Help on trigonometry please!

Question

How do I find sin θ/2 when tan θ = 2 and 0 degrees<θ<pi/2 using the half angle formula?

Answer 1

tanθ = 2: Make a right triangle with horizontal leg of 1, vertical leg of 2, and hypotenuse of sqrt(1²+2²) = sqrt(5). From this triangle, you can come up with every possible trig function for θ.

sin(θ/2) = sqrt((1-cosθ)/2) we dispense with - solution because we know that the angle is 1st quadrant

sin(θ/2) = sqrt((1-1/sqrt(5))/2) = .526

You can also just take sin((tan^-1(2))/2)

I hope that helps.

Answer 2

The half angle formula says sin(θ/2) = ±√((1-cos(x))/2). Since we are in quadrant 1 (0 < θ < π/2), we know that the sin(θ/2) will be positive so the equation will be:

sin(θ/2) = √((1-cos(x))/2)

We are given the tan(θ) = 2, We can say tan(θ) = 2/1 so, since tan(θ) = opposite/adjacent, the opposite side of the triangle must be 2 and the adjacent side of the triangle must be 1, like this:

So, we can use the Pythagorean Theorem to solve for the hypotenuse:

c = √(1² + 2²) = √5

Since cos(θ) = adjacent/hyp it equals 1/√5. Rationalizing the denominator gives:

1/√5

1/√5 •√5/√5

or cos(θ) = √5/5

Plugging that into the half angle formula, we get:

sin(θ/2) = √((1-cos(x))/2)

sin(θ/2) = √((1-√5/5)/2)

Answer 3

tanA = opposite side/adjacent side = 2/1. That leaves the hypotenuse as h^2 = 4+1 =5

so h=square root of 5

sinA = opposite/hypotenuse = 2/sqr5 = (2/5)sqr5 = about 4.4/5 = 8.8/10 = about .88

A = about 60 degrees = pi/3

sin 60 = 0.866

sinA/2 = sqr(1-cosA)/2

cosA =1/sqr5

sinA/2 = sqr((1-1/sqr5)/2)

=sqr(1/2 - 1/2sqr5)

=sqr(1/2 -sqr5/10)

=sqr(5/10 - sqr5/10)

= sqr((5-sqr5)/10)= sqr((5-2.2)/10) = sqr(2.8/10) = sqr (.28) = close to .5

A/2 =about 30 degrees = pi/6, sinA=sin30 = 1/2

0<A<pi/2=90 degrees

A=about 2 x 30 = 60 degrees = pi/3

Answer 4