This is the question:
Let $t = \tan(\theta/2)$. Express the following functions in terms of $t$:
- $\tan \theta$
- $\sin\theta$
- $\cos\theta$
I know that for part (1), $$\tan\theta = \frac{2t}{1-t^2}$$ How do I get parts (2) and (3)?
If $\tan\theta = \frac{2t}{1-t^2}$ then I would multiply by $\cos \theta$ to get $$\sin \theta = 2t\frac{\cos \theta}{1-t^2}$$ but that doesn't look right.
Since $\tan(\theta) = \dfrac{2t}{1-t^2}$, draw a picture using the fact that $\tan(\theta) = \frac{\mbox{opposite}}{\mbox{adjacent}}$ in a right triangle.
From here, find the hypotenuse using the Pythagoren Theorem. Then, use the definition for $\sin(\theta)$ and $\cos(\theta)$ in a right triangle.
Here's a straightforward algebraic approach:
$$\sin\theta = 2\sin(\theta/2)\cos(\theta/2) = 2\tan(\theta/2)\cos^2(\theta/2) =\\ \ \ \ \ = \frac{2\tan(\theta/2)}{\sec^2(\theta/2)} = \frac{2\tan(\theta/2)}{1 + \tan^2(\theta/2)} = \frac{2t}{1+ t^2}$$
You can construct a similar argument for $\cos\theta$.
Recall that $\tan(\alpha+\beta)=\dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$. If $\alpha$ is equal to $\beta$ then $\alpha+\beta$ is $2\alpha$, so we have $\tan(2\alpha)=\dfrac{2\tan\alpha}{1-\tan^2\alpha}$. Apply that to the case where $\alpha=\dfrac\theta2$. You get $$ \tan\theta = \frac{2\tan\frac\theta2}{1-\tan^2\frac\theta2} = \frac{2t}{1-t^2}. $$
Similarly $$ \sin\theta=2\sin\frac\theta2\cos\frac\theta2. \tag 1 $$ If $$ t=\tan\frac\theta2=\frac{\text{opp}}{\text{adj}}=\frac{\text{opp}}1 $$ so the opposite side is $t$ and the adjacent side is $1$, then the hypotenuse is $\sqrt{1+t^2}$ so $$ \sin\frac\theta2=\frac t{\sqrt{1+t^2}} \text{ and }\cos\frac\theta2=\frac{1}{\sqrt{1+t^2}} $$ so $(1)$ becomes $$ 2\frac t{1+t^2}. $$ A similar thing handles the cosine.
All this works in the first quadrant, but for other quadrants you need to think about $\pm$ issues, and then you'll see it's valid in other quadrants too.
$$z_{\frac{\theta}2} = \cos\frac{\theta}2 + i \sin \frac{\theta}2 = (1+t^2)^{-\frac12}(1+it) $$ by deMoivre's theorem $$ z_{\theta} = z_{\frac{\theta}2}^2=(1+t^2)^{-1}\left((1-t^2)+2it\right) = \cos\theta+i\sin\theta $$ hence, equating real parts: $$ \cos \theta = \frac{1-t^2}{1+t^2} $$ and imaginary parts: $$ \sin \theta = \frac{2t}{1+t^2} $$ then taking the ratio: $$ \tan \theta = \frac{2t}{1-t^2} $$
here is a way to see how and why you get the expressions for $\sin t, \cos t$ and $\tan t$ using the unit circle. it is easier to write $m = \tan(t/2).$ midpoint of the line segment(chord) connecting the points $ (x,y)=(\cos t, \sin t)$ and $(1, 0 )$ which is $((x+1)/2, y/2)$ on the the radius with a slope $m.$
now we have two equations for $x$ and $y.$ they are $$ x^2 + y^2 = 1, y = m(x+1)$$ solving the quadratic equation for $x$ we get $x = {1 - m^2 \over 1 + m^2}, -1.$ the answer $x = -1$ corresponds to $t = \pi, m = \infty$ and $y = {2m \over 1 + m^2}, 0.$ finally putting both cases we get $$ \cos t = {1 - m^2 \over 1 + m^2}, \sin t = {2m \over 1 + m^2}.$$
Let Θ be a point $(x,y) = (\cos θ, \sin θ)$ on the trigonometric circle, and F be the point $(−1,0)$. Except the $x=−1$ case (where $θ=π$ modulo 2π and $\tan\frac{θ}{2}$ is infinite) the vector going from F to Θ is at angle $\frac{θ}{2}$ from $x$ direction (modulo π). This follows from geometry of a chord of the circle:
This image shows $\tan\frac{θ}{ 2} = \frac{3}{ 2}$; rendered from File:Tan_a,2_as_chord.svg.
Hence: $$t = \frac{y}{1+x}.\tag 1$$ Note that the “tan” function has period π, and the difference between direction of $(1+x,\ y)$ and $\frac{θ}{2}$ may be ignored.
On the other hand, equation of the trigonometric circle is $$x^2 + y^2 = 1.$$ By the difference of squares identity: $$y^2 = (1-x)(1+x).\tag 2$$
Compute the square of (1) and apply (2): $$t^2 = \frac{1-x}{1+x},$$ then multiply by $1+x$: $$t^2(1+x) = 1-x.$$ Rearranging, $$t^2 - 1 + (t^2 + 1)x = 0,$$ which gives $$\cos θ = x = \frac{1 - t^2}{1 + t^2}.$$
The other two functions of $θ$ can be found as $y$ and $y/x$.
