3.3.3 Trigonometric Identities

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:

$$\begin{array}{}\text{(1)}& {\sin }^2\theta +{\cos }^2\theta =1\end{array}$$

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

If we divide the both sides of Equation (1) by ${\cos }^2\theta$, we get
$$\begin{array}{}\text{(2)}& {\tan }^2\theta +1={\sec }^2\theta \end{array}$$ Recall that $\sec \theta =1/\cos \theta$. Similarly if we divide both sides of Equation (1) by ${\sin }^2\theta$, we get
$$\begin{array}{}\text{(3)}& 1+{\cot }^2\theta ={\csc }^2\theta \end{array}$$ Recall that $\csc \theta =1/\sin \theta$.

Even-Odd Identities

Comparing the angles $\theta$ and $-\theta$ in Figure 1 clearly shows that
$$\begin{array}{}\text{(4)}& \sin (-\theta )=-\sin \theta ,\cos (-\theta )=\cos \theta \end{array}$$ Consequently
$$\tan (-\theta )=\frac{\sin (-\theta )}{\cos (-\theta )}=\frac{-\sin \theta }{\cos \theta }=-\tan \theta$$

$$\begin{array}{}\text{(5)}& \tan (-\theta )=-\tan \theta \end{array}$$

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.

Addition and Subtraction Formulas

The following two identities are called the addition formulas for sine and cosine. Let $\theta$ and $\varphi$ be any two angles. Then
$$\begin{array}{}\text{(6)}& \sin (\theta +\varphi )=\sin \theta \cos \varphi +\cos \theta \sin \varphi \end{array}$$ and
$$\begin{array}{}\text{(7)}& \cos (\theta +\varphi )=\cos \theta \cos \varphi -\sin \theta \sin \varphi \end{array}$$

Show the proofs

Hide the proofs

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

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 $\mathrm{\angle }NPR=\theta$

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 $\mathrm{\angle }NPR$ are perpendicular to the two arms of $\mathrm{\angle }HOR$ and both are acute angles, then $\mathrm{\angle }NPR=\mathrm{\angle }HOR=\theta$.

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

$$\begin{array}{rl}\sin (\theta +\varphi )& =\frac{HP}{OP}\\ & =\frac{HN+NP}{OP}\\ & =\frac{HN}{OP}+\frac{NP}{OP}\\ & =\frac{MR}{OP}+\frac{NP}{OP}\\ & =\frac{MR}{OR}\frac{OR}{OP}+\frac{NP}{PR}\frac{PR}{OP}\\ & =\sin \theta \cos \varphi +\cos \theta \sin \varphi .\end{array}$$

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

The addition formula for tangent is
$$\begin{array}{}\text{(8)}& \tan (\theta +\varphi )=\frac{\tan \theta +\tan \varphi }{1-\tan \theta \tan \varphi }\end{array}$$

Show the proof

Hide the proof

The corresponding subtraction identities read
$$\begin{array}{}\text{(9)}& \begin{array}{rl}\sin (\theta -\varphi )& =\sin \theta \cos \varphi -\cos \theta \sin \varphi \\ \cos (\theta -\varphi )& =\cos \theta \cos \varphi +\sin \theta \sin \varphi \\ \tan (\theta -\varphi )& ={\displaystyle \frac{\tan \theta -\tan \varphi }{1+\tan \theta \tan \varphi }}\end{array}\end{array}$$

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

$$\begin{array}{rl}\cos (\theta -\varphi )& =\cos (\theta +(-\varphi ))\\ & =\cos \theta \cos (-\varphi )-\sin (\theta )\sin (-\varphi )\\ & =\cos \theta \cos (\varphi )-\sin \theta (-\sin \varphi )\\ & =\cos \theta \cos \varphi +\sin \theta \sin \varphi \end{array}$$

