Trigonometric Functions and Identities
5.0 Trigonometric functions of sum and difference of two angles
5.0 Trigonometric functions of sum and difference of two angles
1. $\cos (A + B) = \cos A\cos B - \sin A\sin B$
Proof: Consider the unit circle with centre at the origin ${O(0,0)}$. Let $A$ be the angle $P_4OP_1$ and $B$ be the angle $P_1OP_2$. Then $(A+B)$ is the angle $P_4OP_2$. Also let us assume $(– B)$ be the angle $P_4OP_3$.
Therefore, $P_1$, $P_2$, $P_3$ and $P_4$ will have the coordinates $P_1(cosA,sinA)$, $P_2 [cos(A + B), sin(A + B)]$, $P_3[cos (– B), sin (–B)]$ and $P_4(1,0)$ as shown in figure.
Consider the two congurent triangles $P_1OP_3$ and $P_2OP_4$.
Therefore,
$P_1P_3$ and $P_2P_4$ are equal. By using distance formula, we get
$$\begin{equation} \begin{aligned} {P_1}{P_3}^2 = {\left[ {\cos A - \cos ( - B)} \right]^2} + {\left[ {\sin A - \sin ( - B)} \right]^2} \\ {\text{ }}{\text{ }} = {\left( {\cos A - \cos B} \right)^2} + {\left( {\sin A + \sin B} \right)^2} \\ {\text{ }} = {\cos ^2}A + {\cos ^2}B - 2\cos A\cos B + {\sin ^2}A + {\sin ^2}B + 2\sin A\sin B \\ \because {\sin ^2}A + {\cos ^2}A = 1{\text{ and }}{\sin ^2}B + {\cos ^2}B = 1 \\ {P_1}{P_3}^2 = 2 - 2(\cos A\cos B - \sin A\sin B) \\\end{aligned} \end{equation} $$ and $$\begin{equation} \begin{aligned} {P_2}{P_4}^2 = {\left[ {1 - \cos (A + B)} \right]^2} + {\left[ {0 - \sin (A + B)} \right]^2} \\ {\text{ }}{\text{ }} = 1 - 2\cos (A + B) + {\cos ^2}(A + B) + {\sin ^2}(A + B) \\ {\text{ }} = 2 - 2\cos (A + B) \\\end{aligned} \end{equation} $$ As all the four points $P_1$, $P_2$, $P_3$ and $P_4$ lie on a circle and $P_1P_3=P_2P_4\ (radius)$. On squaring both sides, we get $$\begin{equation} \begin{aligned} {P_1}{P_3}^2 = {P_2}{P_4}^2 \\ \Rightarrow 2 - 2(\cos A\cos B - \sin A\sin B) = 2 - 2\cos (A + B) \\\end{aligned} \end{equation} $$ Therefore, we get $$\cos (A + B) = \cos A\cos B - \sin A\sin B$$
2. $\cos (A - B) = \cos A\cos B + \sin A\sin B$
Proof: To solve this, replace $B$ by $-B$ in the previous identity $1$, we get
$$\begin{equation} \begin{aligned} \cos (A - B) = \cos A\cos \left( { - B} \right) - \sin A\sin \left( { - B} \right) \\ \because \cos ( - B) = \cos B{\text{ and }}\sin ( - B) = - sin(B) \\ \Rightarrow \cos (A - B){\text{ = }}\cos A\cos B + \sin A\sin B \\\end{aligned} \end{equation} $$
3. $\cos \left( {\frac{\pi }{2} - x} \right) = \sin x$
Proof: Put ${A = \frac{\pi }{2}}$ and $B=-x$ in identity $2$, we get $$\begin{equation} \begin{aligned} \cos \left( {\frac{\pi }{2} - x} \right) = \cos \frac{\pi }{2}\cos \left( { - x} \right) - \sin \frac{\pi }{2}\sin \left( { - x} \right) \\ \because \cos ( - x) = \cos x{\text{ and }}\sin ( - x) = - sin(x) \\ \Rightarrow \cos \left( {\frac{\pi }{2} - x} \right){\text{ = }}\cos \frac{\pi }{2}\cos \left( x \right) + \sin \frac{\pi }{2}\sin \left( x \right) \\ \Rightarrow \cos \left( {\frac{\pi }{2} - x} \right){\text{ = }}\left( {0 \times \cos x} \right) + \left( {1 \times \sin x} \right) \\ \therefore \cos \left( {\frac{\pi }{2} - x} \right){\text{ = }}\sin x \\\end{aligned} \end{equation} $$
4. $\sin \left( {\frac{\pi }{2} - x} \right) = \cos x$
