If sin(a) = 3/5, find the value of cos(a) and sin2a?
回答 (10)
2018-04-30 8:40 pm
sinα=3/5
There are two values of a in the trigonometric circumference for which sina = 3/5.
The first belongs to the 1st quadrant
The second belongs to the 2nd quadrant
1. Figure out cosine.
cos²α=1-sin²α
cosα=±√(1-(3/5)²=±4/5
Note:
i) cosα=4/5. in this case α belongs to the 1st quadrant
ii) cosα=-4/5. in this case α belongs to the 2nd quadrant
2. We proceed to calculate sin(2α) in the two cases
i) sin(2α)=2*sinα*cosα=2*(3/5)*(4/5)=
=24/25
ii) sin(2α)=2*sinα*cosα=2*(3/5)*(-4/5)=
=-24/25
There are two values of a in the trigonometric circumference for which sina = 3/5.
The first belongs to the 1st quadrant
The second belongs to the 2nd quadrant
1. Figure out cosine.
cos²α=1-sin²α
cosα=±√(1-(3/5)²=±4/5
Note:
i) cosα=4/5. in this case α belongs to the 1st quadrant
ii) cosα=-4/5. in this case α belongs to the 2nd quadrant
2. We proceed to calculate sin(2α) in the two cases
i) sin(2α)=2*sinα*cosα=2*(3/5)*(4/5)=
=24/25
ii) sin(2α)=2*sinα*cosα=2*(3/5)*(-4/5)=
=-24/25
2018-04-30 8:21 pm
sin(a) = opp/hyp (in a right triangle)
opp = 3
hyp = 5
adj = sqrt(5^2-3^2) = sqrt(25-9) = sqrt(16) = 4
cos(a) = adj/hyp = 4/5
sin 2a = 2 sin(a) cos(a) = (2)(3/5)(4/5) = 24/25
opp = 3
hyp = 5
adj = sqrt(5^2-3^2) = sqrt(25-9) = sqrt(16) = 4
cos(a) = adj/hyp = 4/5
sin 2a = 2 sin(a) cos(a) = (2)(3/5)(4/5) = 24/25
2018-05-02 7:40 am
sin a = opp/hyp
Using the Pythagorean Theorem, we find the adjacent side to be 4.
This means that cos a = adj/hyp.
So, cos a = 4/5.
You know that sin 2a = 2 sin a cos a.
Plug what you know into the trig identity.
sin 2a = 2 (3/5)(4/5) = 24/25
sin 2a = 24/25
Using the Pythagorean Theorem, we find the adjacent side to be 4.
This means that cos a = adj/hyp.
So, cos a = 4/5.
You know that sin 2a = 2 sin a cos a.
Plug what you know into the trig identity.
sin 2a = 2 (3/5)(4/5) = 24/25
sin 2a = 24/25
2018-05-01 2:41 am
cos (a) = 4/5
sin 2a = 2 sin a cos a
sin 2a = 2 (3/5)(4/5) = 24/25
sin 2a = 2 sin a cos a
sin 2a = 2 (3/5)(4/5) = 24/25
2018-04-30 11:41 pm
Sin(A) = o / h = 3/5
Hence by Pythagoras
5^2 = 3^2 + a^2
a^2 = 5^2 - 3^2
Factor
a^2 = (5 - 3)(5 + 3)
a^2 = 2(8) = 16
a = sqrt(16) = 4 The adjacent side
Hence Cos(A) = a/h = 4/5
Using the Trig. Identity
Sin(2A) = Sin(A)Cos(A) + Cos(A)Sin(A) = 2Sin(A)Cos(A)
Substitute
Sin(2A) = 2 x 3/5 x 4/5
Sin(2A) = 24/25
Hence by Pythagoras
5^2 = 3^2 + a^2
a^2 = 5^2 - 3^2
Factor
a^2 = (5 - 3)(5 + 3)
a^2 = 2(8) = 16
a = sqrt(16) = 4 The adjacent side
Hence Cos(A) = a/h = 4/5
Using the Trig. Identity
Sin(2A) = Sin(A)Cos(A) + Cos(A)Sin(A) = 2Sin(A)Cos(A)
Substitute
Sin(2A) = 2 x 3/5 x 4/5
Sin(2A) = 24/25
2018-04-30 9:30 pm
sin(a) = 3/5
cos(a) = 4/5
sin(2a) = 24/25
cos(a) = 4/5
sin(2a) = 24/25
2018-04-30 8:51 pm
cos(a) = ±√(5²-3²)/5 = ±⅘ (+ve if a in Q1, -ve if in Q2)
sin²(a) = (3/5)² = 9/25
or (given the ambiguity)
sin(2a) = 2 × sin(a) × cos(a) = ±24/25 (again, +ve if a in Q1, -ve if in Q2)
sin²(a) = (3/5)² = 9/25
or (given the ambiguity)
sin(2a) = 2 × sin(a) × cos(a) = ±24/25 (again, +ve if a in Q1, -ve if in Q2)
2018-04-30 8:51 pm
sin^2(a) + cos^2(a) = 1
for all angles
(3/5)^2 + (?)^2 = 1
(?)^2 = 1 - 9/25 = 16/25
The thing about squares, is that they are laways positive, even if the root is negative.
