Note: Law of Sines
cos11π/12 cos(x/2) = ± √((1+cosx)/2) we use this Half-Angle Formula to evaluate it.
Since in this question, 11π/12 is in the Quadrant II, so cos(11π/12) should be negative. cos11π/12 = – √((1+cos(11π/6))/2) = – √((1+(√3/2))/2)
to simplify it, both sides multiply by 2, – √(((1+(√3/2))(2))/((2)(2))) = – √((2+(√3))/4)
You should know what cos(11π/6) is, and you just plug in the number and you should get the answer.
(√3 – i)-10
We are using De Moivre’s Theorem to solve this problem.
De Moivre’s Theorem:
If z=r(cosθ + i sinθ), then for any integer n, zn=rn(cos(nθ) + i sin(nθ)).
So , we have z = √3 – i, and we would like to evaluate z-10 = (√3 – i)-10.
First, we need to express z = (√3 – i) into polar form. r = √(〖(√3)〗^2+1^2 )=2 tanθ = -1/√3 θ = 5π/6
So, z=2(cos(5π/6) + i sin(5π/6))
Apply De Moivre’s Theorem, z-10 = (√3 – i)-10 =2-10 (cos(10*5π/6) + i sin(10*5π/6)) = ......
And, I think you should be able to get the answer for (√3 – i)-10.
19. Sketch triangle and find other five ratios of θ. sin θ = 3/5
Step 1, we sketch a right triangle to specify the angle θ.
Step 2, since we are given sin θ = 3/5, mark the opposite and hypotenuse as the graph below.
Step 3, find out the missing length by using Pathagorean Theorem. 32 + adjacent2 = 52 adjacent = 4.
Step 4, we can have all the other five trigonometric ratios: sin θ = 3/5 Csc θ = 5/3 cos θ = 4/5 sec θ = 5/4
4sinƟcosƟ + 2√3 sinƟ – 2√3 cosƟ – 3 = 0
(4sinƟcosƟ + 2√3 sinƟ) – (2√3 cosƟ + 3) = 0 make it into two groups. Attention! Be careful with your sign!
2sinƟ(2cosƟ+√3) – √3(2cosƟ+√3) = 0 for each group, factor out the common factor
(2cosƟ+√3)(2sinƟ – √3) = 0 we have (2cosƟ+√3) as my common factor, so I can factor it out.
2cosƟ+√3 = 0, 2sinƟ – √3 = 0 cosƟ = – √3/2, sinƟ = √3/2
Then, you can draw the unit circle to find the solution for Ɵ.
Since in this question, 11π/12 is in the Quadrant II, so cos(11π/12) should be negative. cos11π/12 = – √((1+cos(11π/6))/2) = – √((1+(√3/2))/2)
to simplify it, both sides multiply by 2, – √(((1+(√3/2))(2))/((2)(2))) = – √((2+(√3))/4)
You should know what cos(11π/6) is, and you just plug in the number and you should get the answer.
(√3 – i)-10
We are using De Moivre’s Theorem to solve this problem.
De Moivre’s Theorem:
If z=r(cosθ + i sinθ), then for any integer n, zn=rn(cos(nθ) + i sin(nθ)).
So , we have z = √3 – i, and we would like to evaluate z-10 = (√3 – i)-10.
First, we need to express z = (√3 – i) into polar form. r = √(〖(√3)〗^2+1^2 )=2 tanθ = -1/√3 θ = 5π/6
So, z=2(cos(5π/6) + i sin(5π/6))
Apply De Moivre’s Theorem, z-10 = (√3 – i)-10 =2-10 (cos(10*5π/6) + i sin(10*5π/6)) = ......
And, I think you should be able to get the answer for (√3 – i)-10.
19. Sketch triangle and find other five ratios of θ. sin θ = 3/5
Step 1, we sketch a right triangle to specify the angle θ.
Step 2, since we are given sin θ = 3/5, mark the opposite and hypotenuse as the graph below.
Step 3, find out the missing length by using Pathagorean Theorem. 32 + adjacent2 = 52 adjacent = 4.
Step 4, we can have all the other five trigonometric ratios: sin θ = 3/5 Csc θ = 5/3 cos θ = 4/5 sec θ = 5/4
4sinƟcosƟ + 2√3 sinƟ – 2√3 cosƟ – 3 = 0
(4sinƟcosƟ + 2√3 sinƟ) – (2√3 cosƟ + 3) = 0 make it into two groups. Attention! Be careful with your sign!
2sinƟ(2cosƟ+√3) – √3(2cosƟ+√3) = 0 for each group, factor out the common factor
(2cosƟ+√3)(2sinƟ – √3) = 0 we have (2cosƟ+√3) as my common factor, so I can factor it out.
2cosƟ+√3 = 0, 2sinƟ – √3 = 0 cosƟ = – √3/2, sinƟ = √3/2
Then, you can draw the unit circle to find the solution for Ɵ.