Question

This is finding exact values of sin theta/2 and tan theta/2. I’m really confused and now don’t have a clue on how to do this, please help

Answer 1

The exact values - sin(θ/2) = √[(1 - (- 5/13)) / 2] = 3/√(13)

tan(θ/2) = √[(1 - (- 5/13)) / (1 + (- 5/13))] = 3/2 .

Initially,

Sin(θ) / cos(θ) equals tan(θ)

Hence we know that cos(θ) < 0 since 90° < θ < 180°, which indicates that is in the second quadrant. (At that point, we also know the sin sign (θ), but it is not relevant.)

By dividing each component of the inequality by two, we can determine that 45° < θ/2 < 90°, meaning that the half-angle lies in the first quadrant and that cos(θ/2) and sin(θ/2) are both greater than 0.

Now, consider the identities of half-angles.

$1 + cos(θ)) / 2 equals cos²(θ/2).

sin²(θ/2) = cos(θ) - (1 - 2) /

and using the square roots that are positive, we have

1 + cos(θ)) / 2 = √[cos(θ/2)]

cos(θ) / 2 - 1 = √[sin(θ/2)]

Next

As θ/2) / cos(θ/2) = √, we get tan(θ/2).(1 + cos(θ)) / ((1 - cos(θ))]

See how we are not dependent on sin(θ)?

Remember the Pythagorean identity now:

1 = cos²(θ) + sin²(θ)

Subtracting cos²(θ) from both sides yields

Cos²(θ) = 1/tan²(θ) + 1

Since we know that cos(θ) is negative, we can take the negative square root by solving for cos²(θ).

1 + tan²(θ) = cos²(θ)

1/√[1 + tan²(θ)] = cos(θ)

Enter tan(θ) = - 12/5 and use that value to find cos(θ):

- 1/√[1 + (-12/5)²] is cos(θ). = - 5/13

Answer 2

First,

tan(θ) = sin(θ) / cos(θ)

and given that 90° < θ < 180°, meaning θ lies in the second quadrant, we know that cos(θ) < 0. (We also then know the sign of sin(θ), but that won't be important.)

Dividing each part of the inequality by 2 tells us that 45° < θ/2 < 90°, so the half-angle falls in the first quadrant, which means both cos(θ/2) > 0 and sin(θ/2) > 0.

Now recall the half-angle identities,

cos²(θ/2) = (1 + cos(θ)) / 2

sin²(θ/2) = (1 - cos(θ)) / 2

and taking the positive square roots, we have

cos(θ/2) = √[(1 + cos(θ)) / 2]

sin(θ/2) = √[(1 - cos(θ)) / 2]

Then

tan(θ/2) = sin(θ/2) / cos(θ/2) = √[(1 - cos(θ)) / (1 + cos(θ))]

Notice how we don't need sin(θ) ?

Now, recall the Pythagorean identity:

cos²(θ) + sin²(θ) = 1

Dividing both sides by cos²(θ) gives

1 + tan²(θ) = 1/cos²(θ)

We know cos(θ) is negative, so solve for cos²(θ) and take the negative square root.

cos²(θ) = 1/(1 + tan²(θ))

cos(θ) = - 1/√[1 + tan²(θ)]

Plug in tan(θ) = - 12/5 and solve for cos(θ) :

cos(θ) = - 1/√[1 + (-12/5)²] = - 5/13

Finally, solve for sin(θ/2) and tan(θ/2) :

sin(θ/2) = √[(1 - (- 5/13)) / 2] = 3/√(13)

tan(θ/2) = √[(1 - (- 5/13)) / (1 + (- 5/13))] = 3/2