Express-sin-33-in-surd-form-

Question Number 12028 by tawa last updated on 09/Apr/17

Express : \sin (33) in surd form .

Answered by sandy_suhendra last updated on 10/Apr/17

let A = 18 °

5 A = 90 °

3 A = 90 ° - 2 A

\cos 3 A = \cos (90 ° - 2 A)

{4 \cos}^{3} A - 3 cosA = \sin 2 A

{4 \cos}^{3} A - 3 cosA = 2 sinAcosA

{4 \cos}^{2} A - 3 = 2 sinA

4 - {4 \sin}^{2} A - 3 = 2 sinA

{4 \sin}^{2} A + 2 sinA - 1 = 0

sinA = \frac{- 2 + \sqrt{4 + 4.4 (- 1)}}{2.4}

\sin 18 ° = \frac{- 2 + 2 \sqrt{5}}{8} = \frac{\sqrt{5} - 1}{4}

\cos 18 ° = \sqrt{1 - \sin^{2} 18 °}

= \sqrt{1 - {(\frac{\sqrt{5} - 1}{4})}^{2}}

= \sqrt{\frac{10 + 2 \sqrt{5}}{16}} = \frac{1}{4} \sqrt{10 + 2 \sqrt{5}}

\sin 15 ° = \sqrt{\frac{1 - \cos 30 °}{2}}

= \sqrt{\frac{1 - \frac{1}{2} \sqrt{3}}{2}}

= \sqrt{\frac{4 - 2 \sqrt{3}}{8}} = \sqrt{\frac{{(\sqrt{3} - 1)}^{2}}{8}}

= \frac{\sqrt{3} - 1}{2 \sqrt{2}}

\cos 15 ° = \sqrt{\frac{1 + \cos 30 °}{2}} = \frac{\sqrt{3} + 1}{2 \sqrt{2}}

\sin 33 °

= \sin (18 ° + 15 °)

= \sin 18 ° . \cos 15 ° + \cos 18 ° . \sin 15 °

= \frac{\sqrt{5} - 1}{4} \times \frac{\sqrt{3} + 1}{2 \sqrt{2}} + \frac{1}{4} \sqrt{10 + 2 \sqrt{5}} \times \frac{\sqrt{3} - 1}{2 \sqrt{2}}

= \frac{1}{8 \sqrt{2}} [(\sqrt{5} - 1) (\sqrt{3} + 1) + \sqrt{10 + 2 \sqrt{5}} . (\sqrt{3} - 1)]

Commented by tawa last updated on 10/Apr/17

Wow, God bless you sir .

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