Question

Find 2 cos(x/2) sin(x/2) when x = -π/6.

I know that cos(x/2) = +- sqrt(1-cosx)/2 and sin(x/2) = +- sqrt(1+cosx)/2, but how to use this identity I'm not entirely sure. Please help.

Answer 1

both are variations of the relation

<b>cos 2A = cos^2 A - sin^2 A
or
cos 2A = 2 cos^2 A - 1
or
cos 2A = 1 - 2sin^2 A</b>

We use this to either find trig ratios of double an angle or 1/2 an angle
e.g. if we know sin 50° , I could now find ratios for 25° or 100°

e.g. you should know the basic trig ratios of angles like 45°, 30° and 60°
but what is the sin 15° in exact form ?

cos 30 = 1 - 2sin^2 15
√3/2 = 1 - 2sin^2 15°
2 sin^2 15° = 1 - √3/2 = (2 - √3)/2
sin^2 15° = (2-√3)/4
sin 15° = ±√(2-√3)/2 , but I know 15° is in quad I
so sin 15° = √(2-√3)/2
( check this with our calculator)

if we replace A with x/2
in
cos 2A = 2 cos^2 A - 1
cos x = 2 cos^2 (x/2) - 1
2cos^2 (x/2) = 1 - cos x
cos^2 (x/2) = (1-cosx)/2
cos (x/2) = ±√( (1-cosx)/2 )
which is your first formula

notice that you need brackets around (1-cosx)/2 before you take the square root.

your second formula can be derived in a similar way from the third version of cos 2A

Answer 2

Hello...

2cosθsinθ = sin2θ

so, what's sin -π/3?

Answer 3

Hello ...
I mean sin -π/6

That is, because

2 cos θ/2 sin θ/2 = sinθ

Answer 4

Yup, I was so totally distracted by the second part of his post, that I completely missed and ignored the actual question.
Another senior moment!!!

but... wasn't that a great explanation??

reminds me of my teaching years, when during an exam a student replied to a certain question:
"I have absolutely no idea what the solution would be, but I know this: ", and then proceeded to write out all the verses of "Ode to a Grecian Urn"

Answer 5

Love it!