Review question

# Can we write $\sin\theta$ and $\cos\theta$ in terms of $\tan(\theta/2)$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6696

## Solution

Prove that, if $t=\tan\tfrac{1}{2}\theta$, then $\sin\theta=\frac{2t}{1+t^2}...$

#### Approach 1: Starting with the left hand side

We can rewrite $\sin\theta$ by using the double angle formula: $\sin\theta= \sin\bigl(2(\tfrac{1}{2}\theta)\bigr)= 2\sin\tfrac{1}{2}\theta \cos\tfrac{1}{2}\theta.$

In order to get $t$ (which we want in the numerator) into the identity, we can rewrite $\sin\tfrac{1}{2}\theta=\tan\tfrac{1}{2}\theta \cos\tfrac{1}{2}\theta,$ which is $t\cos\tfrac{1}{2}\theta$ (where we have used $t=\tan\tfrac{1}{2}\theta$). Therefore, $\sin\theta=2t\cos^2\tfrac{1}{2}\theta.$

Since we want $t$ in the denominator as well, we write $\cos^2\tfrac{1}{2}\theta=\frac{1}{\sec^2\tfrac{1}{2}\theta}.$ Using the identity $1+\tan^2\tfrac{1}{2}\theta=\sec^2\tfrac{1}{2}\theta,$ we deduce that $\cos^2\tfrac{1}{2}\theta=\frac{1}{1+\tan^2\tfrac{1}{2}\theta} =\frac{1}{1+t^2}.$ Therefore $\sin\theta=\frac{2t}{1+t^2}.$

#### Approach 2: Starting with the right hand side

We will rewrite the right hand side in terms of $\sin\dfrac{1}{2}\theta$ and $\cos\dfrac{1}{2}\theta$. We have \begin{align*} \frac{2t}{1+t^2}&= \frac{2\dfrac{\sin\frac{1}{2}\theta}{\cos\tfrac{1}{2}\theta}} {1+\dfrac{\sin^2\frac{1}{2}\theta}{\cos^2\frac{1}{2}\theta}}\\ &=\frac{2\sin\tfrac{1}{2}\theta \cos\tfrac{1}{2}\theta} {\cos^2\tfrac{1}{2}\theta+\sin^2\tfrac{1}{2}\theta}, \end{align*}

where we reached the second line by multiplying the numerator and denominator of the large fraction by $\cos^2\dfrac{1}{2}\theta$.

Since $\sin^2\tfrac{1}{2}\theta+\cos^2\tfrac{1}{2}\theta=1$, the denominator is $1$ and hence $\frac{2t}{1+t^2}=2\sin\tfrac{1}{2}\theta \cos\tfrac{1}{2}\theta=\sin\theta,$ as required.

$\text{...and}\quad\cos\theta=\frac{1-t^2}{1+t^2}.$

As in Approach 1, we write $\tan\theta=\tan\left(2\left(\dfrac{1}{2}\theta\right)\right)$, and use the double-angle formula for $\tan$ to get $\tan\theta=\frac{2\tan\tfrac{1}{2}\theta}{1-\tan^2\tfrac{1}{2}\theta}=\frac{2t}{1-t^2}.$

We also know that $\tan\theta=\dfrac{\sin\theta}{\cos\theta}$, so that $\cos\theta=\dfrac{\sin\theta}{\tan\theta}$.

Therefore, using our expressions for $\sin\theta$ and $\tan\theta$ in terms of $t$, we deduce $\cos\theta=\frac{\left(\dfrac{2t}{1+t^2}\right)}{\left(\dfrac{2t}{1-t^2}\right)}=\frac{1-t^2}{1+t^2}.$

We can also prove that $\cos\theta=\dfrac{1-t^2}{1+t^2}$ using Approaches 1 and 2 above.

By expressing $\dfrac{3+\cos\theta}{\sin\theta}$ in terms of $t$, or otherwise, show that this expression cannot have any value between $-2\sqrt{2}$ and $+2\sqrt{2}$.

#### Method 1

We have \begin{align*} \frac{3+\cos\theta}{\sin\theta} &=\frac{3+\dfrac{1-t^2}{1+t^2}}{\dfrac{2t}{1+t^2}}\\ &=\frac{3(1+t^2)+1-t^2}{2t}\\ &=\frac{4+2t^2}{2t}\\ &=\frac{2+t^2}{t}. \end{align*}

We need to show therefore that $\dfrac{2+t^2}{t}=t+\dfrac{2}{t}$ cannot have any value in the given range, where $t$ is any real number except for zero.

