The trigonometric identities are the equations involving trigonometric functions that are true for all angles for which both sides of the equations are defined. There are many many trigonometric identities, but in this section, we review a number of them which are more useful and you need to memorize.

### Pythagorean Identities

Consider an angle $\theta$ in standard position and let $P$ be the intersection of the terminal side of $\theta$ and the unit circle. Recall that the $x$- and $y$- components of $P$ are $\cos\theta$ and $\sin\theta$, respectively. Because the point $P(x,y)$ is on the unit circle, we have $x^{2}+y^{2}=1$, which translates into the most important trig identity:

$\bbox[#F2F2F2,5px,border:2px solid black]{\sin^{2}\theta+\cos^{2}\theta=1}\tag{1}$

Remark that it is standard to denote $(\sin\theta)^{2}$, the square of the number $\sin\theta$, by the notation $\sin^{2}\theta$. Similarly
$\cos^{2}\theta=(\cos\theta)^{2},\qquad\tan^{2}\theta=(\tan\theta)^{2},\qquad\cdots$

If we divide the both sides of Equation (1) by $\cos^{2}\theta$, we get
$\bbox[#F2F2F2,5px,border:2px solid black]{\tan^{2}\theta+1=\sec^{2}\theta}\tag{2}$ Recall that $\sec\theta=1/\cos\theta$. Similarly if we divide both sides of Equation (1) by $\sin^{2}\theta$, we get
$\bbox[#F2F2F2,5px,border:2px solid black]{1+\cot^{2}\theta=\csc^{2}\theta} \tag{3}$ Recall that $\csc\theta=1/\sin\theta$.

Example 1
Find the values of $\sin\theta,\tan\theta$, and $\cot\theta$ given $\cos\theta=\frac{2}{3}$ and $\theta$ is in the fourth quadrant angle.
Solution
It follows from identity $\sin^{2}\theta+\cos^{2}\theta=1$ (Equation 1) that
$\sin^{2}\theta+\left(\frac{2}{3}\right)^{2}=1\Rightarrow\sin\theta=\pm\sqrt{\frac{5}{9}}.$ Because $\theta$ is in the fourth quadrant, we know its sine is negative. Thus we choose the negative sign for sine
$\sin\theta=-\sqrt{\frac{5}{9}}=-\frac{\sqrt{5}}{3}.$ It follows from the definition of tangent that
$\tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{-\dfrac{\sqrt{5}}{3}}{\dfrac{2}{3}}=-\frac{\sqrt{5}}{2},$ and from $\cot\theta=1/\tan\theta$ that
$\cot\theta=\frac{1}{-\frac{\sqrt{5}}{2}}=-\frac{2}{\sqrt{5}}.$
Example 2
Find the values of $\cos x,\tan x$, and $\cot x$ if $\sin x=-\frac{8}{17}$ and $x$ is in the third quadrant.
Solution
Note that we can denote angles by anything that we like: $\theta,\phi,x,y,A,B,\cdots$. Here the angle is denoted by $x$.

Because $x$ is in the third quadrant, we know its sine and cosine are negative but its tangent and cotangent are positive. Using the identity $\sin^{2}x+\cos^{2}x=1$, we get
$\left(-\frac{8}{17}\right)^{2}+\cos^{2}x=1\Rightarrow\cos^{2}x=-\sqrt{1-\frac{64}{289}}=-\frac{15}{17}.$ Because $\tan x=\sin x/\cos x$ and $\cot x=\cos x/\sin x$, we get
$\tan x=\frac{-\frac{8}{17}}{-\frac{15}{17}}=\frac{8}{15},\qquad\cot x=\frac{1}{\tan x}=\frac{15}{8}.$

