Question


Answer 1

105\degree is not special angle whose trig value can be determined at first glance.

Since the angle 105\degree is in the 2nd quadrant, the sine value of 105\degree is equal to that of its supplementary angle, that is,

\sin 105\degree =\sin (180\degree -75\degree ) =\sin 75\degree

Now the question has been converted to determine the sines value of an acute angle.

Furthermore, 75\degree is found to be the sum of 45\degree and 30\degree whose trig values are known values. By using sum identity for sines function,

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

the final value can be determined.


\sin 105\degree =\sin (180\degree -75\degree ) =\sin 75\degree

=\sin(45\degree +30\degree )

=\sin 45\degree \cos 30\degree +\cos 45\degree \sin 30\degree

=\dfrac{\sqrt{2} }{2} \cdotp \dfrac{\sqrt{3} }{2} +\dfrac{\sqrt{2} }{2} \cdotp \dfrac{1}{2}

=\dfrac{\sqrt{6}+\sqrt{2} }{4}

Now we have determined the exact value of \sin 105\degree

Steven Zheng posted 1 month ago


Answer 2

Since the angle of 105\degree can be expressed as 90\degree +15\degree, \sin 105\degree could be transformed to \cos 15\degree by using cofunction identity and symmetry identity.

\sin (90\degree +15\degree)

=\sin (90\degree -(-15\degree))

=\cos(-15\degree)

=\cos 15\degree

It's obvious that 15\degree is the half of 30\degree. \cos 15\degree can be further determined by using half angle identity.

\cos \dfrac{\alpha }{2}=\pm\sqrt{\dfrac{1+\cos \alpha }{2} }

Since \cos 15\degree is greater than 0, the negative value is cancelled.

\cos 15\degree=\sqrt{\dfrac{1+\cos30\degree }{2} }

=\sqrt{\dfrac{1+\dfrac{\sqrt{3} }{2} }{2} }

=\dfrac{\sqrt{2+\sqrt{3}} }{4}

=\dfrac{\sqrt{(\sqrt{6}+\sqrt{2} )^2 } }{4}

=\dfrac{\sqrt{6}+\sqrt{2} }{4}

So the value of \sin 105\degree is \dfrac{\sqrt{6}+\sqrt{2} }{4}


Steven Zheng posted 1 week ago

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