Question 15 Marks
$\Bigg|\sin\text{x}\sin\Big(\frac{\pi}{3}-\text{x}\Big)\sin\Big(\frac{\pi}{3}+\text{x}\Big)\Bigg|\not<\frac{1}{4}$ for all values of x.
Answer$\big|\sin\theta\sin(60^\circ-\theta)\sin(60+\theta)\big|$
$=\big|\sin(\sin^260-\sin^2\theta)\big|$
$=\big\{\text{since}\ \sin(\text{A+B})\sin(\text{A}-\text{B})=\sin^2\text{A}-\sin^2\text{B}\big\}$
$\Big|\sin\theta\Big(\frac{3}{4}-\sin^2\theta\Big)\Big|$
$=\Big|\frac{1}{4}\sin\theta(3-4\sin^2\theta)\Big|$
$=\Big|\frac{1}{4}\sin3\theta\Big|$
$=\frac{1}{4}|\sin3\theta|$
$\not<\frac{1}{4}$
$\big\{\sin\theta\sin3\theta||\not<|\big\}$
so,
$\big|\sin\theta\sin(60^\circ-\theta)\sin(60^\circ+\theta)\big|\leq\frac{1}{4}$
View full question & answer→Question 25 Marks
If $\text{a}\cos2\text{x}+\text{b}\sin2\text{x}=\text{c}$ has $\alpha$ and $\beta$ as its roots, then prove that,
$\tan\alpha+\tan\beta=\frac{2\text{b}}{\text{a+c}}$
Answer$\cos2\theta=\frac{1-\tan^2\theta}{1+\tan^2\theta}$
$\sin^2\theta=\frac{2\tan\theta}{1+\tan^2\theta}$
substitute these valuse in the given equation, it reduces to
$\text{a}(1-\tan^2)+\text{b}(2\tan\theta)=\text{c}(1+\tan^2\theta)$
$(\text{c+a})\tan^2\theta+2\text{b}\tan\theta+\text{C-a}=0$
As $\alpha$ and $\beta$ are roots
sum of roots, $\tan\alpha+\tan\beta=\frac{2\text{b}}{\text{c+a}}$
View full question & answer→Question 35 Marks
Prove that
$\cos\frac{2\pi}{15}\cos\frac{4\pi}{15}\cos\frac{8\pi}{15}\cos\frac{16\pi}{15}=\frac{1}{16}$
Answer$\sin(\text{a+b})=\sin(\text{a})\cos(\text{b})+\sin(\text{b})\cos(\text{a})\ .....(1)$
$\text{for}\ \text{a}=\text{b},\sin(2\text{a})=2\sin(\text{a})\cos(\text{a})\ .....(2)$
$\text{let}\ \text{a}=16\frac{\text{pi}}{15}\ .....(3)$
$(\text{so}2\text{a}=32\frac{\text{pi}}{15})$
then using (3) in (2),we have
$\sin(2\text{a})=2\sin(\text{a})\cos(\text{a})$
$=2(2\sin(\frac{\text{a}}{2})\cos(\frac{\text{a}}{2}))\cos(\text{a})$
$=2(2(2\sin(\frac{\text{a}}{4})\cos\frac{\text{a}}{4}))\cos(\frac{\text{a}}{2}))\cos(\text{a})$
$=2(2(2(2\sin(\frac{\text{a}}{8})\cos(\frac{\text{a}}{8}))\cos(\frac{\text{a}}{4}))\cos(\frac{\text{a}}{2}))\cos(\text{a})$
$=16\sin(\frac{\text{a}}{8})(\cos(\frac{\text{a}}{8})\cos(\frac{\text{a}}{4})\cos(\frac{\text{a}}{2})\cos(\text{a})$
now note,
