if cos theta=4/5, 0</=theta=<pi/2, find sin theta?

Question

(4/5)^2+sin^2=1
16/25+sin^2=1
sin^2=9/25
sin=3/5, or pi/300

i know that's not right though because of the range, and also it just doesn't sound right...

Answer 1

Given
cos θ= 4/5
0 ≤ θ ≤ π/2

Find
sin θ = ?

---------
The given statement
0 ≤ θ ≤ π/2
simply means that θ is in the 1st quadrant and sin θ is positive.

We know that when given one of the trigonometric functions of θ, then there is a way of finding the 5 remaining trig functions of θ.

We also know that:
cos² θ + sin² θ = 1

We can solve for sin θ using this identity.
sin² θ = 1 - cos² θ

Square root both sides
sin θ = √(1 - cos² θ)

Substitute cos θ = 4/5
sin θ = √[1 - (4/5)²]

Simplify
sin θ = √(25/25 - 16/25)

Subtract
sin θ = √(9/25)

Thus,
sin θ = ± 3/5

But since we know that sin θ is positive, then the only possible value is
sin θ = 3/5

Hope this helps...
^_^

Answer 2

you are in right direction.
sin theta = 3/5 or -3/5
but as theta is between 0 and pi/2 sin theta is positive so 3/5

Answer 3

Draw a triangle where it is constructed with lines of unit lengthes 3,4 &5.

so one will have a90degree angle Now take angle between 4 & 5 and take it as theta & U will see theta<90 & has the values U said

:)

Answer 4

Ok...looks like you are moving in the right direction

sin^2(theta) + cos^2(theta) = 1

sin^2(theta) + (0.80)^2 = 1

sin^2(theta) = 1 - (0.8)^2

sin^2(theta) = 0.36
sin(theta) = +/- sqrt(0.36)

sin(theta) = 0.6 (note the limits of theta, so pick the positive root)

theta = inv_sin(0.6)

36.86 deg or 0.643 radians

Good luck

Answer 5

Cos theta=4/5

theta=acos(4/5) where acos=cos inverse
theta=pi/5(approximately, actually pi()/4.882)
sin(pi/5)=0.6.

Answer 6

a^2 + b^2 = c^2
4^2 + b^2 = 5^2
16 + b^2 = 25
b^2 = 9
b = 3

sinA = (3/5)

"A" would be .64350110879

pi/300 wouldn't be the answer.

Answer 7

Cos@= 4/5 => base =4x and hyp = 5x,
Thus perp =sq rt of 25xsq - 16xsq =3x
Thus Sin@ = perp/hyp = 3x/5x = 3/5.....QED

Answer 8

Okay I dont know how right your answer is but I do want to comend you on at least trying to figure it out instead of just asking for the answer.