Example 3
Find the values of $\sin x$ and $\cos x$, given $\tan x=-2$ and $x$ is in the second quadrant.
Solution
Because $x$ is in the second quadrant, we know $\sin x>0$ and $\cos x<0$. Using the formula (2)
$1+\tan^{2}x=\sec^{2}x=\frac{1}{\cos^{2}x}$ we have
$1+(-2)^{2}=\frac{1}{\cos^{2}x}\Rightarrow\cos^{2}x=\frac{1}{5}\Rightarrow\cos x=-\frac{1}{\sqrt{5}}$ Now using the identity $\sin^{2}x+\cos^{2}x=1$, we find
$\sin^{2}x+\frac{1}{5}=1\Rightarrow\sin x=\frac{2}{\sqrt{5}}.$

### Even-Odd Identities

Comparing the angles $\theta$ and $-\theta$ in Figure 1 clearly shows that
$\bbox[#F2F2F2,5px,border:2px solid black]{\sin(-\theta)=-\sin\theta,\qquad\cos(-\theta)=\cos\theta} \tag{4}$ Consequently
$\tan(-\theta)=\frac{\sin(-\theta)}{\cos(-\theta)}=\frac{-\sin\theta}{\cos\theta}=-\tan\theta$

$\bbox[#F2F2F2,5px,border:2px solid black]{\tan(-\theta)=-\tan\theta} \tag{5}$ Figure 1

Similar identities for cotangent, secant, and cosecant can be derived but they are less important.
$\cot(-\theta)=\frac{\cos(-\theta)}{\sin(-\theta)}=\frac{\cos\theta}{-\sin\theta}=-\cot\theta$ $\sec(-\theta)=\frac{1}{\cos(-\theta)}=\frac{1}{\cos\theta}=\sec\theta$ $\csc(-\theta)=\frac{1}{\sin(-\theta)}=\frac{1}{-\sin\theta}=-\csc\theta$

• The above equations show that $\cos\theta$ and $\sec\theta$ are even functions and the other four functions are odd.

The following two identities are called the addition formulas for sine and cosine. Let $\theta$ and $\phi$ be any two angles. Then
$\bbox[#F2F2F2,5px,border:2px solid black]{\sin(\theta+\phi)=\sin\theta\cos\phi+\cos\theta\sin\phi} \tag{6}$ and
$\bbox[#F2F2F2,5px,border:2px solid black]{\cos(\theta+\phi)=\cos\theta\cos\phi-\sin\theta\sin\phi} \tag{7}$

#### Show the proofs

The identities (6) and (7) hold for all angles $\theta$ and $\phi$, but here we provide a proof for the restricted case in which $\theta$ and $\phi$ are both positive angles such that $\theta+\phi<90^{\circ}$.

To prove this restricted case, we consult Figure 2. Let $PR$ be a line perpendicular to line $OQ$ defined by angle $\theta$. Draw $N$ on $PH$ such that $NR$ is parallel to the $x$-axis. Now $\angle NPR=\theta$ Figure 2

Proof 1: There is a theorem in geometry saying that if the two arms of an angle are perpendicular to the two arms of another angle, then the angles are either equal or supplementary. Because two arms of $\angle NPR$ are perpendicular to the two arms of $\angle HOR$ and both are acute angles, then $\angle NPR=\angle HOR=\theta$.

Proof 2: $\triangle OMR$ is a right triangle $\Rightarrow\angle ORM=90^{\circ}-\theta$. $\angle MRN=90^{\circ}$ and $\angle ORM=90^{\circ}-\theta\Rightarrow\angle ORN=\theta$. $\angle ORP=\angle ORN+\angle NRP=90^{\circ}$ and $\angle ORN=\theta\Rightarrow\angle NRP=90^{\circ}-\theta.$ $\triangle RNP$ is a right triangle $\Rightarrow\angle NRP=90^{\circ}-\angle NPR$. Because $\angle NRP=90^{\circ}-\theta$ , we have $\angle NPR=\theta$.

