Key to "Show that−2≤cos⁡θ(sin⁡θ+sin2θ+3)≤2for all value of θ."

Treboldiqtw

Answered question

2022-05-03

Show that

- 2 \leq \cos θ (\sin θ + \sqrt{\sin^{2} θ + 3}) \leq 2

for all value of

θ

Camryn Diaz

Beginner2022-05-04Added 14 answers

Let

\cos θ (\sin θ + \sqrt{\sin^{2} θ + 3}) = y

\Rightarrow \sin θ + \sqrt{\sin^{2} θ + 3} = y \sec θ

\Rightarrow \sqrt{\sin^{2} θ + 3} = y \sec θ - \sin θ

Squaring we get

\sin^{2} θ + 3 = y^{2} (1 = \tan^{2} θ) + \sin^{2} θ - 2 y \tan θ

\Rightarrow y^{2} (\tan^{2} θ) - 2 y (\tan θ) + y^{3} - 3 = 0

\tan θ

is real, the discriminant must be

\geq 0 \Rightarrow (y - 2) (y + 2) \leq 0

Landyn Whitney

Beginner2022-05-05Added 19 answers

My Solution: Given

f (θ) = \cos θ (\sin θ + \sqrt{\sin^{2} θ + 3})

Now let

y = \sin θ \cdot \cos θ + \cos θ \cdot \sqrt{\sin^{2} θ + 3}

Now using the Cauchy-Schwarz inequality, we get

(\sin^{2} θ + \cos^{2} θ) \cdot [\cos^{2} θ + {(\sin^{2} θ + 3)}^{2}] \geq {[\sin θ \cdot \cos θ + \cos θ \cdot \sqrt{\sin^{2} θ + 3}]}^{2}

So we get

y^{2} \leq (\sin^{2} θ + \cos^{2} θ) \cdot [\cos^{2} θ + \sin^{2} θ + 3] = 2^{2}

So we get

- 2 \leq y \leq 2 \Rightarrow y \in [- 2, 2]

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