# How do I use a half-angle formula to find the exact value of 5π/12?

It's supposed to be sin 5π/12 and I am so sorry for that mistake

### 5 Answers

- IndikosLv 72 weeks agoBest Answer
We use the identity sin^2 (x) = ( 1 - cos(2x))/2

Since sin^2 (5π/12) = ( 1 - cos(2 * 5π/12 ) / 2

= ( 1 - cos(5π/6) / 2

= ( 1 - cos( π - π / 6))/2

= ( 1 - ( - cos(π /6) )/2

= ( 1 + √3 / 2 ) / 2

= ( 1/2 + √3 / 4)

So sin( 5π/12) = √ ( 1/2 + √3 / 4)

we reject -ve value since we are in first quadrant

We can now get cos ( 5π/12 ) using sin^2 u + cos^2 u = 1

then we can get tan(5π/12) ( = sin(5π/12) / cos( 5π/12 ) )

- MorningfoxLv 72 weeks ago
The exact value of 5π /12 is (5/12) pi. Are you looking for the sine, cosine, and tangent of that angle?

- 2 weeks ago
sin(5pi/12) =

sin((5pi/6)/2) =

sqrt((1/2) * (1 - cos(5pi/6)))cos(5pi/12) = cos((5pi/6)/2) = sqrt((1/2) * (1 + cos(5pi/6)))tan(5pi/12) = sin(5pi/12)/cos(5pi/12) =

sqrt((1/2) * (1 - cos(5pi/6))) / sqrt((1/2) * (1 + cos(5pi/6))) =

sqrt((1 - cos(5pi/6)) / (1 + cos(5pi/6))) =

sqrt((1 - cos(5pi/6))^2 / (1 - cos(5pi/6)^2)) =

(1 - cos(5pi/6)) / sqrt(sin(5pi/6))^2) =

(1 - cos(5pi/6)) / sin(5pi/6)

csc(5pi/12) = 1/sin(5pi/12)

sec(5pi/12) = 1/cos(5pi/12)

cot(5pi/12) = sin(5pi/6) / (1 - cos(5pi/6))

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