# Determine the angle a for arcsina + arcos1/2 = pi.

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arc sina+ arxco(1/2) = pi.

We know that sin(A+B) = sinAcosB+cosAsinB .

Therfore taking sine of both sides ofthe given equation, we get:

sin (arcsina) cos(arccos(1/2) +cos(arc sina) sin( arc cos(1/2) ) = sin pi = 0.

a* (1/2) - sqrt(1-a^2)sqrt{1-(1/2)^2} = 0.

a/2 +sqrt(1-a^2)*(sqrt3/2) = 0

a /2= -(1/2)sqrt(1-a^2)*sqrt3

a = -sqrt(1-a^2)*sqrt3

a^2 = 3(1-a^2)

a^2 = 3-3a^2

4a^2=3

a^2 = 3/4

a = +sqrt3/2. Or a = -sqrt3/2

Therefore angles are : Therfore arc sqrt sinsqrt3/2 = 120 and arc cos(1/2) = 60.

Also a = 240 degree and arc cos (1/2) = -60 deg

We'll re-write arccos (1/2) = pi/3

We'll re-write the equation:

arcsina + pi/3 = pi

We'll subtract pi/3 both sides:

arcsin a = pi - pi/3

arcsin a = (3pi-pi)/3

arcsin a = 2pi/3

We'll take the function sine both side.

sin(arcsin a) = sin 2pi/3

According to the rule, sin(arcsin a) = a. Based on the rule, we'll have sin(arcsin a) = a.

a = sin 2(pi/3)

sin 2(pi/3) = 2sin(pi/3)*cos(pi/3)

sin pi/3 = sqrt3/2

cos pi/3 = 1/2

We'll substitute the values of the functions sine and cosine in the formula 2sin(pi/3)*cos(pi/3) and we'll get:

sin 2(pi/3) = 2sqrt3/2*2

We'll simplify and we'll get:

sin 2(pi/3) = sqrt3/2

**So, a = sqrt3/2.**