﻿ Find the exact value of sin 33\degree

#### Question

Find the exact value of sin 33\degree

Collected in the board: Trigonometry

Steven Zheng posted 1 year ago

Since 33 is the sum of 18 and 15,

33=18+15

the exact value of \sin 33\degree could be determined by the trigonometric values of 18\degree and 15\degree using the sum identity for sines function.

There are multiple ways to obtain the trig values of 18\degree and 15\degree. The exact value of \sin 18\degree could be determined by using golden triangle, an isosceles triangle with an vertex angle of 36\degree. Trig values of 15\degree could be determined by half angle identity or geometric method. Here are trig values we have obtained to derive the exact value of \sin 33\degree.

\sin18\degree =\dfrac{-1+\sqrt{5} }{4}

\cos 18\degree =\dfrac{\sqrt{10+2\sqrt{5} } }{4}

\sin 15\degree =\dfrac{\sqrt{6}-\sqrt{2}}{4}

\cos 15\degree =\dfrac{\sqrt{6}+\sqrt{2} }{4}

Using the sum identity for sines function,

\sin (\alpha +\beta ) = \sin \alpha\cos \beta +\cos \alpha \sin \beta

\sin 33\degree

= \sin (18° +15° )

= \sin 18°\cos 15° +\cos 18° \sin 15°

=\dfrac{-1+\sqrt{5} }{4} \cdotp \dfrac{\sqrt{6}+\sqrt{2}}{4} +\dfrac{\sqrt{10+2\sqrt{5}}}{4}\cdotp\dfrac{\sqrt{6}-\sqrt{2}}{4}

=\dfrac{1}{16}[(-1+\sqrt{5} )(\sqrt{6}+\sqrt{2})+(\sqrt{10+2\sqrt{5}})(\sqrt{6}-\sqrt{2})]

Using the excel formula to confirm the value

=((SQRT(5)-1)*(SQRT(6)+SQRT(2))+(SQRT(10+2*SQRT(5)))*(SQRT(6)-SQRT(2)))/16

the value of sin 33° is approximately 0.544394131

Steven Zheng posted 1 year ago

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