Question 11 Mark
If $\sin\text{A}+\sin\text{B}=\alpha\text{ and }\cos\text{A}+\cos\text{B}=\beta,$
than write the value of $\tan\Big(\frac{\text{A+B}}{2}\Big).$
AnswerWe have, $\sin\text{A}+\sin\text{B}=\alpha...(\text{i})$ $\text{and },\cos\text{A}+\cos\text{B}=\beta...(\text{ii})$ Now, $\frac{\sin\text{A}+\sin\text{B}}{\cos\text{A}+\cos\text{B}}=\frac{\alpha}{\beta}$ $\Rightarrow\ \frac{2\sin\Big(\frac{\text{A+B}}{2}\Big)\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)}{2\cos\Big(\frac{\text{A+B}}{2}\Big)\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)}=\frac{\alpha}{\beta}$$\Rightarrow\ \frac{\sin\Big(\frac{\text{A+B}}{2}\Big)}{\cos\Big(\frac{A+B}{2}\Big)}=\frac{\alpha}{\beta}$
$\Rightarrow\ \tan\Big(\frac{\text{A+B}}{2}\Big)=\frac{\alpha}{\beta}$
View full question & answer→Question 21 Mark
Write the value of $\frac{\sin\text{A}+\sin3\text{A}}{\cos\text{A}+\cos3\text{A}}.$
AnswerWe have,
$\frac{\sin\text{A}+\sin3\text{A}}{\cos\text{A}+\cos3\text{A}}=\frac{\sin3\text{A}+\sin\text{A}}{\cos3\text{A}+\cos\text{A}}$
$=\ \frac{2\sin\Big(\frac{3\text{A}+\text{A}}{2}\Big)\cos\Big(\frac{3\text{A}-\text{A}}{2}\Big)}{2\cos\Big(\frac{3\text{A}+\text{A}}{2}\Big)\cos\Big(\frac{3\text{A}-\text{A}}{2}\Big)}$
$=\ \frac{\sin2\text{A}\cos\text{A}}{\cos2\text{A}\cos\text{A}}$
$=\ \frac{\sin2\text{A}}{\cos2\text{A}}$
$=\ \tan2\text{A}$
$\therefore\ \frac{\sin\text{A}+\sin3\text{A}}{\cos\text{A}+\cos3\text{A}}=\tan2\text{A}.$
View full question & answer→Question 31 Mark
If $\text{A}+\text{B}=\frac{\pi}{3}\text{ and }\cos\text{A}+\cos\text{B}=1,$ than find the value of $\cos\frac{\text{A}-\text{B}}{2}.$
AnswerWe have,
$\text{A+B}=\frac{\pi}{3}$
$\text{and },\cos\text{A}+\cos\text{B}=1$
Now,
$\cos\text{A}+\cos\text{B}=1$
$\Rightarrow\ 2\cos\Big(\frac{\text{A+B}}{2}\Big)\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)=1$
$\Rightarrow\ 2\cos\Big(\frac{1}{2}\times\frac{\pi}{3}\Big)\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)=1$ $\Big[\because\ \text{A+B}=\frac{\pi}{3}\Big]$
$\Rightarrow\ 2\cos\frac{\pi}{6}\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)=1$
$\Rightarrow\ 2\times\frac{\sqrt3}{2}\times\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)=1$
$\Rightarrow\ \sqrt3\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)=1$
$\Rightarrow\ \cos\Big(\frac{\text{A}-\text{B}}{2}\Big)=\frac{1}{\sqrt3}$
$\text{Hence},\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)=\frac{1}{\sqrt3}.$
View full question & answer→Question 41 Mark
Express the following as the product of sines and cosines:
$\sin5\text{x}-\sin\text{x}$
Answer$\sin5\text{x}-\sin\text{x}$ $\Big[\because\ \sin\text{C}+\sin\text{D}=2\sin\frac{\text{C+D}}{2}\sin\frac{\text{C}-\text{D}}{2}\Big]$