Proof: Using identity 3, we can write $$\begin{equation} \begin{aligned} \sin \left( {\frac{\pi }{2} - x} \right) = \cos \left[ {\frac{\pi }{2} - \left( {\frac{\pi }{2} - x} \right)} \right] \\ {\text{ = }}\sin \left( {\frac{\pi }{2} - x} \right) = \cos x \\\end{aligned} \end{equation} $$
5. $\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B$
Proof: We can write $$\sin \left( {A + B} \right) = \cos \left[ {\frac{\pi }{2} - (A + B)} \right] = \cos \left[ {\left( {\frac{\pi }{2} - A} \right) - B} \right]$$ Now, we apply the identity $2$, we get $$\begin{equation} \begin{aligned} \sin \left( {A + B} \right) = \cos \left( {\frac{\pi }{2} - A} \right)\cos B + \sin \left( {\frac{\pi }{2} - A} \right)\sin B \\ \sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B \\\end{aligned} \end{equation} $$
6. $\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B$
Proof: Replace $B$ by $-B$ in identity $5$, we get $$\begin{equation} \begin{aligned} \sin \left( {A - B} \right) = \cos \left[ {\frac{\pi }{2} - (A - B)} \right] = \cos \left[ {\left( {\frac{\pi }{2} - A} \right) + B} \right] \\ \sin \left( {A - B} \right) = \cos \left( {\frac{\pi }{2} - A} \right)\cos B - \sin \left( {\frac{\pi }{2} - A} \right)\sin B \\ \sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B \\\end{aligned} \end{equation} $$
7. $\tan (A + B) = \frac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$
Proof: The above identity is true only when the angles $A$, $B$ and $(A+B)$ is not an odd multiple of $\frac{\pi }{2}$. This condition is applied because if we break $tanA$, $tanB$ or $tan(A+B)$ in terms of other trigonometric functions $sin$ and $cos$, then we get function $cos$ in the denominator and if it is an odd multiple of $\frac{\pi }{2}$, then function $cos$ becomes $0$. Therefore, we can write $$\begin{equation} \begin{aligned} \tan (A + B) = \frac{{\sin (A + B)}}{{\cos (A + B)}} \\ \tan (A + B) = \frac{{\sin A\cos B + \cos A\sin B}}{{\cos A\cos B - \sin A\sin B}} \\\end{aligned} \end{equation} $$ Divide numerator and denominator by $cosAcosB$, we get $$\begin{equation} \begin{aligned} \tan (A + B) = \frac{{\frac{{sinA\cos B + \cos A\sin B}}{{\cos A\cos B}}}}{{\frac{{\cos A\cos B - \sin A\sin B}}{{\cos A\cos B}}}} \\ \tan (A + B) = \frac{{\frac{{sinA\cos B}}{{\cos A\cos B}} + \frac{{\cos A\sin B}}{{\cos A\cos B}}}}{{\frac{{\cos A\cos B}}{{\cos A\cos B}} - \frac{{\sin A\sin B}}{{\cos A\cos B}}}} \\ \tan (A + B) = \frac{{\tan A + \tan B}}{{1 - \tan A\tan B}} \\\end{aligned} \end{equation} $$
8. $\tan (A - B) = \frac{{\tan A - \tan B}}{{1 + \tan A\tan B}}$
Proof: Replace $B$ by $-B$ in identity $7$, we get $$\begin{equation} \begin{aligned} \tan (A + ( - B)) = \frac{{\tan A + \tan ( - B)}}{{1 - \tan A\tan ( - B)}} \\ \because \tan ( - B) = - \tan B \\ \tan (A - B) = \frac{{\tan A - \tan B}}{{1 + \tan A\tan B}} \\\end{aligned} \end{equation} $$
9. $\cot (A + B) = \frac{{\cot A\cot B - 1}}{{\cot B + \cot A}}$
Proof: The above identity is true only when the angles $A$, $B$ and $(A+B)$ is not a multiple of $\pi$. Since , none of the angles $A$, $B$ and $(A+B)$ is multiple of $\pi$, we can say that $sinA$, $sinB$ and $sin(A+B)$ are non-zero. Therefore, we can write $$\cot (A + B) = \frac{{\cos (A + B)}}{{\sin (A + B)}} = \frac{{\cos A\cos B - \sin A\sin B}}{{\sin A\cos B + \cos A\sin B}}$$ Divide numerator and denominator by $sinA.sinB$, we get $$\begin{equation} \begin{aligned} \cot (A + B) = \frac{{\frac{{\cos A\cos B}}{{\sin A\sin B}} - \frac{{\sin A\sin B}}{{\sin A\sin B}}}}{{\frac{{\sin A\cos B}}{{\sin A\sin B}} + \frac{{\cos A\sin B}}{{\sin A\sin B}}}} \\ \cot (A + B) = \frac{{\cot A\cot B - 1}}{{\cot B + \cot A}} \\\end{aligned} \end{equation} $$