Double Angle Formulas

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

$$\begin{array}{}\text{(10)}& \sin 2\theta =2\sin \theta \cos \theta \end{array}$$

$$\begin{array}{}\text{(11)}& \cos 2\theta ={\cos }^2\theta -{\sin }^2\theta \end{array}$$ Because ${\sin }^2\theta +{\cos }^2\theta =1$, in the equation we can replace ${\cos }^2\theta$ by $1-{\sin }^2\theta$
$$\begin{array}{rl}\cos 2\theta & =(1-{\sin }^2\theta )-{\sin }^2\theta \\ & =1-2{\sin }^2\theta \end{array}$$
$$\begin{array}{}\text{(11)}& \cos 2\theta =1-2{\sin }^2\theta \end{array}$$ or we can replace ${\sin }^2\theta$ by $1-{\cos }^2\theta$ to obtain
$$\begin{array}{rl}\cos 2\theta & ={\cos }^2\theta -(1-{\cos }^2\theta )\\ & =2{\cos }^2\theta -1\end{array}$$
$$\begin{array}{}\text{(12)}& \cos 2\theta =1-2{\sin }^2\theta \end{array}$$ or we can replace ${\sin }^2\theta$ by $1-{\cos }^2\theta$ and obtain
$$\begin{array}{}\text{(13)}& \cos 2\theta =2{\cos }^2\theta -1\end{array}$$ 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:
$$\begin{array}{}\text{(14)}& {\sin }^2\theta =\frac{1-\cos 2\theta }{2}\end{array}$$ $$\begin{array}{}\text{(15)}& {\cos }^2\theta =\frac{1+\cos 2\theta }{2}\end{array}$$

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
$$\sin \theta =\frac{\text{side opposite }\theta }{\text{hypotenuse}}=\frac{BC}{AC}=\frac{\text{side adjacent }{\displaystyle \frac{\pi }{2}}-\theta }{\text{hypotenuse}}=\cos (\frac{\pi }{2}-\theta )$$ $$\cos \theta =\frac{\text{side adjacent }\theta }{\text{hypotenuse}}=\frac{AB}{AC}=\frac{\text{side opposite }{\displaystyle \frac{\pi }{2}}-\theta }{\text{hypotenuse}}=\sin (\frac{\pi }{2}-\theta )$$

$$\tan \theta =\frac{\text{side opposite }\theta }{\text{side adjacent}\theta }=\frac{BC}{AB}=\frac{\text{side adjacent }{\displaystyle \frac{\pi }{2}}-\theta }{\text{side opposite }\frac{\pi }{2}-\theta }=\cot (\frac{\pi }{2}-\theta )$$ 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
$$\text{trig}(x)=\text{co-trig}(\frac{\pi }{2}-\theta )$$

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)

$$\begin{array}{}\text{(16)}& \begin{array}{rl}\sin ({\displaystyle \frac{\pi }{2}}-\theta )& =\cos \theta \\ \cos ({\displaystyle \frac{\pi }{2}}-\theta )& =\sin \theta \\ \tan ({\displaystyle \frac{\pi }{2}}-\theta )& =\cot \theta \\ \cot ({\displaystyle \frac{\pi }{2}}-\theta )& =\tan \theta \end{array}\end{array}$$ and consequently we have
$$\sec (\frac{\pi }{2}-\theta )=\csc \theta ,\csc (\frac{\pi }{2}-\theta )=\sec \theta .$$

Simplifications When π/2 or π Is Involved

Using the addition formulas, we can show
$$\begin{array}{rl}\sin (\theta +{\displaystyle \frac{\pi }{2}})& =\cos \theta \\ \cos (\theta +{\displaystyle \frac{\pi }{2}})& =-\sin \theta \\ \text{(17)}& \tan (\theta +{\displaystyle \frac{\pi }{2}})& =-\cot \theta \end{array}$$

$$\begin{array}{}\text{(18)}& \sin (\theta +\pi )=-\sin \theta ,\cos (\theta +\pi )=-\sin \theta ,\tan (\theta +\pi )=\tan \theta \end{array}$$

$$\begin{array}{}\text{(19)}& \sin (\pi -\theta )=\sin \theta ,\cos (\pi -\theta )=-\cos \theta ,\tan (\pi -\theta )=-\tan (\pi -\theta )\end{array}$$

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:
$$\sin (\theta -\pi /2)=\sin \theta \stackrel{0}{\cancel{\cos \frac{\pi }{2}}}-\cos \theta \underset{=1}{\underbrace{\sin \frac{\pi }{2}}}=-\cos \theta .$$