$=\ 2\cos\Big(\frac{5\text{x}+\text{x}}{2}\Big)\sin\Big(\frac{5\text{x}-\text{x}}{2}\Big)$
$=\ 2\sin2\text{x}\cos3\text{x}$
View full question & answer→Question 51 Mark
$\text{If }\cos(\text{A+B})\sin(\text{C}-\text{D})=\cos(\text{A}-\text{B})\sin(\text{C+D}),$
than write the value $\tan\text{A}\tan\text{B}\tan\text{C}.$
AnswerWe have,
$\cos(\text{A+B})\sin(\text{C}-\text{D})=\cos(\text{A}-\text{B})\sin(\text{C+D})$
$\Rightarrow\ \frac{\cos(\text{A+B})}{\cos(\text{A}-\text{B})}=\frac{\sin(\text{C+D})}{\sin(\text{C}-\text{D})}...(\text{i})$
$\Rightarrow\ \frac{\cos(\text{A+B})}{\cos(\text{A}-\text{B})}+1=\frac{\sin(\text{C+D})}{\sin(\text{C}-\text{D})}+1$
$\Rightarrow\ \frac{\cos(\text{A+B})+\cos(\text{A}-\text{B})}{\cos(\text{A}-\text{B})}=\frac{\sin(\text{C+D})+\sin(\text{C}-\text{D})}{\sin(\text{C}-\text{D})}...(\text{ii})$
Again,
$\frac{\cos(\text{A+B})}{\cos(\text{A}-\text{B})}=\frac{\sin(\text{C+D})}{\sin(\text{C}-\text{D})}$ [By equation (i)]
$\Rightarrow\ \frac{\cos(\text{A+B})}{\cos(\text{A}-\text{B})}-1=\frac{\sin(\text{C+D})}{\sin(\text{C}-\text{D})}-1$
$\Rightarrow\ \frac{\cos(\text{A+B})-\cos(\text{A}-\text{B})}{\cos(\text{A}-\text{B})}=\frac{\sin(\text{C+D})-\sin(\text{C}-\text{D})}{\sin(\text{C}-\text{D})}...(\text{iii})$
Dividing equation (ii) by equation (iii), we get
$\frac{\cos(\text{A+B})+\cos(\text{A}-\text{B})}{\cos(\text{A+B})-\cos(\text{A}-\text{B})}=\frac{\sin(\text{C+D})+\sin(\text{C}-\text{D})}{\sin(\text{C+D})-\sin(\text{C}-\text{D})}$
$\Rightarrow\ \frac{2\cos\Big\{\frac{\text{A+B+A}-\text{B}}{2}\Big\}\cos\Big\{\frac{\text{A+B}-\text{A+B}}{2}\Big\}}{-2\sin\Big\{\frac{\text{A+B+A}-\text{B}}{2}\Big\}\sin\Big\{\frac{\text{A+B}-\text{A+B}}{2}\Big\}}\\ \ \ \ \ =\frac{2\sin\Big\{\frac{\text{C+D+C}-\text{D}}{2}\Big\}\cos\Big\{\frac{\text{C+D}-\text{C+D}}{2}\Big\}}{2\sin\Big\{\frac{\text{C+D}-\text{C+D}}{2}\Big\}\cos\Big\{\frac{\text{C+D+C}-\text{D}}{2}\Big\}}$
$\Rightarrow\ \frac{\cos\text{A}\cos\text{B}}{-\sin\text{A}\sin\text{B}}=\frac{\sin\text{C}\sin\text{D}}{\sin\text{D}\sin\text{C}}$
$\Rightarrow\ \frac{1}{-\tan\text{A}\tan\text{B}}=\frac{\tan\text{C}}{\tan\text{D}}$
$\Rightarrow\ -\tan\text{D}=\tan\text{A}\tan\text{B}\tan\text{C}$
$\therefore\ \tan\text{A}\tan\text{B}\tan\text{C}=-\tan\text{D}.$
View full question & answer→Question 61 Mark
Prove that:
$2\cos\frac{5\pi}{12}\cos\frac{\pi}{12}=\frac{1}{2}$
Answer$\text{LHS}=2\cos\frac{5\pi}{12}\cos\frac{\pi}{12}$
$\because\ 2\cos\text{A}\cos\text{B}=\cos(\text{A+B})+\cos(\text{A}-\text{B})$
$=\ \cos\Big(\frac{5\pi}{12}+\frac{\pi}{12}\Big)+\cos\Big(\frac{5\pi}{12}-\frac{\pi}{12}\Big)$