10. $\cot (A - B) = \frac{{\cot A\cot B + 1}}{{\cot B - \cot A}}$
Proof: Replace $B$ by $-B$ in identity $9$, we get $$\begin{equation} \begin{aligned} \cot (A + ( - B)) = \frac{{\cot A\cot ( - B) - 1}}{{\cot ( - B) + \cot A}} \\ \because \cot ( - B) = - \cot B \\ \cot (A - B) = \frac{{ - \cot A\cot B - 1}}{{ - \cot B + \cot A}} \\ \cot (A - B) = \frac{{ - (\cot A\cot B + 1)}}{{ - (\cot B - \cot A)}} \\ \cot (A - B) = \frac{{\cot A\cot B + 1}}{{\cot B - \cot A}} \\\end{aligned} \end{equation} $$
11. $\cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta = 2{\cos ^2}\theta - 1 = 1 - 2{\sin ^2}\theta = \frac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}$
Proof: From identity $(1)$ we have $$\cos (A + B) = \cos A\cos B - \sin A\sin B$$ Replace $B$ by $A$ we get $$\begin{equation} \begin{aligned} \cos (A + A) = \cos A\cos A - \sin A\sin A \\ \cos 2A = {\cos ^2}A - {\sin ^2}A \\\end{aligned} \end{equation} $$ Put $A=\theta$, we get $$\cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta ...(1)$$ As we know $${\cos ^2}\theta + {\sin ^2}\theta = 1$$ Put the value of ${sin^2}\theta$ in equation $(1)$, we get $$\begin{equation} \begin{aligned} \cos 2\theta = {\cos ^2}\theta - \left( {1 - {{\cos }^2}\theta } \right) \\ \cos 2\theta = {\cos ^2}\theta - 1 + {\cos ^2}\theta \\ \cos 2\theta = 2{\cos ^2}\theta - 1 \\\end{aligned} \end{equation} $$ Now, Put the value of ${cos^2}\theta$ in equation $(1)$, we get $$\begin{equation} \begin{aligned} \cos 2\theta = \left( {1 - {{\sin }^2}\theta } \right) - {\sin ^2}\theta \\ \cos 2\theta = 1 - 2{\sin ^2}\theta \\\end{aligned} \end{equation} $$ Now, we can write equation $(1)$ as $$\cos 2\theta = \frac{{{{\cos }^2}\theta - {{\sin }^2}\theta }}{1} = \frac{{{{\cos }^2}\theta - {{\sin }^2}\theta }}{{{{\cos }^2}\theta + {{\sin }^2}\theta }}$$ Divide both numerator and denominator by ${cos^2}\theta$, we get $$\cos 2\theta = \frac{{\frac{{{{\cos }^2}\theta }}{{{{\cos }^2}\theta }} - \frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}}}{{\frac{{{{\cos }^2}\theta }}{{{{\cos }^2}\theta }} + \frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}}} = \frac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}$$
12. $\sin 2\theta = 2\sin \theta \cos \theta = \frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}$
Proof: From identity $(5)$ we have $$\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B$$ Replace $B$ by $A$ we get $$\begin{equation} \begin{aligned} \sin \left( {A + A} \right) = \sin A\cos A + \cos A\sin A \\ \sin 2A = 2\sin A\cos A \\\end{aligned} \end{equation} $$ Put $A=\theta$, we get $$\sin 2\theta = 2\sin \theta \cos \theta ...(1)$$ Now, we can write equation $(1)$ as $$\sin 2\theta = \frac{{2\sin \theta \cos \theta }}{1} = \frac{{2\sin \theta \cos \theta }}{{{{\sin }^2}\theta + {{\cos }^2}\theta }}$$ Divide each term by ${{{\cos }^2}\theta }$ we get $$\sin 2\theta = \frac{{\frac{{2\sin \theta \cos \theta }}{{{{\cos }^2}\theta }}}}{{\frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }} + \frac{{{{\cos }^2}\theta }}{{{{\cos }^2}\theta }}}} = \frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}$$