\begin{align*}
\sin(\theta+\phi) & =\frac{HP}{OP}\\
& =\frac{HN+NP}{OP}\\
& =\frac{HN}{OP}+\frac{NP}{OP}\\
& =\frac{MR}{OP}+\frac{NP}{OP}\\
& =\frac{MR}{OR}\frac{OR}{OP}+\frac{NP}{PR}\frac{PR}{OP}\\
& =\sin\theta\cos\phi+\cos\theta\sin\phi.
\end{align*}

The proof of the addition formula for cosine goes as follows:
$OP=1$ $\frac{PR}{OP}=\sin\phi\Rightarrow PR=\sin\phi$ $\frac{OR}{OP}=\cos\phi\Rightarrow OR=\cos\phi$ $\frac{OM}{OR}=\cos\theta\Rightarrow OM=OR\cdot\cos\theta=\cos\phi\cos\theta$ $\frac{NR}{PR}=\sin\theta\Rightarrow NR=PR\cdot\sin\theta=\sin\phi\sin\theta$ but $NR=HM\Rightarrow HM=\sin\phi\sin\theta$
\begin{align*}
\cos(\theta+\phi) & =\frac{OH}{OP}=OH\\
& =OM-OH\\
& =\cos\phi\cos\theta-\sin\phi\sin\theta
\end{align*}

The addition formula for tangent is
$\bbox[#F2F2F2,5px,border:2px solid black]{\tan(\theta+\phi)=\frac{\tan\theta+\tan\phi}{1-\tan\theta\tan\phi}}\tag{8}$

#### Show the proof

It follows from the addition formulas (6) and (7) by a relatively simple argument:
\begin{align*}
\tan(\theta+\phi) & =\frac{\sin(\theta+\phi)}{\cos(\theta+\phi)}\\
& =\frac{\sin\theta\cos\phi+\cos\theta\sin\phi}{\cos\theta\cos\phi-\sin\theta\sin\phi}
\end{align*}
Now if we divide both the numerator and the denominator by $\cos\theta\cos\phi$, we obtain
\begin{align*}
\tan(\theta+\phi) & =\frac{\dfrac{\sin\theta}{\cos\theta}+\dfrac{\sin\phi}{\cos\phi}}{1-\dfrac{\sin\theta}{\cos\theta}\dfrac{\sin\phi}{\cos\phi}}\\
& =\frac{\tan\theta+\tan\phi}{1-\tan\theta\tan\phi}.
\end{align*}

\bbox[#F2F2F2,5px,border:2px solid black]{ \begin{align} \sin(\theta-\phi)&=\sin\theta\cos\phi-\cos\theta\sin\phi\\[10pt] \cos(\theta-\phi)&=\cos\theta\cos\phi+\sin\theta\sin\phi\\[10pt] \tan(\theta-\phi)&=\dfrac{\tan\theta-\tan\phi}{1+\tan\theta\tan\phi} \end{align}}\tag{9}

These identities can be obtained from the addition identities by substituting $-\phi$ for $\phi$ and using the identities (4). For example,
\begin{align*}
\sin(\theta-\phi) & =\sin(\theta+(-\phi))\\
& =\sin\theta\cos(-\phi)+\sin(-\phi)\cos(\theta)\\
& =\sin\theta\cos\phi-\sin\phi\cos\theta
\end{align*}

\begin{align*}
\cos(\theta-\phi) & =\cos(\theta+(-\phi))\\
& =\cos\theta\cos(-\phi)-\sin(\theta)\sin(-\phi)\\
& =\cos\theta\cos(\phi)-\sin\theta(-\sin\phi)\\
& =\cos\theta\cos\phi+\sin\theta\sin\phi
\end{align*}

### Double Angle Formulas

If we replace $\phi$ by $\theta$ in the addition formulas (Equations 6 and 7), we get

$\bbox[#F2F2F2,5px,border:2px solid black]{\sin2\theta=2\sin\theta\cos\theta}\tag{10}$