$=\ \cos\Big(\frac{\pi}{2}\Big)+\cos\Big(\frac{\pi}{3}\Big)$
$=\ 0+\frac{1}{2}=\frac{1}{2}=\text{RHS}$
View full question & answer→Question 71 Mark
If $(\cos\alpha+\cos\beta)^2+(\sin\alpha+\sin\beta)^2=\lambda\cos^2\Big(\frac{\alpha-\beta}{2}\Big),$ write the value of $\lambda.$
AnswerWe have,
$\text{LHS}=(\cos\alpha+\cos\beta)^2+(\sin\alpha+\sin\beta)^2$
$=\ \Big[2\cos\Big(\frac{\alpha+\beta}{2}\Big)\cos\Big(\frac{\alpha-\beta}{2}\Big)\Big]^2+\Big[2\sin\Big(\frac{\alpha+\beta}{2}\Big)\cos\Big(\frac{\alpha-\beta}{2}\Big)\Big]^2$
$=\ 4\cos^2\Big(\frac{\alpha+\beta}{2}\Big)\cos^2\Big(\frac{\alpha-\beta}{2}\Big)+4\sin^2\Big(\frac{\alpha+\beta}{2}\Big)\cos^2\Big(\frac{\alpha-\beta}{2}\Big)$
$=\ 4\cos^2\Big(\frac{\alpha-\beta}{2}\Big)\Big[\cos^2\Big(\frac{\alpha+\beta}{2}\Big)+\sin^2\Big(\frac{\alpha+\beta}{2}\Big)\Big]$
$=\ 4\cos^2\Big(\frac{\alpha-\beta}{2}\Big)$ $[\because\ \cos^2\theta+\sin^2\theta=1]$
$\Rightarrow\ (\cos\alpha+\cos\beta)^2+(\sin\alpha+\sin\beta)^2=4\cos^2\Big(\frac{\alpha-\beta}{2}\Big)$
It is given that,
$(\cos\alpha+\cos\beta)^2+(\sin\alpha+\cos\beta)^2=\lambda\cos^2\Big(\frac{\alpha-\beta}{2}\Big)$
On comparing, we get
$\lambda=4$
$\therefore\ \lambda=4$
View full question & answer→Question 81 Mark
Prove that:
$2\sin\frac{5\pi}{12}\sin\frac{\pi}{12}=\frac{1}{2}$
Answer$\text{LHS}=2\sin\frac{5\pi}{12}\sin\frac{\pi}{12}$
$\because\ 2\sin\text{A}\sin\text{B}=\cos(\text{A}-\text{B})-\cos(\text{A+B})$
$\Rightarrow\ 2\sin\frac{5\pi}{12}\sin\frac{\pi}{12}=\cos\Big(\frac{5\pi}{12}-\frac{\pi}{12}\Big)-\cos\Big(\frac{5\pi}{12}+\frac{\pi}{12}\Big)$
$=\ \cos\Big(\frac{4\pi}{12}\Big)-\cos\Big(\frac{6\pi}{12}\Big)$
$=\ \cos\Big(\frac{\pi}{3}\Big)-\cos\Big(\frac{6\pi}{12}\Big)$
$=\ \frac{1}{2}-0=\frac{1}{2}=\text{RHS}$
View full question & answer→Question 91 Mark
Write the value of $\sin\frac{\pi}{15}\sin\frac{4\pi}{15}\sin\frac{3\pi}{10}.$
Answer$\sin12^\circ\sin48^\circ\sin54^\circ$
$=\ \frac{1}{2}(2\sin12^\circ\sin48^\circ)\sin54^\circ$
$=\ \frac{1}{2}[\cos(48^\circ-12^\circ)-\cos(48^\circ+12^\circ)]\sin54^\circ$ $\{\text{Since }2\sin\text{A}\cos\text{B}=\cos(\text{A}-\text{B})-\cos(\text{A+B})\}$
$=\ \frac{1}{2}[\cos36^\circ-\cos60^\circ]\sin54^\circ$
$=\ \frac{1}{2}\Big[\cos36^\circ\sin54^\circ-\frac{1}{2}\sin45^\circ\Big]$
$=\ \frac{1}{4}[2\cos36^\circ\sin54^\circ-\sin45^\circ]$
$=\ \frac{1}{4}[\sin(54^\circ+36^\circ)+\sin(54^\circ-36^\circ)-\sin54^\circ]$ $\{\text{Since }2\sin\text{A}\cos\text{B}=\sin(\text{A}+\text{B})+\sin(\text{A}-\text{B})\}$
$=\ \frac{1}{4}[\sin90^\circ+\sin18^\circ-\sin54^\circ]$
$=\ \frac{1}{4}[\sin90^\circ-\sin54^\circ+\sin18^\circ]$
$=\ \frac{1}{4}\Big[1-\frac{\sqrt5+1}{4}+\frac{\sqrt5-1}{4}\Big]$