13. $\tan 2\theta = \frac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}$
Proof: From identity $(7)$, we have $$\tan (A + B) = \frac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$$ Replace $B$ by $A$, we get $$\tan 2A = \frac{{\tan A + \tan A}}{{1 - \tan A\tan A}} = \frac{{2\tan A}}{{1 - {{\tan }^2}A}}$$ Put $A=\theta$ we get $$\tan 2\theta = \frac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}$$
14. $\sin 3\theta = 3\sin \theta - 4{\sin ^3}\theta $
Proof: We can write it as $$\sin 3\theta = \sin \left( {2\theta + \theta } \right)$$ Now apply identity $(5)$ i.e., $$\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B$$ Put $A=2\theta$ and $B=\theta$, we get $$\sin 3\theta = \sin 2\theta \cos \theta + \cos 2\theta \sin \theta $$ Now apply identity $(11)$ and $(12)$, we get $$\begin{equation} \begin{aligned} \sin 3\theta = \left( {2\sin \theta \cos \theta } \right)\cos \theta + \left( {1 - 2{{\sin }^2}\theta } \right)\sin \theta \\ \sin 3\theta = 2\sin \theta {\cos ^2}\theta + \sin \theta - 2{\sin ^3}\theta \\ \sin 3\theta = 2\sin \theta \left( {1 - {{\sin }^2}\theta } \right) + \sin \theta - 2{\sin ^3}\theta \\ \sin 3\theta = 2\sin \theta - 2{\sin ^3}\theta + \sin \theta - 2{\sin ^3}\theta \\ \sin 3\theta = 3\sin \theta - 4{\sin ^3}\theta \\\end{aligned} \end{equation} $$
15. $\cos 3\theta = 4{\cos ^3}\theta - 3\cos \theta $
Proof: We can write it as $$\cos 3\theta = \cos \left( {2\theta + \theta } \right)$$ Now apply identity $(1)$ i.e., $$\cos (A + B) = \cos A\cos B - \sin A\sin B$$ Put $A=2\theta$ and $B=\theta$, we get $$\begin{equation} \begin{aligned} \cos (2\theta + \theta ) = \cos 2\theta \cos \theta - \sin 2\theta \sin \theta \\ \cos 3\theta = \left( {2{{\cos }^2}\theta - 1} \right)\cos \theta - \left( {2\sin \theta \cos \theta } \right)\sin \theta \\ \cos 3\theta = 2{\cos ^3}\theta - \cos \theta - 2{\sin ^2}\theta \cos \theta \\ \cos 3\theta = 2{\cos ^3}\theta - \cos \theta - 2(1 - {\cos ^2}\theta )\cos \theta \\ \cos 3\theta = 2{\cos ^3}\theta - \cos \theta - 2\cos \theta + 2{\cos ^3}\theta \\ \cos 3\theta = 4{\cos ^3}\theta - 3\cos \theta \\\end{aligned} \end{equation} $$
16. $\tan 3\theta = \frac{{3\tan \theta - {{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }}$
Proof: We can write it as $$\tan 3\theta = \tan (2\theta + \theta )$$ Now we apply identity $(7)$ i.e., $$\tan (A + B) = \frac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$$ Put $A=2\theta$ and $B=\theta$, we get $$\tan (2\theta + \theta ) = \frac{{\tan 2\theta + \tan \theta }}{{1 - \tan 2\theta \tan \theta }}$$ Apply identity $(13)$, we get $$\begin{equation} \begin{aligned} \tan (2\theta + \theta ) = \frac{{\frac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }} + \tan \theta }}{{1 - \frac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}.\tan \theta }} \\ \tan (3\theta ) = \frac{{\frac{{2\tan \theta + \tan \theta - {{\tan }^3}\theta }}{{1 - {{\tan }^2}\theta }}}}{{\frac{{1 - {{\tan }^2}\theta - 2{{\tan }^2}\theta }}{{1 - {{\tan }^2}\theta }}}} = \frac{{3\tan \theta - {{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }} \\\end{aligned} \end{equation} $$
