Help with finding the exact value?

Question

My math teacher gave us a lovely home assignment to finish on our own. I'm absolutely stuck on a couple. If you could point me in the right direction on one, I could probably get the others.
Could you help me with this, please?

sin(ArcCosine1/2+ArcSine3/5)

Answer 1

No doubt you've been given the formula
sin(a + b) = sina cosb + cosa sinb
Applying that here, the second term is easy because
cos(arcos1/2) = 1/2 and sin(arcsin3/5) = 3/5

If you sketch a rightangled triangle with hypotenuse 2 and adjacent side 1, the included angle is arcos1/2 and the other side is sqrt3, so you can see that sin(arcos1/2) is (sqrt3)/2. Similarly you can see that cos(arcsin3/5) is 4/5, and so you've finished.

Answer 2

I wish I was given fun questions like these when I first took math! Here we go.

sin(arccos(1/2) + arcsin(3/5))

Your first step would be to use the sine addition formula. It goes
sin(a + b) = sin(a)cos(b) + sin(b)cos(a)

Therefore

sin ( arccos(1/2) + arcsin(3/5) ) =
sin [ arccos(1/2) ] cos[ arcsin(3/5) ] +
sin[ arcsin(3/5) ] cos[ arccos(1/2) ]

Let's solve these one at a time.

First, to solve sin(arccos(1/2)), what you have to do is equate the inside to a variable, y.

Let y = arccos(1/2). Taking the cos of both side would leave the 1/2 on the right hand side, and we'd get

cos(y) = 1/2

At this point, what you need to do is draw a right angle triangle, with y as the angle. Note that cos = adj/hyp (by the SOHCAHTOA trig stuff), so draw a right angle triangle with an angle y, the adjacent side 1, and the hypotenuse 2. This makes the opposite side sqrt(2^2 - 1^2) = sqrt(3).

Since cos(y) = 1/2, we can solve for sin(y), since, after all
sin(y) = sin(arccos(1/2)) which is what we want to solve for
sin(y) = opp/hyp = sqrt(3) / 2

WHEW! That's our first value: sin(arccos(1/2)) = sqrt(3)/2

Next up, cos(arcsin(3/5)).
Let y = arcsin(3/5). Then
sin(y) = 3/5
[opp = 3, hyp = 5, adj = sqrt(5^2 - 3^2) = sqrt(16) = 4]
so cos(y) = cos(arcsin(3/5)) = adj/hyp = 4/5]

Second value down: cos(arcsin(3/5)) = 4/5

sin(arcsin(3/5)) = 3/5, and
cos(arccos(1/2)) = 1/2 {since they are inverses of each other}.

So we have:

sin ( arccos(1/2) + arcsin(3/5) ) =
sin [ arccos(1/2)] cos[arcsin(3/5) ] +
sin [ arcsin(3/5)] cos[arccos(1/2) ]

sin(arccos(1/2) + arcsin(3/5)) = [sqrt(3)/2] [4/5] + [3/5][1/2]
= 2sqrt(3)/5 + 3/10

If we wanted we can put this under a common denominator, 10. Yeah, let's do that. Make our solution nice and neat.

= [4sqrt(3) + 3]/10

Answer 3

arc cos 1/2=60°
sin A+B=sin A cos B+cos A sin B
sin(60°+arcsin 3/5)=sin 60° cos arcsin 3/5 +cos 60° sin arcsin 3/5

arcsin 3/5: this is a 3, 4, 5 right triangle so arc sin 3/5=arc cos 4/5

sin (60°+arcsin 3/5)=√3/2 cos arccos 4/5+1/2* 3/5
=√3/2 *4/5+.3=.3+2√3/5=sin (arccos1/2 arcsin 3/5)

Answer 4

If we express each part by the Taylor series of order 5, we have:

arccos(x) = -3x^5 - x^3 - x + pi
.................... 40...... 6.......... 2

arcsin(x) = .3x^5 + x^3 + x
.................... 40....6

evaluating arccos with x=1/2 and arcsin with x=3/5

arccos(1/2) = -3*(1/2)^5 - (1/2)^3 - (1/2) + pi
........................ 40........... 6................. 2

arcsin(3/5) = .3(3/5)^5 + (3/5)^3 + (3/5)
......................... 40.........6

arccos(1/2) = pi/2 - 2009/3840

arcsin(3/5) = 80229/12500

the sum of both is:

arccos(1/2) + arcsin(3/5)= pi/2 - 2009/3840 + 80229/12500

arccos(1/2) + arcsin(3/5)= pi/2 + 1423859/12000000

the taylor series of the sin(x) = x^5/120 - x^3/6 + x

sin(pi/2+1423859/12000000) = ...

this is a way to express it, but he numbers will make too big.
maybe this can help you, the sin taylor series of the expresion give an exact result, I don´t know if thi is the way you need, I can help you in something

Answer 5

First let's simplify.

arccos(1/2) = π/3

Now let's solve for cos x when sin x = 3/5.
cos x = √(1 - sin² x) = √(1 - (3/5)²) = √(1 - 9/25)
cos x = √(16/25) = 4/5

Now we are ready to apply the angle addition formula for sine.

sin(α+β) = (sin α)(cos β) + (cos α)(sin β)

Plugging in for α and β we have:

sin[arccos(1/2)+arcsin(3/5)] = sin[π/3 + arcsin(3/5)]
= (sin π/3)(cos arcsin(3/5)) + (cos π/3)(sin arcsin(3/5))
= (sin π/3)(cos arccos(4/5)) + (cos π/3)(sin arcsin(3/5))
= ((√3)/2)(4/5) + (1/2)(3/5) = (4√3 + 3)/10

Answer 6

remember: sin(A+B)=sinAcosB+sinBcosA. therefor:
sin(ArcCos1/2+ArcSin3/5)=Sin(ArcCos1/2)Cos(ArcSin3/5)+Sin(ArcSin3/5)Cos(ArcCos1/2)=Sin(ArcCos1/2)Cos(ArcSin3/5)+3/5*1/2=Sin(ArcCos1/2)Cos(ArcSin3/5)+3/10.
Now:

(SinA)(SinA)+(CosA)(CosA)=1
Sin(90-A)=CosA
Cos(90-A)=SinA

So:
Sin(ArcCos1/2)=SQRTE(1-(1/2)*(1/2))=SQRTE(3/4)=SQRTE(3)/2

Cos(ArcSin3/5)=SQRTE(1-(3/5)*(3/5))=SQRTE(16/25)=4/5

And we get:
SQRTE(3)/2*(4/5)+3/10=(4SQRTE(3)+3)/10

Hope I've helped.
seems like kinda mistake with the 1/2 she gave.

Answer 7

sin(arccos1/2+arcsin3/5)
sinarccos1/2cosarcsin3/5+cosarccos1/2sinarcsin3/5
=sinarcsinrt3/2cosarc4/5+1/2*3/5
=rt3/2*4/5+3/10
=(4rt3+3)/10