$\Big\{\text{Since }\sin18^\circ=\frac{\sqrt5-1}{4},\sin54^\circ\frac{\sqrt5+1}{4}\Big\}$
$=\ \frac{1}{4}\Big[\frac{4-\sqrt5-1+\sqrt5-1}{4}\Big]$
$=\ \frac{1}{4}\Big[\frac{2}{4}\Big]$
$=\ \frac{1}{8}$
Thus,
$\sin12^\circ\sin48^\circ\sin54^\circ=\frac{1}{8}$
View full question & answer→Question 101 Mark
Express the following as the sum or difference of sines and cosines:
$2\sin3\text{x}\cos\text{x}$
Answer$2\sin3\text{x}\cos\text{x}$
$=\sin(3\text{x}+\text{x})+\sin(3\text{x}-\text{x})$$[\because2\sin\text{A}\cos\text{B}=\sin(\text{A+B})+\sin(\text{A}-\text{B})]$
$=\sin4\text{x}+\sin2\text{x}$
View full question & answer→Question 111 Mark
Write the value of the expression $\frac{1-4\sin10^\circ\sin70^\circ}{2\sin10^\circ}.$
AnswerWe have,
$\frac{1-4\sin10^\circ\sin70^\circ}{2\sin10^\circ}$
$=\ \frac{1-2(2\sin10^\circ\sin70^\circ)}{2\sin10^\circ}$
$=\ \frac{1-2\big[\cos(70^\circ-10^\circ)-\cos(70^\circ+10^\circ)\big]}{2\sin10^\circ}$
$=\ \frac{1-2(\cos60^\circ-\cos80^\circ)}{2\sin(90^\circ-10^\circ)}$
$=\ \frac{1-2\big(\frac{1}{2}-\cos80^\circ\big)}{2\cos80^\circ}$
$=\ \frac{1-\frac{2}{2}+2\cos80^\circ}{2\cos80^\circ}$
$=\ \frac{1-1+2\cos80^\circ}{\cos80^\circ}$
$=\ \frac{2\cos80^\circ}{2\cos80^\circ}$
$=\ 1$
View full question & answer→Question 121 Mark
If $\cos\text{A}=\text{m}\cos\text{B},$ than write the value of $\cot\frac{\text{A+B}}{2}\cot\frac{\text{A}-\text{B}}{2}.$
AnswerWe have,
$\cos\text{A}=\text{m}\cos\text{B}$
$\frac{\cos\text{A}}{\cos\text{B}}=\text{m}$
$\Rightarrow\ \frac{\cos\text{A}}{\cos\text{B}}+1=\text{m}+1$
$\Rightarrow\ \frac{\cos\text{A}+\cos\text{B}}{\cos\text{B}}=\text{m}+1...(\text{i})$
Again,
$\frac{\cos\text{A}}{\cos\text{B}}=\text{m}$
$\Rightarrow\ \frac{\cos\text{A}}{\cos\text{B}}-1=\text{m}-1$
$\Rightarrow\ \frac{\cos\text{A}-\cos\text{B}}{\cos\text{B}}=\text{m}-1...(\text{ii})$
Dividing equation (i) by equation (ii), we get
$\frac{\frac{\cos\text{A}+\cos\text{B}}{\cos\text{B}}}{\frac{\cos\text{A}-\cos\text{B}}{\cos\text{B}}}=\frac{\text{m}+1}{\text{m}-1}$
$\Rightarrow\ \frac{\cos\text{A}+\cos\text{B}}{\cos\text{A}-\cos\text{B}}=\frac{\text{m}+1}{\text{m}-1}$
$\Rightarrow\ \frac{2\cos\Big(\frac{\text{A+B}}{2}\Big)\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)}{-2\sin\Big(\frac{\text{A+B}}{2}\Big)\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)}=\frac{\text{m}+1}{\text{m}-1}$
$\Rightarrow\ -\cot\Big(\frac{\text{A+B}}{2}\Big)\cot\Big(\frac{\text{A}-\text{B}}{2}\Big)=\frac{\text{m}+1}{\text{m}-1}$
$\Rightarrow\ \cot\Big(\frac{\text{A+B}}{2}\Big)\cot\Big(\frac{\text{A}-\text{B}}{2}\Big)=\frac{1+\text{m}}{1-\text{m}}$
