How do you use half angle identities and the fact that sin(pi/6)=1/2 to find cos(-pi/12) and sin (-pi/12)?

Question

Victoria Adebayo · Accepted Answer

Please see below.

We have , #sin(pi/6)=sin30^circ=1/2# Now , #theta=pi/12=15^circto1^(st)Quadrant# #"Using"color(blue)" half angle formula :"# #color(blue)(sin^2 (theta/2)=(1-costheta)/2=(1-sqrt(1-sin^2theta))/2# Let , #theta=30^circ=>theta/2=15^circto1^(st)Quadrant# #:.sin^2 15^circ=(1-sqrt(1-sin^2 30^circ))/2# #:.sin^2 15^circ=(1-sqrt(1-1/4))/2# #:.sin^2 15^circ=(1-sqrt3/2)/2=(2-sqrt3)/4=(4-2sqrt3)/8# #:.sin^2 15^circ=(3+1-2sqrt3)/8#=#((sqrt3)^2-2(sqrt3)(sqrt1)+(sqrt1)^2)/8# #:.sin^2 15^circ=(sqrt3-1)^2/(4xx2)# #:.sin 15^circ=(sqrt3-1)/(2sqrt2)......to[because1^(st)Quadrant]# #:.sin 15^circ=(sqrt3-1)/(2sqrt2) xxsqrt2/sqrt2=(sqrt6-sqrt2)/4# #color(green)(-pi/12=-15^circtoIV^(th)Quadrant =>sin(-pi/12) < 0# #:.color(red)(sin(-pi/12)=-sin(pi/12)=-(sqrt6-sqrt2)/4# For #cos(-pi/12)# ,please see next answer.

Josiah Anderson · Answer

Please see below.

We have , #sin(pi/6)=sin30^circ=1/2# Now , #theta=pi/12=15^circto1^(st)Quadrant# #"Using"color(blue)" half angle formula :"# #color(blue)(cos^2 (theta/2)=(1+costheta)/2=(1+sqrt(1-sin^2theta))/2# Let , #theta=30^circ=>theta/2=15^circto1^(st)Quadrant# #:.cos^2 15^circ=(1+sqrt(1-sin^2 30^circ))/2# #:.cos^2 15^circ=(1+sqrt(1-1/4))/2# #:.cos^2 15^circ=(1+sqrt3/2)/2=(2+sqrt3)/4=(4+2sqrt3)/8# #:.cos^2 15^circ#=#(3+1+2sqrt3)/8#=#((sqrt3)^2+2(sqrt3)(sqrt1)+(sqrt1)^2)/8# #:.cos^2 15^circ=(sqrt3+1)^2/(4xx2)# #:.cos15^circ=(sqrt3+1)/(2sqrt2)......to[because1^(st)Quadrant]# #:.cos15^circ=(sqrt3+1)/(2sqrt2) xxsqrt2/sqrt2=(sqrt6+sqrt2)/4# #color(green)(-pi/12=-15^circtoIV^(th)Quadrant =>cos(-pi/12) > 0# #:.color(red)(cos(-pi/12)=+cos(pi/12)=(sqrt6+sqrt2)/4# For #sin(-pi/12)# , please see next answer.

Delilah Aguilar · Answer

undefined To find $ \cos(-\frac{\pi}{12}) $ and $ \sin(-\frac{\pi}{12}) $ using half-angle identities and the fact that $ \sin(\frac{\pi}{6}) = \frac{1}{2} $, first recognize that $ -\frac{\pi}{12} $ is half of $ \frac{\pi}{6} $. Therefore, we can use the half-angle identities:

$$ \cos(\frac{\theta}{2}) = \sqrt{\frac{1 + \cos(\theta)}{2}} $$
$$ \sin(\frac{\theta}{2}) = \sqrt{\frac{1 - \cos(\theta)}{2}} $$

Given $ \sin(\frac{\pi}{6}) = \frac{1}{2} $, we know $ \cos(\frac{\pi}{6}) = \sqrt{1 - (\frac{1}{2})^2} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} $.

Now, we can use these values to find $ \cos(-\frac{\pi}{12}) $ and $ \sin(-\frac{\pi}{12}) $. Since $ -\frac{\pi}{12} $ is negative, we use the following trigonometric identities:

$$ \cos(-\theta) = \cos(\theta) $$
$$ \sin(-\theta) = -\sin(\theta) $$

Substitute $ \theta = \frac{\pi}{6} $ into these identities:

$$ \cos(-\frac{\pi}{12}) = \cos(\frac{\pi}{12}) $$
$$ \sin(-\frac{\pi}{12}) = -\sin(\frac{\pi}{12}) $$

Then, apply the half-angle identities to $ \frac{\pi}{12} $:

$$ \cos(\frac{\pi}{12}) = \sqrt{\frac{1 + \cos(\frac{\pi}{6})}{2}} $$
$$ \sin(\frac{\pi}{12}) = \sqrt{\frac{1 - \cos(\frac{\pi}{6})}{2}} $$

Substitute the values we found earlier:

$$ \cos(\frac{\pi}{12}) = \sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}} $$
$$ \sin(\frac{\pi}{12}) = \sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}} $$