Question

Use a half–angle identity to find the exact value of sin 105°

Answer 1

sin² u = (1 - cos(2u))/2

sin² 105º = (1 - cos(210º))/2

. . . . . . . . 1 - (-√(3)/2)
sin² 105º = ------------------
. . . . . . . . . . . 2

. . . . . . . . 1 +√(3)/2
sin² 105º = --------------
. . . . . . . . . . . 2

. . . . . . . . 2 +√(3)
sin² 105º = ----------- = .9330
. . . . . . . . . . . 4

sin² 105º = .9330

sin 105º = .9659 <==ANSWER

Answer 2

Half Angle Identity for Sine:\[\sin{u\over2}=\pm{\sqrt{1-\cos u}\over2}\]

Answer 3

where \(u\) is the angle

Answer 4

In my attempt to solve this using the half angle identity I got stuck at \[-\sqrt{(1+7\pi/6)}\]over 2

Answer 5

\[\sin105=\pm\sqrt{1-\cos210\over2}\]\[=\pm\sqrt{1-(\cos180\cos30-sin180\sin30)\over2}\]\[=\pm\sqrt{{1-\left(-{\sqrt{3}\over2}\right)-\left(0\right)}\over2}\]\[=\pm\sqrt{1+\sqrt{3}\over2}\]\[\sin105=\pm0.9659258263\]

Answer 6

thank you! My book might be wrong but it says that the answer is \[\frac{ \sqrt{2+\sqrt{3}} }{ 2 }\]

Answer 7

actually, that's correct as well:\[{=\pm\sqrt{1+{\sqrt{3}\over2}\over2}}~~~\implies~~~{=\pm\sqrt{{2\over4}+{\sqrt{3}\over4}}}\]\[{=\pm\sqrt{2+\sqrt{3}\over4}}~~~\implies~~~{=\pm{\sqrt{2+\sqrt{3}}\over2}}\]

Answer 8

oh im sorry my last step on my first post was supposed to be the square root of 1 plus the square root of 3 over 2, not just the square root of the 3, my bad..but yes your book is correct ad i've shown how :)

Answer 9

Thank you very much!

Answer 10

no problem :)