$\therefore\ \cot\Big(\frac{\text{A+B}}{2}\Big)\cot\Big(\frac{\text{A}-\text{B}}{2}\Big)=\frac{1+\text{m}}{1-\text{m}}$
View full question & answer→Question 131 Mark
Express the following as the sum or difference of sines and cosines:
$2\cos7\text{x}\cos3\text{x}$
Answer$2\cos7\text{x}\cos3\text{x}$
$\because\ 2\cos\text{A}\cos\text{B}=\cos(\text{A+B})+\cos(\text{A}-\text{B})$
$\Rightarrow\ 2\cos7\text{x}\cos3\text{x}=\cos(7\text{x}+3\text{x})+\cos(7\text{x}-3\text{x})$
$=\ \cos10\text{x}+\cos4\text{x}$
View full question & answer→Question 141 Mark
Prove that:
$2\sin\frac{5\pi}{12}\cos\frac{\pi}{12}=\frac{\sqrt3+2}{2}$
Answer$\text{LHS}=2\sin\frac{5\pi}{12}\cos\frac{\pi}{12}$
$\because\ 2\sin\text{A}\sin\text{B}=\sin(\text{A+B})+\sin(\text{A}-\text{B})$
$=\ \Big(\frac{5\pi}{12}+\frac{\pi}{12}\Big)+\sin\Big(\frac{5\pi}{12}-\frac{\pi}{12}\Big)$
$=\ \sin\frac{\pi}{2}+\sin\frac{\pi}{3}$
$=\ 1+\frac{\sqrt3}{2}=\frac{2+\sqrt3}{2}=\text{RHS}(\text{Taking LCM})$
View full question & answer→Question 151 Mark
Express the following as the product of sines and cosines:
$\sin2\text{x}+\cos4\text{x}$
Answer$\sin2\text{x}+\cos4\text{x}$
$=\ \sin2\text{x}+\sin(90-4\text{x})$
$=\ 2\sin\frac{(2\text{x}+90-4\text{x})}{2}\cos\frac{2\text{x}-90+4\text{x}}{2}$
$=\ 2\sin\Big(\frac{\pi}{4}+\text{x}\Big)\cos\Big(\frac{\pi}{4}-3\text{x}\Big)$
View full question & answer→Question 161 Mark
If $\sin2\text{A}=\lambda\sin2\text{B},$ then write the value of $\frac{\lambda+1}{\lambda-1}.$
AnswerWe have,
$\sin2\text{A}=\lambda\sin2\text{B}$
$\Rightarrow\ \frac{\sin2\text{A}}{\sin2\text{B}}=\lambda$
$\Rightarrow\ \frac{\sin2\text{A}}{\sin2\text{B}}+1=\lambda+1$
$\Rightarrow\ \frac{\sin2\text{A}+\sin2\text{B}}{\sin2\text{B}}=\lambda+1...(\text{i})$
Again,
$\sin2\text{A}=\lambda\sin2\text{B}$
$\Rightarrow\ \frac{\sin2\text{A}}{\sin2\text{B}}=\lambda$
$\Rightarrow\ \frac{\sin2\text{A}}{\sin2\text{B}}-1=\lambda-1$
$\Rightarrow\ \frac{\sin2\text{A}-\sin2\text{B}}{\sin2\text{B}}=\lambda-1...(\text{ii})$
Dividing equation (i) by equation (ii), we get
$\frac{\sin2\text{A}+\sin2\text{B}}{\sin2\text{A}-\sin2\text{B}}=\frac{\lambda+1}{\lambda-1}$
$\Rightarrow\ \frac{2\sin\Big(\frac{2\text{A}+2\text{B}}{2}\Big)\cos\Big(\frac{2\text{A}-2\text{B}}{2}\Big)}{2\sin\Big(\frac{2\text{A}-2\text{B}}{2}\Big)\cos\Big(\frac{2\text{A}+2\text{B}}{2}\Big)}=\frac{\lambda+1}{\lambda-1}$
$\Rightarrow\ \frac{\sin(\text{A+B})\cos(\text{A}-\text{B})}{\sin(\text{A}-\text{B})\cos(\text{A}+\text{B})}=\frac{\lambda+1}{\lambda-1}$
$\Rightarrow\ \frac{\sin(\text{A+B})\cos(\text{A}-\text{B})}{\cos(\text{A+B})\sin(\text{A}-\text{B})}=\frac{\lambda+1}{\lambda-1}$
$\Rightarrow\ \frac{\tan(\text{A+B})}{\tan(\text{A}-\text{B})}=\frac{\lambda+1}{\lambda-1}$