4 = (+2)^2
4 = (-2)^2
Therefore, the cosine (?) could be
4/5 or -4/5
both are good.
In the first case, the angle would be in the first quadrant (sine is positive AND cosine is positive)
In the second case, the angle would be in the 2nd quadrat (sine is positive, cosine is negative).
-----
sin(2a) = 2 sin(a)cos(a)
in the first quadrant:
sin(2a) = 2 [ (3/5)(4/5) ] = 2[12/25] = 24/25
check:
if sin(a) = 3/5 in the first quadrant, then a = 0.6435011... radians (approx. 36.87 deg.)
then
sin(2a) = sin(1.2870022) = 0.96 = 24/25 = sin(73.74 deg.)
You do the 2nd quadrant
for all angles
(3/5)^2 + (?)^2 = 1
(?)^2 = 1 - 9/25 = 16/25
The thing about squares, is that they are laways positive, even if the root is negative.
4 = (+2)^2
4 = (-2)^2
Therefore, the cosine (?) could be
4/5 or -4/5
both are good.
In the first case, the angle would be in the first quadrant (sine is positive AND cosine is positive)
In the second case, the angle would be in the 2nd quadrat (sine is positive, cosine is negative).
-----
sin(2a) = 2 sin(a)cos(a)
in the first quadrant:
sin(2a) = 2 [ (3/5)(4/5) ] = 2[12/25] = 24/25
check:
if sin(a) = 3/5 in the first quadrant, then a = 0.6435011... radians (approx. 36.87 deg.)
then
sin(2a) = sin(1.2870022) = 0.96 = 24/25 = sin(73.74 deg.)
You do the 2nd quadrant
2018-04-30 8:38 pm
By Pythagoras:
sin^2(a) + cos^2(a) = 1
(3/5)^2 + cos^2(a) = 1
(9/25) + cos^2(a) = 1
cos^2(a) = 1 - (9/25)
cos^2(a) = 16/25
cos(a) = +/- sqrt(16/25)
cos(a) = +/- 4/5
cos(a) = +/- 0.8
sin(2a) = 2*sin(a)*cos(a)
sin(2a) = 2 * (3/5) * (+/- 4/5)
sin(2a) = +/- 24/25
sin(2a) = +/- 0.96
sin^2(a) + cos^2(a) = 1
(3/5)^2 + cos^2(a) = 1
(9/25) + cos^2(a) = 1
cos^2(a) = 1 - (9/25)
cos^2(a) = 16/25
cos(a) = +/- sqrt(16/25)
cos(a) = +/- 4/5
cos(a) = +/- 0.8
sin(2a) = 2*sin(a)*cos(a)
sin(2a) = 2 * (3/5) * (+/- 4/5)
sin(2a) = +/- 24/25
sin(2a) = +/- 0.96
2018-04-30 8:18 pm
a = sin^-1(3/5) = 36.87
cos(36.87) = 0.8
sin(2*36.87) = 0.96
cos(36.87) = 0.8
sin(2*36.87) = 0.96
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