how to take out sin(15) degree explaination plz. with formulae sin^2tita = 1- cos2tita upon 2
Hey !
2sin ^2( theta)= 1- cos (2 theta)
let theta=15
then,
2sin ^2(15)= 1- cos (2 x 15)
sin ^2(15)= (1- cos (30) )/2
= (1- (root(3)/2))/2
sin (15) =root ( (1- ( root(3)/2) ) / 2 )
= ( root(6) - root(2) )/4
= 0.25
.
This is a very easy question. You just need to put theta value on both side. Put theta=15 and solve the value of sin
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Hello,
I found: sin(15°)=0.258
Explanation:
Using the Formula given in question:
sin2(x)=[1−cos(2x)] /2
with x=15° and 2x=30°
you get:
sin2(15°)=[1−cos(30°)]/2
knowing that: cos(30°)=√3/2 :
sin2(15°)=[1−√3 / 2]/2
sin2(15°)=(2−√3) 4=0.067
So:
sin(15°)=±√0.067=±0.258
We choose the positive one.
Hope you will understand :)
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