$\therefore\ \frac{\lambda+1}{\lambda-1}=\frac{\tan(\text{A+B})}{\tan(\text{A}-\text{B})}.$
View full question & answer→Question 171 Mark
Write the value of $\sin\frac{\pi}{12}\sin\frac{5\pi}{12}.$
AnswerWe have,
$\sin\frac{\pi}{12}\sin\frac{5\pi}{12}=\frac{1}{2}\Big[2\sin\frac{\pi}{12}+\sin\frac{5\pi}{12}\Big]$
$=\ \frac{1}{2}\Big[\cos\Big(\frac{\pi}{12}-\frac{5\pi}{12}\Big)-\cos\Big(\frac{\pi}{12}+\frac{5\pi}{12}\Big)\Big]$
$=\ \frac{1}{2}\Big[\cos\Big(\frac{-4\pi}{12}\Big)-\cos\Big(\frac{6\pi}{12}\Big)\Big]$
$=\ \frac{1}{2}\Big[\cos\frac{\pi}{3}-\cos\frac{\pi}{2}\Big]$
$=\ \frac{1}{2}\Big[\frac{1}{2}-0\Big]$
$=\ \frac{1}{4}$
$\therefore\ \text{Value of }\sin\frac{\pi}{12}\sin\frac{5\pi}{12}=\frac{1}{4}.$
View full question & answer→Question 181 Mark
Express of the following as the product of sines and cosines:
$\cos12\text{x}-\cos4\text{x}$
Answer$\cos12\text{x}-\cos4\text{x}$
$=\ -2\sin\Big(\frac{12\text{x}+4\text{x}}{2}\Big)\sin\Big(\frac{12\text{x}-4\text{x}}{2}\Big)$$\Big[\because\ \cos\text{D}-\cos\text{C}=-2\sin\frac{\text{C+D}}{2}\sin\frac{\text{C}-\text{D}}{2}\Big]$
$=\ 2\sin8\text{x}\sin4\text{x}$
View full question & answer→Question 191 Mark
Express the following as the sum or difference of sines and cosines:
$2\sin4\text{x}\sin3\text{x}$
Answer$2\sin4\text{x}\sin3\text{x}$
$\because\ 2\sin\text{A}\sin\text{B}=\cos(\text{A}-\text{B})-\cos(\text{A+B})$
$\Rightarrow\ 2\sin4\text{x}\sin3\text{x}=\cos(4\text{x}-3\text{x})-\cos(4\text{x}+3\text{x})$
$=\ \cos\text{x}-\cos7\text{x}$
View full question & answer→Question 201 Mark
Express the following as the product of sines and cosines:
$\cos12\text{x}+\cos8\text{x}$
Answer$\cos12\text{x}+\cos8\text{x}$ $\Big[\because\ \cos\text{C}+\cos\text{D}=2\cos\frac{\text{C+D}}{2}\cos\frac{\text{C}-\text{D}}{2}\Big]$
$=\ 2\cos10\text{x}\cos2\text{x}$
View full question & answer→Question 211 Mark
Express the following as the sum or difference of sines and cosines:
$2\cos3\text{x}\sin2\text{x}$
Answer$2\cos3\text{x}\sin2\text{x}$
$\because\ 2\cos\text{A}\sin\text{B}=\sin(\text{A+B})-\sin(\text{A}-\text{B})$
$\Rightarrow\ 2\cos3\text{x}\sin2\text{x}=\sin(3\text{x}+2\text{x})-\sin(3\text{x}-2\text{x})$
$=\ \sin5\text{x}-\sin\text{x}$
View full question & answer→Question 221 Mark
Express the following as the product of sines and cosines:
$\sin12\text{x}+\sin4\text{x}$
Answer$\sin12\text{x}+\sin4\text{x}$ $\Big[\because\ \sin\text{C}+\sin\text{D}=2\sin\frac{\text{C+D}}{2}\cos\frac{\text{C}-\text{D}}{2}\Big]$
$=\ 2\sin\Big(\frac{12\text{x}+4\text{x}}{2}\Big)\cos\Big(\frac{12\text{x}-4\text{x}}{2}\Big)$
$=\ 2\sin8\text{x}\cos4\text{x}$
View full question & answer→