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.
both are variations of the relation
cos 2A = cos^2 A - sin^2 A
or
cos 2A = 2 cos^2 A - 1
or
cos 2A = 1 - 2sin^2 A
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
Hello...
2cosθsinθ = sin2θ
so, what's sin -π/3?
Hello ...
I mean sin -π/6
That is, because
2 cos θ/2 sin θ/2 = sinθ
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"
Love it!
To find the value of 2cos(x/2)sin(x/2) when x = -π/6, we can substitute the given value of x into the expressions for cos(x/2) and sin(x/2), and then simplify the expression.
First, let's find cos(x/2) and sin(x/2) individually. We know that cos(x/2) = ±√(1+cosx)/2. So, substituting x = -π/6 into this identity, we have:
cos(-π/12) = ±√(1+cos(-π/6))/2
To find cos(-π/12), we can use the fact that cos is an even function, which means cos(-θ) = cos(θ). So, cos(-π/12) = cos(π/12).
Now, to find cos(π/12), we can use the unit circle or a calculator. By evaluating cos(π/12), we find that it is approximately 0.9659258263.
Substituting this value into the expression for cos(x/2), we have:
cos(x/2) = ±√(1+0.9659258263)/2
Simplifying further:
cos(x/2) = ±√(1.9659258263)/2
cos(x/2) = ±√0.9829629131
cos(x/2) = ±0.99145
Next, let's find sin(x/2) using the identity sin(x/2) = ±√(1-cosx)/2. Substituting x = -π/6, we have:
sin(-π/12) = ±√(1-cos(-π/6))/2
Again, using the fact that sin is an odd function, sin(-θ) = -sin(θ), we can rewrite sin(-π/12) as -sin(π/12).
To find sin(π/12), we can use the unit circle or a calculator. By evaluating sin(π/12), we find that it is approximately 0.2588190451.
Substituting this value into the expression for sin(x/2), we have:
sin(x/2) = ±√(1-0.9659258263)/2
Simplifying further:
sin(x/2) = ±√(1-0.9659258263)/2
sin(x/2) = ±√0.0340741737
sin(x/2) = ±0.18408
Finally, we can find 2cos(x/2)sin(x/2) by multiplying the expressions for cos(x/2) and sin(x/2):
2cos(x/2)sin(x/2) = 2 * 0.99145 * 0.18408
Evaluating this expression gives approximately 0.3649824.
Therefore, 2cos(x/2)sin(x/2) when x = -π/6 is approximately 0.3649824.