$\sin(2\text{a})=\sin(2\frac{\text{pi}}{15})$ and $\sin(\frac{\text{a}}{8})=\sin(2\frac{\text{pi}}{15})$
so,
$\cos(\frac{\text{a}}{8})\cos(\frac{\text{a}}{4})\cos(\frac{\text{a}}{2})\cos(\text{a})=\frac{1}{16}$
or, relacing a with $16\frac{\text{pi}}{15},$
$\cos\Big(2\frac{\text{pi}}{15}\Big)\times\cos\Big(4\frac{\text{pi}}{15}\Big)\times\cos\Big(8\frac{\text{pi}}{15}\Big)\times\cos\Big(16\frac{\text{pi}}{15}\Big)=\frac{1}{16}$
View full question & answer→Question 45 Marks
Prove that
$\cos\frac{\pi}{65}\cos\frac{2\pi}{65}\cos\frac{4\pi}{65}\cos\frac{8\pi}{65}\cos\frac{16\pi}{65}\cos\frac{32\pi}{65}=\frac{1}{64}$
Answer$\text{LHS}=\cos\frac{\pi}{65}.\cos\frac{2\pi}{65}.\cos\frac{4\pi}{65}.\cos\frac{8\pi}{65}.\cos\frac{16\pi}{65}.\cos\frac{32\pi}{65}$
Divide and multiply by $2\sin\frac{\pi}{65},$ we get
$=\frac{2.\sin\frac{\pi}{65}}{2\sin\frac{\pi}{65}}.\cos\frac{\pi}{65}.\cos\frac{2\pi}{65}.\cos\frac{4\pi}{65}.\cos\frac{8\pi}{65}.\cos\frac{16\pi}{65}.\cos\frac{32\pi}{65}$
$=\frac{2.\sin\frac{2\pi}{65}}{2.2\sin\frac{\pi}{65}}.\cos\frac{2\pi}{65}.\cos\frac{4\pi}{65}.\cos\frac{6\pi}{65}.\cos\frac{8\pi}{65}.\cos\frac{32\pi}{65}$
$=\frac{2.\sin\frac{4\pi}{65}}{2.4\sin\frac{\pi}{65}}.\cos\frac{4\pi}{65}.\cos\frac{8\pi}{65}.\cos\frac{16\pi}{65}.\cos\frac{32\pi}{65}$
$=\frac{2.\sin\frac{8\pi}{65}}{2.8\sin\frac{\pi}{65}}.\cos\frac{8\pi}{65}.\cos\frac{16\pi}{65}.\cos\frac{16\pi}{65}.\cos\frac{32\pi}{65}$
$=\frac{2.\sin\frac{16\pi}{65}}{2.16\sin\frac{\pi}{65}}.\cos\frac{16\pi}{65}.\cos\frac{32\pi}{65}$
$=\frac{2.\sin\frac{32\pi}{65}}{2.32\sin\frac{\pi}{65}}.\cos\frac{32\pi}{65}$
$=\frac{2.\sin\frac{64\pi}{65}}{2.64\sin\frac{\pi}{65}}$
$=\frac{1}{64}.\frac{\sin\Big(\pi-\frac{\pi}{65}\Big)}{\sin\frac{\pi}{65}}$
$=\frac{1}{64}\frac{\sin\frac{\pi}{65}}{\sin\frac{\pi}{65}}$
$=\frac{1}{64}=\text{RHS}$
View full question & answer→Question 55 Marks
If $0\leq\text{x}\leq\pi$ and x lies in the IInd quadrant such that $\sin\text{x}=\frac{1}{4}.$ Find the values of $\cos\frac{\text{x}}{2},\sin\frac{\text{x}}{2}$ and $\tan\frac{\text{x}}{2}$
Answersince x lies $II^{nd}$ quadrant.
$\Rightarrow\frac{\pi}{2}<\text{x}<\pi$
$\Rightarrow\frac{\pi}{4}<\frac{\text{x}}{2}<\frac{\pi}{2},$ which mencs $\frac{\text{x}}{2}$ lies in $I^{st}$ quad.