(a)	(b)
(c)	(d)

Figure 4

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

The following identities are called the product-to-sum formulas.
$$\begin{array}{}\text{(20)}& \begin{array}{r}\sin \theta \cos \varphi =\frac{1}{2}[\sin (\theta -\varphi )+\sin (\theta +\theta )]\\ \sin \theta \sin \varphi =\frac{1}{2}[\cos (\theta -\varphi )-\cos (\theta +\varphi )]\\ \cos \theta \cos \varphi ={\displaystyle \frac{1}{2}}[\cos (\theta -\varphi )+\cos (\theta +\varphi )]\end{array}\end{array}$$ 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
$$\begin{array}{rl}\frac{1}{2}[\sin (\theta -\varphi )+\sin (\theta +\theta )]& =\frac{1}{2}[\sin \theta \cos \varphi -\cos \theta \sin \varphi +\sin \theta \cos \varphi +\cos \theta \sin \varphi ]\\ & =\frac{1}{2}[2\sin \theta \cos \varphi ]\\ & =\sin \theta \cos \varphi .\end{array}$$
Similarly you can prove the following identities, called sum-to-product formulas, by expanding the right hand sides
$$\begin{array}{}\text{(21)}& \begin{array}{r}\sin \theta +\sin \varphi =2\sin {\displaystyle \frac{\theta +\varphi }{2}}\cos {\displaystyle \frac{\theta -\varphi }{2}}\\ \sin \theta -\sin \varphi =2\cos {\displaystyle \frac{\theta +\varphi }{2}}\sin {\displaystyle \frac{\theta -\varphi }{2}}\\ \cos \theta +\cos \varphi =2\cos {\displaystyle \frac{\theta +\varphi }{2}}\cos {\displaystyle \frac{\theta -\varphi }{2}}\\ \cos \theta -\cos \varphi =-2\sin {\displaystyle \frac{\theta +\varphi }{2}}\sin {\displaystyle \frac{\theta -\varphi }{2}}\end{array}\end{array}$$

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 $C$.

Figure 5

Law of cosines:
$$\begin{array}{}\text{(22)}& \begin{array}{r}{a}^2=b^2+c^2-2bc\cos A\\ b^2=a^2+c^2-2ac\cos B\\ c^2=a^2+b^2-2ab\cos C\end{array}\end{array}$$

Law of sines:

$$\begin{array}{}\text{(23)}& \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}\end{array}$$

Show the proofs

Hide 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

$$\begin{array}{rlr}{c}^2& =B{H}^2+A{H}^2& (\text{in right }\mathrm{△}AHB)\\ & =(BC-HC{)}^2+A{H}^2\\ & =B{C}^2-2BC\cdot HC+H{C}^2+A{H}^2& (\text{expanding }(BC-HC{)}^2)\\ & =B{C}^2-2BC\cdot HC+A{C}^2& (\text{in right }\mathrm{△}AHC:A{H}^2+H{C}^2=A{C}^2)\\ & =a^2-2a\cdot HC+b^2& (BC=a,AC=b)\\ & =a^2-2ab\cos C+b^2& (HC=AC\cdot \cos C=b\cos C)\\ & =a^2+b^2-2ab\cos C\end{array}$$
If angle $C$ is obtuse (Figure 6(b)), then $BH=BC+HC$, but $HC=b\cos (\pi -C)=-b\cos C$. Therefore
$$\begin{array}{rl}{c}^2& =(BC+HC{)}^2+A{H}^2\\ & =(a+(-b\cos C){)}^2+A{H}^2\end{array}$$
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.

(a)	(b)

Figure 6

To prove the law of sines, in Figure 6(a) we note that
$$\begin{array}{rlr}AH& =c\sin B& (\text{in }\mathrm{△}BHA)\\ & =b\sin C& (\text{in }\mathrm{△}CHA)\end{array}$$
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
$$\begin{array}{rl}AH& =c\sin B\\ & =b\sin (\pi -C)\end{array}$$
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.

• For more identities see https://en.wikipedia.org/wiki/Proofs_of_trigonometric_identities.