$\bbox[#F2F2F2,5px,border:2px solid black]{\cos2\theta=\cos^{2}\theta-\sin^{2}\theta}\tag{11}$ Because $\sin^{2}\theta+\cos^{2}\theta=1$, in the equation we can replace $\cos^{2}\theta$ by $1-\sin^{2}\theta$
\begin{align*}
\cos2\theta & =(1-\sin^{2}\theta)-\sin^{2}\theta\\
& =1-2\sin^{2}\theta
\end{align*}
$\bbox[#F2F2F2,5px,border:2px solid black]{\cos2\theta=1-2\sin^{2}\theta}\tag{11}$ or we can replace $\sin^{2}\theta$ by $1-\cos^{2}\theta$ to obtain
\begin{align*}
\cos2\theta & =\cos^{2}\theta-(1-\cos^{2}\theta)\\
& =2\cos^{2}\theta-1
\end{align*}
$\bbox[#F2F2F2,5px,border:2px solid black]{\cos2\theta=1-2\sin^{2}\theta} \tag{12}$ or we can replace $\sin^2\theta$ by $1-\cos^2\theta$ and obtain
$\bbox[#F2F2F2,5px,border:2px solid black]{\cos 2\theta=2\cos^2\theta-1}\tag{13}$ The formulas (10), (11), (12), and (13) are known as double angle formulas for sine and cosine. You need to know only one of the three forms for the double angle formulas for cosine but be able to derive the other two from the identity $\sin^{2}\theta+\cos^{2}\theta=1$.

### Half Angle Formulas

If we solve Equations (12) and (13) for $\sin^{2}\theta$ and $\cos^{2}\theta$, we obtain the following equations which are called half angle formulas; we will use them later on for integration:
$\bbox[#F2F2F2,5px,border:2px solid black]{\sin^{2}\theta=\frac{1-\cos2\theta}{2}}\tag{14}$ $\bbox[#F2F2F2,5px,border:2px solid black]{\cos^{2}\theta=\frac{1+\cos2\theta}{2}} \tag{15}$