Now,
$\sin\text{x}=\frac{1}{4}=\frac{\text{p}}{\text{h}}\Rightarrow\text{p}=1\Rightarrow\text{b}=\sqrt{5}$
so,
$\cos\text{x}=\frac{\text{b}}{\text{h}}=\frac{-\sqrt{5}}{4}$ (-ve due to $II^{nd}$ quad)
Thus,
$\cos\frac{\text{x}}{2}=\sqrt{\frac{1+\cos\text{x}}{2}}=\sqrt{\frac{1-\frac{\sqrt{15}}{14}}{2}}=\frac{\sqrt{{4-\sqrt{15}}{}}}{8}$
$\sin\frac{\text{x}}{2}=\sqrt{\frac{1-\cos\text{x}}{2}}=\sqrt{\frac{1-\frac{\sqrt{15}}{4}}{2}}=\frac{\sqrt{{4+\sqrt{15}}{}}}{8}$
$\tan\frac{x}{2}=\frac{\sin\frac{\text{x}}{2}}{\cos\frac{\text{x}}{2}}=\frac{\sqrt{\frac{4+\sqrt{15}}{8}}}{\sqrt{\frac{4-\sqrt{15}}{8}}}=\sqrt{\frac{4+\sqrt{15}}{4-\sqrt{15}}}$
$=\sqrt{\frac{(4+\sqrt{15})(4+\sqrt{15})}{(4-\sqrt{15})(4+\sqrt{15})}}$
$=4+\sqrt{15}$
View full question & answer→Question 65 Marks
$\Bigg|\cos\text{x}\cos\Big(\frac{\pi}{3}-\text{x}\Big)\cos\Big(\frac{\pi}{3}+\text{x}\Big)\Bigg|\leq\frac{1}{4}$ for all values of x.
Answer$\big|\cos\theta\cos(60^\circ-\theta)\cos(60^\circ+\theta)\big|$
$=\big|\cos\theta(\cos^260^\circ-\sin^2\theta)\big|$
$\big\{\text{since}\cos(\text{A}-\text{B})\cos(\text{A+B})-\cos^2\text{A}-\sin^2\text{B}\big\}$
$=\Big|\cos\theta\Big(\frac{1}{4}-\sin^2\theta\Big)\Big|$
$=\Big|\cos\theta\frac{1}{4}(1-4\sin^2\theta)\Big|$
$=\Big|\frac{1}{4}\cos\theta(1-4)(1-\cos^2\theta)\Big|$
$=\Big|\frac{1}{4}\cos\theta(-3+4\cos^2\theta)\Big|$
$=\Big|\frac{1}{4}(4\cos3\theta-3\cos\theta)\Big|$
$=\Big|\frac{1}{4}\cos3\theta\Big|$
$\leq\frac{1}{4}$
$\big\{\text{since|}\cos3\theta|\leq1\big\}$
so,
$\Big|\cos\theta\cos(60^\circ-\theta)\cos(60^\circ+\theta)\Big|\leq\frac{1}{4}$
View full question & answer→Question 75 Marks
Prove that:
$\cos6^\circ\cos42^\circ\cos66^\circ\cos78^\circ=\frac{1}{16}$
Answer$\text{LHS}=\cos6^\circ\cos42^\circ\cos66^\circ\cos78^\circ$
$=\frac{1}{4}(2\cos6^\circ\cos66^\circ)(2\cos42^\circ\cos78^\circ)$
$=\frac{1}{4}(\cos72^\circ+\cos60^\circ)(\cos120^\circ+\cos36^\circ)$ $[\because2\cos\text{A}\cos\text{B}=\cos(\text{A+B})+\cos(\text{A}-\text{B})]$
$=\frac{1}{4}\big\{\cos(90^\circ-72^\circ)+\frac{1}{2}\big\}\Big\{-\frac{1}{2}+\frac{\sqrt{5}+1}{4}\Big\}$
$=\frac{1}{4}\big(\sin18^\circ+\frac{1}{2}\big)\Big(-\frac{1}{2}+\frac{\sqrt{5}+1}{4}\Big)$