The last part is because $t=\tan\tfrac{1}{2}\theta$ can take any real value, but we cannot allow $t=0$ in this fraction, just as we cannot allow $\sin\theta=0$ in the original fraction. We have
$\sin\theta \neq 0 \implies \tan\frac{1}{2}\theta \neq 0.$

Let $t+\dfrac{2}{t} = k$. Then $t^2-kt+2=0$, and for this to have real roots, the discriminant must be non-negative.

This means that $k^2 \geq 8$, and so $k \geq 2\sqrt{2}$, or $k \leq -2\sqrt{2}$, as required.

### Method 2

If we sketch a graph of $y=t+\dfrac{2}{t}$, we see that it has an asymptote at $t=0$, is positive for $t>0$ and negative for $t<0$.

For $t>0$, it tends to $\infty$ as $t\to\infty$ and as $t\to 0$; for $t<0$, it tends to $-\infty$ as $t\to-\infty$ and as $t\to0$. So the graph looks something like this:

We can find the stationary points using calculus.

Writing $y=t+\dfrac{2}{t}$, we have $\dfrac{dy}{dt}=1-\dfrac{2}{t^2}$, so$\dfrac{dy}{dt}=0$ has solutions $t=\pm\sqrt{2}$.

When $t=\sqrt{2}$, $y=\sqrt{2}+\dfrac{2}{\sqrt{2}}=2\sqrt{2}$; when$t=-\sqrt{2}$, $y=-2\sqrt{2}$.

So referring back to our sketch graph, we see that $y$ cannot take anyvalues between $-2\sqrt{2}$ and $+2\sqrt{2}$.

### Method 3

Another calculus-free approach is to use the arithmetic mean–geometric mean (AM–GM) inequality.

This says that if $a_1$ and $a_2$ are positive real numbers, then their arithmetic mean is greater than or equal to their geometric mean.

In other words, $\dfrac{a_1 + a_2}{2} \geq \sqrt{a_1a_2}$ where $a_1, a_2 \geq 0$, with equality if and only if $a_1 = a_2$.

In our case, if we take $n=2$ and set $a_1=t$ and $a_2=\frac{2}{t}$, we find that $\frac{t+\frac{2}{t}}{2}\ge \sqrt{t \times\frac{2}{t}}=\sqrt{2},$ with equality if and only if $t=\dfrac{2}{t}$.

Thus for positive values of $t$, $t+\dfrac{2}{t}\ge 2\sqrt{2}$ with equality if and only if $t=\dfrac{2}{t}$, that is, if $t=\sqrt{2}$.

For negative values of $t$, let $t=-u$, with $u$ positive. Then $t+\dfrac{2}{t}=-u-\dfrac{2}{u}=-\left(u+\dfrac{2}{u}\right)$.

Since $u+\dfrac{2}{u}\ge2\sqrt{2}$, it follows that for negative values of $t$, $t+\dfrac{2}{t}\le -2\sqrt{2}$ with equality if and only if $t=-\sqrt{2}$.

Thus $t+\dfrac{2}{t}$ cannot take any values between $-2\sqrt{2}$ and$+2\sqrt{2}$.

### Method 4

As a final method, if $t$ is positive, we can write $t+\dfrac{2}{t}$ as $t+\frac{2}{t}=\left(\sqrt{t}-\sqrt{\frac{2}{t}}\right)^2+2\sqrt{2}.$ Since a square is always positive or zero, the right hand side must be at least $2\sqrt{2}$, with equality if and only if$\sqrt{t}=\sqrt{\dfrac{2}{t}}$.

We can deal with the case of $t$ negative as above.

What could make us think to write $t+\dfrac{2}{t}$ in such a bizarre way? And what made us think that it would be useful?

To prove the AM–GM inequality for $n=2$, we first observe that $(x-y)^2=x^2+y^2-2xy,$ so that $x^2+y^2=(x-y)^2+2xy\ge 2xy$ with equality if and only if $x=y$.

If we now write $x=\sqrt{\vphantom{0}a_1}$ and$y=\sqrt{\vphantom{0}a_2}$, we get $a_1+a_2\ge 2\sqrt{\vphantom{0}a_1a_2},$ which is, on dividing by $2$, the AM–GM inequality for $n=2$.

So in this approach, we have simply replicated this proof without mentioning the AM–GM inequality!