The equality I'm trying to prove looks like that:tan⁡40∘+3=4sin⁡40∘

Let's take LHS as followsLHS=tan⁡40∘+3=tan⁡40∘+tan⁡60∘=sin⁡40∘cos⁡40∘+sin⁡60∘cos⁡60∘=sin⁡40∘cos⁡60∘+cos⁡40∘sin⁡60∘cos⁡40∘cos⁡60∘=sin⁡(40∘+60∘)cos⁡(90∘−50∘)(12)=2×2sin⁡50∘cos⁡50∘sin⁡50∘=4cos⁡50∘=4sin⁡40∘=RHS

Resolved: "The equality I'm trying to prove looks like..."

Marzadri9lyy

Answered question

2022-04-02

The equality I'm trying to prove looks like that:

{\tan 40}^{\circ} + \sqrt{3} = 4 {\sin 40}^{\circ}

Answer & Explanation

Cason Singleton

Beginner2022-04-03Added 13 answers

Let's take LHS as follows

L H S = {\tan 40}^{\circ} + \sqrt{3} = {\tan 40}^{\circ} + {\tan 60}^{\circ}

= \frac{{\sin 40}^{\circ}}{{\cos 40}^{\circ}} + \frac{{\sin 60}^{\circ}}{{\cos 60}^{\circ}}

= \frac{{\sin 40}^{\circ} {\cos 60}^{\circ} + {\cos 40}^{\circ} {\sin 60}^{\circ}}{{\cos 40}^{\circ} {\cos 60}^{\circ}}

= \frac{\sin (40^{\circ} + 60^{\circ})}{\cos (90^{\circ} - 50^{\circ}) (\frac{1}{2})}

= \frac{2 \times 2 {\sin 50}^{\circ} {\cos 50}^{\circ}}{{\sin 50}^{\circ}}

= 4 {\cos 50}^{\circ}

= 4 {\sin 40}^{\circ} = R H S

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