$=\frac{1}{4}\Big(\frac{\sqrt{5}-1}{4}+\frac{1}{2}\Big)\Big(\frac{\sqrt{5}+1}{4}+\frac{1}{2}\Big)$
$=\frac{1}{4}\Big(\frac{\sqrt{5}-1+2}{4}\Big)\Big(\frac{\sqrt{5}+1-2}{4}\Big)$
$=\frac{1}{64}(\sqrt{5}+1)(\sqrt{5}-1)$
$=\frac{1}{64}(5-1)$
$=\frac{1}{16}=\text{RHS}$
Hence proved
View full question & answer→Question 85 Marks
Prove that:
$\cot\frac{\pi}{8}=\sqrt{2}+1$
AnswerWe know thar,
$\sin\frac{\text{A}}{2}=\pm\sqrt{\frac{1-\cos\text{A}}{2}}$
$\text{put}\ \text{A}=45^\circ$
$\sin22\frac{1^\circ}{2}=\sqrt{\frac{1-\cos45^\circ}{2}}$ $\Big\{\text{since}\sin22\frac{1}{2},\text{is positive}\Big\}$
$=\sqrt{\frac{1-\frac{1}{2}}{2}}$
$\sin22\frac{1^\circ}{2}=\sqrt{\frac{\sqrt{2}-1}{2\sqrt{2}}}$
and
$\cos\frac{\text{A}}{2}\pm\sqrt{\frac{1+\cos\text{A}}{2}}$
$\text{put}\ \text{A}\ 45^\circ$
$\cos22\frac{1^\circ}{2}=\sqrt{\frac{1+\cos45^\circ}{2}}$
$=\sqrt{\frac{1-\frac{1}{2}}{2}}$
$\cos22\frac{1^\circ}{2}=\sqrt{\frac{\sqrt{2}+1}{2\sqrt{2}}}$
Now,
$\cot22\frac{1^\circ}{2}=\frac{\cos22\frac{1^\circ}{2}}{\sin22\frac{1^\circ}{2}}$
$=\sqrt{\frac{\sqrt{2}+1}{2\sqrt{2}}\times\frac{2\sqrt{2}}{\sqrt{2-1}}}$
$=\sqrt{\frac{\sqrt{2}+1}{\sqrt{2}-1}}$
Rationalizing denominator,
$=\sqrt{\frac{\sqrt{2}+1\times\sqrt{2}+1}{\sqrt{2}-1\times\sqrt{2}+1}}$
$=\sqrt{\frac{(\sqrt{2}+1)^2}{2-1}}$
$\cot22\frac{1^\circ}{2}=\sqrt{2}+1$
View full question & answer→Question 95 Marks
Prove that:
$\tan82\frac{1^\circ}{2}=(\sqrt{3}+\sqrt{2})(\sqrt{2}+1)=\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}$
Answer$\tan82\frac{1^\circ}{2}=\tan\Big(90-7\frac{1}{7}\Big)$
$=\cot7\frac{1^\circ}{2}$
$=\cot\text{A}$ if $\text{A}=7\frac{1^\circ}{2}$
Now,
$\cot\text{x}=\frac{\cos\text{x}}{\sin\text{x}}$
$=\frac{2\cos^2\text{x}}{1\sin\text{x}\cos\text{x}}$
$=\frac{1+\cos^2\text{x}}{\sin^2\text{x}}$
$\cot\text{x}=\frac{1+\cos15}{\sin15}$
$=\frac{1+\cos(45-30)}{\sin15}$
$\frac{1+\Big(\frac{1}{\sqrt{2}}\times\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\times\frac{1}{2}\Big)}{\frac{1}{\sqrt{2}}\times\frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}\times\frac{}{}\frac{1}{2}}$
$=\frac{2\sqrt{2}+(\sqrt{3}+1)}{\sqrt{3}-1}\times\frac{\sqrt{3}+1}{\sqrt{3}+1}$