Example 4
Find the exact value of each expression:
(a) $\sin75^{\circ}$
(b) $\cos\frac{\pi}{12}$
(c) $\tan105^{\circ}$
Solution
(a) Using the addition formula (6) we can write
\begin{align*}
\sin75^{\circ} & =\sin(45^{\circ}+30^{\circ})\\
& =\sin45^{\circ}\cos30^{\circ}+\cos45^{\circ}\sin30^{\circ}\\
& =\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}\cdot\frac{1}{2}\\
& =\frac{\sqrt{2}(\sqrt{3}+1)}{4}
\end{align*}
(b) Method 1: Using the subtraction formula (9)
\begin{align*}
\cos\frac{\pi}{12} & =\cos\left(\frac{\pi}{3}-\frac{\pi}{4}\right)\\
& =\cos\frac{\pi}{3}\cdot\cos\frac{\pi}{4}+\sin\frac{\pi}{3}\cdot\sin\frac{\pi}{4}\\
& =\frac{1}{2}\cdot\frac{\sqrt{2}}{2}+\frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2}\\
& =\frac{\sqrt{2}(1+\sqrt{3})}{4}
\end{align*}
Method 2: Using the half angle formula (15)
\begin{align*}
\cos^{2}\frac{\pi}{12} & =\frac{1+\cos\frac{\pi}{6}}{2}\\
& =\frac{1+\frac{\sqrt{3}}{2}}{2}\\
& =\frac{2+\sqrt{3}}{4}
\end{align*}
Because $\pi/12$ is in the first quadrant, its cosine is positive. Thus
$\cos\frac{\pi}{12}=\sqrt{\frac{2+\sqrt{3}}{4}},$ which is the same as $\sqrt{2}(1+\sqrt{3})/4$, because
$\left(\frac{\sqrt{2}(1+\sqrt{3})}{4}\right)^{2}=\frac{2(1+\sqrt{3})^{2}}{16}=\frac{1+2\sqrt{3}+3}{8}=\frac{2+\sqrt{3}}{4}.$ (c) Using the addition formula (8)
\begin{align*}
\tan105^{\circ} & =\tan(60^{\circ}+45^{\circ})\\
& =\frac{\tan60^{\circ}+\tan45^{\circ}}{1-\tan60^{\circ}\cdot\tan45^{\circ}}\\
& =\frac{\sqrt{3}+1}{1-\sqrt{3}}
\end{align*}
We can further simply it if we multiply the numerator and denominator
by $1+\sqrt{3}$:
\begin{align*}
\tan105^{\circ} & =\frac{(1+\sqrt{3})^{2}}{(1-\sqrt{3})(1+\sqrt{3})}\\
& =\frac{(1+\sqrt{3})^{2}}{1-3}\\
& =-\frac{(1+\sqrt{3})^{2}}{2}.
\end{align*}
Because $105^{\circ}$ is in the second quadrant, we expected that its tangent to be negative, and here we observe that $\tan105^{\circ}\approx-3.73205$.
Example 5
Find the exact values of $\sin22.5^{\circ}$ and $\cos22.5^{\circ}$
Solution
Using the half angle formulas (14), (15) and considering the fact that because the angle $22.5^{\circ}$ is in the first quadrant its sine and cosine are positive, we obtain
$\sin^{2}22.5^{\circ}=\frac{1-\cos45^{\circ}}{2}=\frac{1-\frac{\sqrt{2}}{2}}{2}=\frac{2-\sqrt{2}}{4}$ $\Rightarrow\sin22.5^{\circ}=\frac{\sqrt{2-\sqrt{2}}}{2}$ and similarly
$\cos^{2}22.5^{\circ}=\frac{1+\cos45^{\circ}}{2}=\frac{1+\frac{\sqrt{2}}{2}}{2}=\frac{2+\sqrt{2}}{4}$ $\Rightarrow\cos22.5^{\circ}=\frac{\sqrt{2+\sqrt{2}}}{2}.$
Example 6
Show that
(a) $\sin3\theta=-4\sin^{3}\theta+3\sin\theta$
(b) $\cos3\theta=4\cos^{3}\theta-3\sin\theta$
Solution
Using the addition formula (6) and the double angle formulas (10) and (12), we obtain
\begin{align*}
\sin3\theta & =\sin(2\theta+\theta)\\
& =\sin2\theta\ \cos\theta+\cos2\theta\ \sin\theta\\
& =2\sin\theta\cos\theta\ \cos\theta+(1-2\sin^{2}\theta)\sin\theta\\
& =2\sin\theta(1-\sin^{2}\theta)+(1-2\sin^{2}\theta)\sin\theta & {\small (\text{because }\sin^{2}\theta+\cos^{2}\theta=1)}\\
& =-4\sin^{3}\theta+3\sin\theta.
\end{align*}
Similarly
\begin{align*}
\cos3\theta & =\cos(2\theta+\theta)\\
& =\cos2\theta\ \cos\theta-\sin2\theta\ \sin\theta & {\small (\text{by Eq. 7})}\\
& =(2\cos^{2}\theta-1)\cos\theta-2\sin^{2}\theta\cos\theta & {\small (\text{by Eq. 10 and 13})}\\
& =(2\cos^{2}\theta-1)\cos\theta-2(1-\cos^{2}\theta)\cos\theta &{\small (\text{because }\sin^{2}\theta+\cos^{2}\theta=1)}\\
& =4\cos^{3}\theta-3\cos\theta.
\end{align*}

