Respuesta :
[tex]\bf \textit{Half-Angle Identities} \\\\ sin\left(\cfrac{\theta}{2}\right)=\pm \sqrt{\cfrac{1-cos(\theta)}{2}} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ 75\cdot 2\implies 150\qquad \qquad \qquad \cfrac{150}{2}\implies 75 \\\\[-0.35em] ~\dotfill\\\\ sin\left( \cfrac{150^o}{2} \right)=\pm\sqrt{\cfrac{1-cos(150^o)}{2}}\implies sin\left( \cfrac{150^o}{2} \right)=\pm\sqrt{\cfrac{1-\left( -\frac{\sqrt{3}}{2} \right)}{2}}[/tex]
[tex]\bf sin\left( \cfrac{150^o}{2} \right)=\pm\sqrt{\cfrac{1+\frac{\sqrt{3}}{2}}{2}}\implies sin\left( \cfrac{150^o}{2} \right)=\pm\sqrt{\cfrac{\frac{2+\sqrt{3}}{2}}{2}} \\\\\\ sin\left( \cfrac{150^o}{2} \right)=\pm\sqrt{\cfrac{2+\sqrt{3}}{4}}\implies sin\left( \cfrac{150^o}{2} \right)=\pm\cfrac{\sqrt{2+\sqrt{3}}}{\sqrt{4}} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill sin\left( \cfrac{150^o}{2} \right)=\pm\cfrac{\sqrt{2+\sqrt{3}}}{2}~\hfill[/tex]
The value of sin75 by using half angle identity is [tex]sin\dfrac{[150^0]}{[22]}=\pm\dfrac{\sqrt{2+\sqrt{3}}}{2}[/tex]
What is trigonometry?
Trigonometry is the branch of mathematics that set up a relationship between the sides and angle of the right-angle triangles.
Value of sin75 by using half angle identity is calculated as:-
Half angle identity for is,
[tex]sin\dfrac{\theta}{2}=\pm\sqrt{\dfrac{1-cos\theta}{2}[/tex]
[tex]\dfrac{\theta}{2}=\dfrac{150}{2}[/tex]
[tex]sin\dfrac{150}{2}=\pm\sqrt{\dfrac{1-cos150}{2}[/tex]
[tex]sin \dfrac{150}{2}=\pm\sqrt{\dfrac{{1+\dfrac{\sqrt{3}}{2}}}{2}[/tex]
[tex]sin \dfrac{150}{2}=\pm\sqrt{\dfrac{2+\sqrt{3}}{4}[/tex]
[tex]sin\dfrac{[150^0]}{[22]}=\pm\dfrac{\sqrt{2+\sqrt{3}}}{2}[/tex]
Hence, the value of sin75 is [tex]sin\dfrac{[150^0]}{[22]}=\pm\dfrac{\sqrt{2+\sqrt{3}}}{2}[/tex].
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