$=\frac{2\sqrt{2}(\sqrt{3}+1)+(\sqrt{3}+1)^2}{3-1}$
$=\frac{2\sqrt{6}+2\sqrt{2}+4+2\sqrt{3}}{2}$
$\cot\text{x}=\sqrt{6}+\sqrt{2}+2+\sqrt{3}\ .....(1)$
$=\sqrt{2}+2\sqrt{6}+\sqrt{3}$
$=\sqrt{2}(1+\sqrt{2})+\sqrt{3}(\sqrt{2}+1)$
$\cot\text{x}=(\sqrt{2}+1)(\sqrt{2}+\sqrt{3})]\ .....(2)$
From equation (1) and (2)
$\tan82\frac{1^\circ}{2}=\cot7\frac{1^\circ}{2}=\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}$
$=(\sqrt{2}+1)(\sqrt{2}+\sqrt{3})$
View full question & answer→Question 105 Marks
Prove that:
$\sin6\circ\sin42^\circ\sin66^\circ\sin78^\circ=\frac{1}{16}$
Answer$\text{LHS}=\cos6^\circ\cos42^\circ\cos66^\circ\cos78^\circ$$=\frac{1}{4}(2\cos6^\circ\cos66^\circ)(2\cos42^\circ\cos78^\circ)$
$=\frac{1}{4}(\cos72^\circ+\cos60^\circ)(\cos120^\circ+\cos36^\circ)$
$=\frac{1}{4}\Big(\sin18^\circ+\frac{1}{2}\Big)\bigg(-\frac{2}{2}+\frac{\sqrt{5}+1}{4}\bigg)$
$=\frac{1}{4}\bigg(\frac{\sqrt{5}-1}{4}+\frac{1}{2}\bigg)\bigg(\frac{\sqrt{5}+1}{4}-\frac{1}{2}\bigg)$
$=\frac{1}{4}\bigg(\frac{\sqrt{5}-1+2}{4}\bigg)\bigg(\frac{\sqrt{5}+1-2}{4}\bigg)$
$=\frac{1}{64}(\sqrt{5}+1)(\sqrt{5}-1)$
$=\frac{1}{64}(\sqrt{5})^2-1^2)$
$=\frac{1}{64}(5-1)$
$=\frac{1}{16}$
$=\text{RHS}$
View full question & answer→Question 115 Marks
$\cot\text{x}+\cot\Big(\frac{\pi}{3}-\text{x}\Big)+\cot\Big(\frac{\pi}{3}-\text{x}\Big)=3\cot3\text{x}$
Answer$\cot\text{x}\cot(60^\circ+\text{x})=\cot(60^\circ-\text{x})=3\cot2\text{x}$
$\text{LHS}=\cot\text{x}+\cot(60^\circ+\text{x})-\cot(60^\circ-\text{x})$
$=\cot\text{x}+\frac{\cot60^\circ+\cot\text{x}}{1-\cot60^\circ\cot\text{x}}-\frac{\cot60^\circ-\cot\text{x}}{1+\cot60^\circ\cot\text{x}}$
$=\cot\text{x}+\frac{\sqrt{3}+\cot\text{x}}{1-\sqrt{3\cot\text{x}}}-\frac{\sqrt{3}-\cot\text{x}}{1+\sqrt{3\cot\text{x}}}$
$=\cot\text{x}+\Bigg[\frac{\sqrt{3}+3\cot\text{x}+\cot\text{x}+\sqrt{3}\cot^2+\sqrt{3}+3\cot\text{x}+\cot\text{x}-\sqrt{3}\cot^2\text{x}}{(1-\sqrt{3}\cot\text{x}(1+\sqrt{3}\cot\text{x})}\Bigg]$
$=\cot\text{x}+\frac{8\cot\text{x}}{1-3\cot^2\text{x}}$
$=\frac{\cot\text{x}-3\cot^3\text{x}+8\cot\text{x}}{1-3\cot^2\text{x}}$
$=\frac{9\cot\text{x}-3\cot^3\text{x}}{1-3\cot^2\text{x}}$