### Complementary Angle Identities

We say two angles are complementary when they add up to $\pi/2$ (or $90^{\circ})$. The following figure shows two complementary angles $\theta$ and $\pi/2-\theta$ of a right triangle. By definition (Equation 1 in Section 3.3.2), we have
$\sin\theta=\frac{\text{side opposite }\theta}{\text{hypotenuse}}=\frac{BC}{AC}=\frac{\text{side adjacent }\dfrac{\pi}{2}-\theta}{\text{hypotenuse}}=\cos\left(\frac{\pi}{2}-\theta\right)$ $\cos\theta=\frac{\text{side adjacent }\theta}{\text{hypotenuse}}=\frac{AB}{AC}=\frac{\text{side opposite } \dfrac{\pi}{2}-\theta}{\text{hypotenuse}}=\sin\left(\frac{\pi}{2}-\theta\right)$

$\tan\theta=\frac{\text{side opposite }\theta}{\text{side adjacent}\theta}=\frac{BC}{AB}=\frac{\text{side adjacent } \dfrac{\pi}{2}-\theta}{\text{side opposite }\frac{\pi}{2}-\theta}=\cot\left(\frac{\pi}{2}-\theta\right)$ In fact, the prefix “co-” in “cosine”, “cotangent” and “cosecant” stands for complementary. Cosine is the abbreviation for “sine of the complementary angle” and cotangent is the abbreviation for “tangent of the complementary angle.” So we can summarize these equations as
$\bbox[#F2F2F2,5px,border:2px solid black]{\text{trig}(\theta)=\text{co-trig}\left(\frac{\pi}{2}-\theta\right)}$ Figure 3: A and C are complementary angles

Using the addition formulas, we can show that the above identities hold for all angles (acute, not acute, positive, negative)

\bbox[#F2F2F2,5px,border:2px solid black]{ \begin{align} \sin\left(\dfrac{\pi}{2}-\theta\right)&=\cos\theta\\[6pt] \cos\left(\dfrac{\pi}{2}-\theta\right)&=\sin\theta\\[6pt] \tan\left(\dfrac{\pi}{2}-\theta\right)&=\cot\theta\\[6pt] \cot\left(\dfrac{\pi}{2}-\theta\right)&=\tan\theta \end{align}}\tag{16} and consequently we have
$\sec\left(\frac{\pi}{2}-\theta\right)=\csc\theta,\qquad\csc\left(\frac{\pi}{2}-\theta\right)=\sec\theta.$

