How to get from tangent to half angle sine

[mauricev]

Math text says angle is in the third quadrant with a tan of 1. It asks to find the sin angle/2. The answer given is negative sqrt (2 + sqrt(2))/2.

I reasoned that one needs to use the half angle formula for sine, which requires cosine. I reasoned the triangle to be adjacent =1, opposite = 1 and hypotenuse = sqrt(2) from the pythagorean theorem. This makes cosine 1/sqrt(2). Rationalize the denominator and that becomes sqrt(2)/2. Plug into the half angle formula gives negative sqrt (1 - sqrt(2)/2)/2, which is nothing like the given answer.

 

[Harry_the_cat]

If \(\displaystyle tan\theta=1\)and \(\displaystyle \theta\) is in the third quadrant, then \(\displaystyle cos\theta=-\frac{1}{\sqrt2} = -\frac{\sqrt2}{2}\).

Also, if \(\displaystyle tan\theta=1\) and \(\displaystyle 180<\theta<270\) , then \(\displaystyle \theta=225\)

So \(\displaystyle \frac{\theta}{2}= 112.5\) so \(\displaystyle sin {\frac{\theta}{2}}\) is positive so your "answer given" cannot be correct as it is negative.

 

[tkhunny]

It has always been the responsibility of the human to know which quadrant one is talking about. This has only become more important since about 1975 when calculators began to relieve humans from the responsibility of the numerical calculation.

[math]180^{\circ} < \theta < 270^{\circ}[/math] - Quadrant III [math](cos(\theta) < 0, sin(\theta) < 0, tan(\theta) > 0)[/math]
[math]90^{\circ} < \dfrac{\theta}{2} < 135^{\circ}[/math] - Quadrant II [math](cos(\theta) < 0, sin(\theta) > 0, tan(\theta) < 0)[/math]

 

[Deleted member 4993]

Math text says angle is in the third quadrant with a tan of 1. It asks to find the sin angle/2. The answer given is negative sqrt (2 + sqrt(2))/2.

I reasoned that one needs to use the half angle formula for sine, which requires cosine. I reasoned the triangle to be adjacent =1, opposite = 1 and hypotenuse = sqrt(2) from the pythagorean theorem. This makes cosine 1/sqrt(2). Rationalize the denominator and that becomes sqrt(2)/2. Plug into the half angle formula gives negative sqrt (1 - sqrt(2)/2)/2, which is nothing like the given answer.

As pointed out in earlier posts that is incorrect. [if we assume that 0 ≤ Θ/2 ≤ 2π. ]

If Θ is in third quadrant then:

π ≤ Θ ≤ 3π/2

then

π/2 ≤ Θ/2 ≤ 3π/4

Clearly (Θ/2) is in second quadrant (for this case) and sin(Θ/2) ≥ 0 (positive)
[edit: this statement is partially true - see the post below by Dr. P]

 

[Dr.Peterson]

Math text says angle is in the third quadrant with a tan of 1. It asks to find the sin angle/2. The answer given is negative sqrt (2 + sqrt(2))/2.

I reasoned that one needs to use the half angle formula for sine, which requires cosine. I reasoned the triangle to be adjacent =1, opposite = 1 and hypotenuse = sqrt(2) from the pythagorean theorem. This makes cosine 1/sqrt(2). Rationalize the denominator and that becomes sqrt(2)/2. Plug into the half angle formula gives negative sqrt (1 - sqrt(2)/2)/2, which is nothing like the given answer.

There are several issues here.

First, as has been noted, if the given angle is in the third quadrant, then its cosine is negative, so it is [MATH]\cos(\theta) = -\frac{\sqrt{2}}{2}[/MATH].

Second, your formula for the half-angle is wrong; you have the division by 2 outside the radical, rather than inside. It should be [MATH]\sin(\frac{\theta}{2}) = \sqrt{\frac{1-\cos({\theta})}{2}} = \sqrt{\frac{1-\left(-\frac{\sqrt{2}}{2}\right)}{2}}[/MATH]. This can be simplified to the given answer -- almost. Try doing that.

Third, there are really two answers, depending on whether the given angle is taken as 225° or as -135°, so that the half angle can be in the second or fourth quadrant. In the latter case, the negative answer you say you were given is correct. How this is to be resolved depends on the conventions in your text. How does it state the half-angle formulas?