθ·cosθ = 1 + 2n. The variant 1 − 2n belongs to (sinθ − cosθ)².

If sinθ + cosθ = $m$ and sinθ·cosθ = $n$ , then $m^{2}$ equals:

Competition practiceTrigonometry

m^{2} = (s in θ + cos θ)^{2} = s i

n^{2}

θ +

co

s^{2}

θ + 2sinθ·cosθ =

1 + 2 n

. The variant

1 - 2 n

belongs to

(s in θ - cos θ)^{2}

Answer: B)

1 + 2 n

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