### Simplifications When π/2 or π Is Involved

Using the addition formulas, we can show
\begin{align}
\sin\left(\theta+\dfrac{\pi}{2}\right)&=\cos\theta\6pt] \cos\left(\theta+\dfrac{\pi}{2}\right)&=-\sin\theta\\[6pt]\tag{17} \tan\left(\theta+\dfrac{\pi}{2}\right)&=-\cot\theta \end{align} \begin{equation} \sin(\theta+\pi)=-\sin\theta,\quad\cos(\theta+\pi)=-\sin\theta,\quad\tan(\theta+\pi)=\tan\theta\tag{18} \end{equation} \begin{equation} \sin(\pi-\theta)=\sin\theta,\quad\cos(\pi-\theta)=-\cos\theta,\quad\tan(\pi-\theta)=-\tan(\pi-\theta)\tag{19} \end{equation} It looks like that there are so many identities to memorize, but in fact, you need to memorize one thing: if you add \theta to or subtract it from \pi/2 (or 3\pi/2), the trig function will be converted to its co-trig function (sine to cosine, cosine to sine, tangent to cotangent, cotangent to tangent, and vice versa), but if you add \theta to or subtract it from \pi, the trig function will remain the same. We also need to add a plus or minus sign in front of each case. The sign can be easily determined using the unit circle. For example, suppose we want to simplify \sin(\theta+\pi/2). Because we have added \pi/2, the result will be equal to +\cos\theta or -\cos\theta. But which one will it be? Ok, assume \theta is a small positive angle in the first quadrant, so \sin\theta and \cos\theta are both positive. In this case, \theta+\pi/2 will lie in the second quadrant, then we know \sin(\theta+\pi/2) will be positive (Figure 4(a)). So the fact that \cos\theta and \sin(\theta+\theta) have the same sign yields \sin(\theta+\pi/2)=+\cos\theta. Another example. Suppose we wish to simplify \tan(\theta+\pi). Because we have added \pi, the result will be the same trig function; that is, the result will be either +\tan\theta or -\tan\theta. If \theta is in the first quadrant, then \tan\theta>0. In this case \pi+\theta will lie in the third quadrant, and therefore \tan(\pi+\theta)>0 (Figure 4(b)). So \tan(\theta+\pi)=\tan\theta. Let’s consider another example. Suppose we wish to simplify \cos(\theta+\pi). Because we have added \pi, the result will be the same trig function; that is the result will be either +\cos\theta or -\cos\theta. If \theta is in the first quadrant, then \cos\theta>0. In this case \pi+\theta will lie in the third quadrant, and therefore \cos(\pi+\theta)<0 (Figure 4(c)). Because \cos\theta and \cos(\pi+\theta) have opposite signs, we must have \cos(\theta+\pi)=-\cos\theta. As the last example, consider \sin(\theta-\pi/2). Because \pi/2 is involved, sine will be switched to cosine, so the result is either +\cos\theta or -\cos\theta. If \theta is in the first quadrant, then \theta-\pi/2 will lie in the fourth quadrant, and thus \sin(\theta-\pi/2)<0, but \cos\theta>0 (Figure 4(d)). Because \sin(\theta-\pi/2) and \cos\theta have opposite signs, we must have \sin(\theta-\pi/2)=-\cos\theta. Let’s work this out: \[ \sin(\theta-\pi/2)=\sin\theta\ \overset{0}{\cancel{\cos\frac{\pi}{2}}}-\cos\theta\ \underbrace{\sin\frac{\pi}{2}}_{=1}=-\cos\theta.  (a) (b)  (c) (d)

Figure 4

### Product-to-Sum and Sum-to-Product Formulas

The following identities are called the product-to-sum formulas.
\bbox[#F2F2F2,5px,border:2px solid black]{ \begin{align} \sin \theta\cos\phi=\frac{1}{2}\Big[\sin(\theta-\phi)+\sin(\theta+\theta)\Big]\\[10pt] \sin\theta\sin\phi=\frac{1}{2}\Big[\cos(\theta-\phi)-\cos(\theta+\phi)\Big]\\[10pt] \cos\theta\cos\phi=\dfrac{1}{2}\Big[\cos(\theta-\phi)+\cos(\theta+\phi)\Big] \end{align}\tag{20} } To prove these formulas, you just need to use the addition formulas for sine and cosine and expand the right hand sides. For example to prove the first identity, we write
\begin{align*}
\frac{1}{2}\left[\sin(\theta-\phi)+\sin(\theta+\theta)\right] & =\frac{1}{2}\left[\sin\theta\cos\phi-\cos\theta\sin\phi+\sin\theta\cos\phi+\cos\theta\sin\phi\right]\\
& =\frac{1}{2}\left[2\sin\theta\cos\phi\right]\\
& =\sin\theta\cos\phi.
\end{align*}
Similarly you can prove the following identities, called sum-to-product formulas, by expanding the right hand sides
\bbox[#F2F2F2,5px,border:2px solid black]{ \begin{align} \sin\theta+\sin\phi=2\sin\dfrac{\theta+\phi}{2}\ \cos\dfrac{\theta-\phi}{2}\\[10pt] \sin\theta-\sin\phi=2\cos\dfrac{\theta+\phi}{2}\ \sin\dfrac{\theta-\phi}{2}\\[10pt] \cos\theta+\cos\phi=2\cos\dfrac{\theta+\phi}{2}\ \cos\dfrac{\theta-\phi}{2}\\[10pt] \cos\theta-\cos\phi=-2\sin\dfrac{\theta+\phi}{2}\ \sin\dfrac{\theta-\phi}{2} \end{align}\tag{21} }

