### 1 Trigonometric identities

An identity is a relation which is always true. To emphasise this the symbol ‘ $\equiv$ ’ is often used rather than ‘ $=$ ’. For example, ${\left(x+1\right)}^{2}\equiv {x}^{2}+2x+1$ (always true) but ${\left(x+1\right)}^{2}=0$ (only true for $x=-1$ ).

1. Using the exact values, evaluate ${sin}^{2}\theta +{cos}^{2}\theta$ for (i) $\theta =3{0}^{\circ }$ (ii) $\theta =4{5}^{\circ }$

[Note that ${sin}^{2}\theta$ means ${\left(sin\theta \right)}^{2},\phantom{\rule{1em}{0ex}}{cos}^{2}\theta$ means ${\left(cos\theta \right)}^{2}$ ]

2. Choose a non-integer value for $\theta$ and use a calculator to evaluate ${sin}^{2}\theta +{cos}^{2}\theta$ .
1. (i)   ${sin}^{2}3{0}^{\circ }+{cos}^{2}3{0}^{\circ }={\left(\frac{1}{2}\right)}^{2}+{\left(\frac{\sqrt{3}}{2}\right)}^{2}=\frac{1}{4}+\frac{3}{4}=1$

(ii)   ${sin}^{2}4{5}^{\circ }+{cos}^{2}4{5}^{\circ }={\left(\frac{1}{\sqrt{2}}\right)}^{2}+{\left(\frac{1}{\sqrt{2}}\right)}^{2}=\frac{1}{2}+\frac{1}{2}=1$

2. The answer should be $1$ whatever value you choose.
##### Key Point 12

For any value of $\theta$

${sin}^{2}\theta +{cos}^{2}\theta \equiv 1\left(5\right)$

One way of proving the result in Key Point 12 is to use the definitions of $sin\theta$ and $cos\theta$ obtained from the circle of unit radius. Refer back to Figure 22 on page 23.

Recall that   $cos\theta =OQ$ , $\phantom{\rule{1em}{0ex}}sin\theta =OR=QP$ . By Pythagoras’ theorem

$\phantom{\rule{2em}{0ex}}{\left(OQ\right)}^{2}+{\left(QP\right)}^{2}={\left(OP\right)}^{2}=1$

hence ${cos}^{2}\theta +{sin}^{2}\theta =1.$

We have demonstrated the result (5) using an angle $\theta$ in the first quadrant but the result is true for any $\theta$ i.e. it is indeed an identity.

By dividing the identity ${sin}^{2}\theta +{cos}^{2}\theta \equiv 1$ by

1. ${sin}^{2}\theta$
2. ${cos}^{2}\theta$ obtain two further identities.

[Hint: Recall the definitions of $cosec\phantom{\rule{0.3em}{0ex}}\theta ,\phantom{\rule{1em}{0ex}}sec\phantom{\rule{0.3em}{0ex}}\theta ,\phantom{\rule{1em}{0ex}}cot\phantom{\rule{0.3em}{0ex}}\theta$ .]

1. $\phantom{\rule{1em}{0ex}}\frac{{sin}^{2}\theta }{{sin}^{2}\theta }+\frac{{cos}^{2}\theta }{{sin}^{2}\theta }=\frac{1}{{sin}^{2}\theta }$
2. $\frac{{sin}^{2}\theta }{{cos}^{2}\theta }+\frac{{cos}^{2}\theta }{{cos}^{2}\theta }=\frac{1}{{cos}^{2}\theta }$

$\phantom{\rule{1em}{0ex}}\phantom{\rule{2em}{0ex}}1+\phantom{\rule{1em}{0ex}}{cot}^{2}\theta \equiv {cosec}^{2}\theta$ $\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}{tan}^{2}\theta +1\equiv {sec}^{2}\theta$

Key Point 13 introduces two further important identities.

##### Key Point 13
$sin\left(A+B\right)\equiv sinAcosB+cosAsinB\phantom{\rule{2em}{0ex}}\left(6\right)$
$cos\left(A+B\right)\equiv cosAcosB-sinAsinB\phantom{\rule{2em}{0ex}}\left(7\right)$

Further identities can readily be obtained from (6) and (7).

Dividing (6) by (7) we obtain

$\phantom{\rule{2em}{0ex}}tan\left(A+B\right)\equiv \frac{sin\left(A+B\right)}{cos\left(A+B\right)}\equiv \frac{sinAcosB+cosAsinB}{cosAcosB-sinAsinB}$

Dividing every term by $cosA\phantom{\rule{1em}{0ex}}cosB$ we obtain

$\phantom{\rule{2em}{0ex}}tan\left(A+B\right)\equiv \frac{tanA+tanB}{1-tanAtanB}$

Replacing $B$ by $-B$ in (6) and (7) and remembering that $cos\left(-B\right)\equiv cosB,\phantom{\rule{1em}{0ex}}sin\left(-B\right)\equiv -sinB$ we find

$\phantom{\rule{2em}{0ex}}sin\left(A-B\right)\equiv sinAcosB-cosAsinB$

$\phantom{\rule{2em}{0ex}}cos\left(A-B\right)\equiv cosAcosB+sinAsinB$

Using the identities $sin\left(A-B\right)\equiv sinAcosB-cosAsinB$ and

$cos\left(A-B\right)\equiv cosAcosB+sinAsinB$ obtain an expansion for $tan\left(A-B\right)$ :

$\phantom{\rule{2em}{0ex}}tan\left(A-B\right)\equiv \frac{sinAcosB-cosAsinB}{cosAcosB+sinAsinB}$ .

Dividing every term by $cosA\phantom{\rule{1em}{0ex}}cosB$ gives

$\phantom{\rule{2em}{0ex}}tan\left(A-B\right)\equiv \frac{tanA-tanB}{1+tanAtanB}$

The following identities are derived from those in Key Point 13.

##### Key Point 14

$\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}tan\left(A+B\right)\equiv \frac{tanA+tanB}{1-tanAtanB}\left(8\right)$

$\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}sin\left(A-B\right)\equiv sinAcosB-cosAsinB\left(9\right)$

$\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}cos\left(A-B\right)\equiv cosAcosB+sinAsinB\left(10\right)$

$\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}tan\left(A-B\right)\equiv \frac{tanA-tanB}{1+tanAtanB}\left(11\right)$