$=3\Big(\frac{3\cot\text{x}-\cot^3\text{x}}{1-3\cot^2\text{x}}\Big)$
$=3\cot3\text{x}$
$=\text{RHS}$
so,
$\cot\text{x}+\cot(60^\circ+\text{x})-\cot(60^\circ-\text{x})=3\cot3\text{x}$
View full question & answer→Question 125 Marks
If $\sec(\text{x}+\alpha)+\sec(\text{x}-\alpha)=2\sec\text{x},$ prove that $\cos\text{x}-\pm\sqrt{2}\cos\frac{\alpha}{2}$
AnswerWe have,
$\sec(\theta+\alpha)+\sec(\theta-\alpha)=2\sec\theta.$
$\Rightarrow\frac{1}{\cos^2\theta.\cos^2\alpha-\sin^\theta\sin^2\alpha}+\frac{1}{\cos\theta.\cos\alpha-\sin\theta\sin\alpha}=\frac{2}{\cos\theta}$
$\Rightarrow\frac{2\cos\theta\cos\alpha}{\cos^2\cos^2\alpha-\sin^2\sin^2\alpha}=\frac{2}{\cos\theta}$
$\Rightarrow\frac{2\cos\theta\cos\alpha}{\cos^2\theta\cos^2\alpha-(\cos^2\alpha+\sin^2\alpha)}=\frac{1}{\cos\theta}$
$\Rightarrow\cos^2\theta\cos\alpha=\cos^2\theta(\cos^2\alpha+\sin^2\alpha)-\sin^2\alpha$
$\Rightarrow\cos^2\theta(1-\cos\alpha)=\sin^2\alpha$
$\Rightarrow\cos^2\theta=\frac{\sin^2\alpha}{2\sin^2\frac{\alpha}{2}}$
$=\frac{1\sin^2\frac{\alpha}{2}.\cos^2\frac{\alpha}{2}}{2\sin^2\frac{\alpha}{2}}$
$\Rightarrow\cos\theta=\pm\sqrt{2}\cos\frac{\alpha}{2}$
View full question & answer→Question 135 Marks
Prove that:
$\cos\frac{\pi}{15}\cos\frac{2\pi}{15}\cos\frac{4\pi}{15}\cos\frac{7\pi}{15}=\frac{1}{16}$
Answer$\text{LHS}=\cos\frac{\pi}{15}\cos\frac{2\pi}{15}\cos\frac{4\pi}{15}\cos\frac{7\pi}{15}$
$=\frac{2\sin\frac{\pi}{15}.\cos\frac{\pi}{15}}{2\sin\frac{\pi}{15}}.\cos\frac{2\pi}{15}\cos\frac{4\pi}{15}\cos\frac{7\pi}{15}$ $\Big[$ Divide and multiply by $2\sin\frac{\pi}{15}\Big]$
$=\frac{2\sin\frac{\pi}{15}.\cos\frac{\pi}{15}}{2\sin\frac{\pi}{15}}.\cos\frac{2\pi}{15}\cos\frac{4\pi}{15}\cos\frac{7\pi}{15}$
$=\frac{2\sin\frac{4\pi}{15}}{2.4\sin\frac{\pi}{15}}.\cos\frac{4\pi}{15}\cos\frac{7\pi}{15}$
$=\frac{2\sin\frac{8\pi}{15}}{2.8\sin\frac{\pi}{15}}.\cos\frac{7\pi}{15}$
$=\frac{\sin\big(\frac{8\pi}{15}+\frac{7\pi}{15}\big)+\sin\big(\frac{8\pi}{15}-\frac{7\pi}{15}\big)}{16\sin\frac{\pi}{15}}$
$=\frac{\sin\pi+\sin\frac{\pi}{15}}{16\sin\frac{\pi}{15}}$
$=\frac{\sin\frac{\pi}{15}}{16\sin\frac{\pi}{15}}[\because\sin\pi=0]$
$=\frac{1}{16}$