### Laws of Cosines and Sines

There are two laws known as the law of cosines (or cosine rule) and the law of sines (or sine rule) relating the lengths of sides of a triangle (of any shape) to the cosines and sines of its angle. Consider a triangle (Figure 5) with interior angles $A,B,$ and $C$, and sides $a,b$,and $c$ such that side $a$ faces angle $A$, side $b$ faces angle $B$ and side $c$ faces angle $C$.

Law of cosines:
\bbox[#F2F2F2,5px,border:2px solid black]{ \begin{align} a^{2}=b^{2}+c^{2}-2bc\cos A\\[6pt] b^2=a^{2}+c^{2}-2ac\cos B\\[6pt] c^{2}=a^{2}+b^{2}-2ab\cos C \end{align}}\tag{22}

Law of sines:

$\bbox[#F2F2F2,5px,border:2px solid black]{\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}}\tag{23}$

#### Show the proofs

To prove the law of cosine, draw $AH$ perpendicular to $BC$ (if $B$ and $C$ are acute) or $BC$ extended (if $B$ or $C$ is obtuse). If angles $B$ and $C$ are acute (Figure 6(a)), using the Pythagorean theorem we have

\begin{align*}
c^{2} & =BH^{2}+AH^{2} &{\small (\text{in right }\triangle AHB)}\\
& =(BC-HC)^{2}+AH^{2}\\
& =BC^{2}-2BC\cdot HC+HC^{2}+AH^{2}&{\small (\text{expanding } (BC-HC)^2)}\\
& =BC^{2}-2BC\cdot HC+AC^{2}& {\small (\text{in right } \triangle AHC: AH^2+HC^2=AC^2)}\\
& =a^{2}-2a\cdot HC+b^{2}&{\small{(BC=a,AC=b)}}\\
& =a^{2}-2ab\cos C+b^{2}&{\small (HC=AC\cdot\cos C=b\cos C)}\\
& =a^{2}+b^{2}-2ab\cos C
\end{align*}
If angle $C$ is obtuse (Figure 6(b)), then $BH=BC+HC$, but $HC=b\cos(\pi-C)=-b\cos C$. Therefore
\begin{align*}
c^{2} & =(BC+HC)^{2}+AH^{2}\\
& =(a+(-b\cos C))^{2}+AH^{2}
\end{align*}
Now if we expand this expression and follow the same steps that we did when all the angles were acute, we will realize that the law of cosine also holds true for this case.

To prove the law of sines, in Figure 6(a) we note that
\begin{align*}
AH & =c\sin B&{\small (\text{in } \triangle BHA)}\\
& =b\sin C &{\small(\text{in } \triangle CHA)}
\end{align*}
Therefore,
$c\sin B=b\sin C\Rightarrow\frac{\sin B}{b}=\frac{\sin C}{c}$ In a similar way, by drawing $CM$ perpendicular to $AB$ we can show that
$\frac{\sin A}{a}=\frac{\sin B}{b}.$ If $C$ is obtuse as in Figure 6(b) then
\begin{align*}
AH & =c\sin B\\
& =b\sin(\pi-C)
\end{align*}
but $\sin(\pi-C)=\sin C$. So the law of sines also holds true if an angle is obtuse.

Applications: These rules are used to find the lengths of sides or the angles of a triangle. If (a) two angles and one side are given or (b) two sides and a non-included angle are given, we can use the sine rule to find the remaining side(s) or angle(s).

If (a) three sides are given or (b) two sides and the angle between them are given, we may use the cosine rule.

• Note that the sum of all three (interior) angles of any triangle is $180^{\circ}$ (or $\pi$ radians). So if two angles are given, we know the third one too.