$=\text{RHS}$
View full question & answer→Question 145 Marks
$\sin^3\text{x}+\sin^3\big(\frac{2\pi}{3}+\text{x}\big)+\sin^3\big(\frac{4\pi}{3}+\text{x}\big)=-\frac{3}{4}\sin3\text{x}$
Answer$\text{LHS}=\sin^3\text{x}+\sin^3\Big(\frac{2\pi}{3}+\text{x}\Big)+\sin^3\Big(\frac{4\pi}{3}+\text{x}\Big)$
$\Big\{$ We know that $\sin^3\text{x}=\frac{3\sin\text{x}-\sin3\text{x}}{4}\Big\}$
$=\Big(\frac{3\sin\text{x}-\sin3\text{x}}{4}\Big)+\Bigg\{\frac{3\sin\big(\frac{2\pi}{4}+\text{x}\big)-\sin3\big(\frac{2\pi}{3}+\text{x}\big)}{4}\Bigg\}\\+\Bigg\{\frac{3\sin\big(\frac{4\pi}{3}+\text{x}\big)-\sin3\big(\frac{4\pi}{3}+\text{x}\big)}{4}\Bigg\}$
$=\Big[\frac{3\sin\text{x}-\sin3\text{x}}{4}\Big]+\Bigg[\frac{3\sin\big[\pi\big(\frac{2\pi}{3}+\text{x}\big)\big]-\sin(2\pi+3\text{x})}{4}\Bigg]\\+\Bigg[\frac{3\sin\big[\pi\big(\frac{\pi}{3}+\text{x}\big)\big]-\sin(4\pi+3\text{x})}{4}\Bigg]$
$=\frac{1}{4}\Big\{[3\sin\text{x}-\sin3\text{x}]+\Big[3\sin\Big(\frac{\pi}{3}-\text{x}\Big)-\sin3\text{x}\Big]\\-\Big[3\sin\Big(\frac{\pi}{3}+\text{x}\Big)+\sin3\text{x}\Big]\Big\}$
$=\frac{1}{4}\Big[3\sin\text{x}-\sin3\text{x}+3\sin\Big(\frac{\pi}{3}-\text{x}\Big)-3\sin\Big(\frac{\pi}{3}+\text{x}\Big)-\sin3\text{x}\sin3\text{x}\Big]$
$=\frac{1}{4}\Big[3\sin\text{x}-3\sin3\text{x}+3\Big(\sin\Big(\frac{\pi}{3}-\text{x}\Big)-\sin\Big(\frac{\pi}{3}+\text{x}\Big)\Big)\Big]$
$=\frac{1}{4}\Bigg[3\sin\text{x}-3\sin3\text{x}+3\Bigg\{2\cos\frac{\frac{\pi}{3}-\text{x}+\frac{\pi}{3}+\text{x}}{2}\sin\frac{\frac{\pi}{3}-\text{x}\frac{\pi}{3}-\text{x}}{2}\Bigg\}\Bigg]$
$=\frac{1}{4}\Big[3\sin\text{x}-3\sin2\text{x}+6\cos\frac{\pi}{3}\sin(-\text{x})\Big]$
$\frac{1}{4}[3\sin\text{x}-3\sin3\text{x}-3\sin\text{x}]$
$=-\frac{3}{4}\sin3\text{x}$
$=\text{RHS}$
$\text{LHS}=\text{RHS}$
View full question & answer→Question 155 Marks
If $2\tan\frac{\alpha}{2}=\tan\frac{\beta}{2},$ prove that $\cos\alpha=\frac{3+5\cos\beta}{5+3\cos\beta}$
AnswerWe have,
$2\tan\frac{\alpha}{2}=\tan\frac{\beta}{2}$
$\Rightarrow\frac{\tan\frac{\alpha}{2}}{\tan\frac{\beta}{2}}=\frac{1}{2}$
Let $\tan\frac {\alpha}{2}=\text{k}$ and $\tan\frac{\beta}{2}=2\text{k}$
Then,
$\cos\alpha=\frac{1-\tan^2\frac{\alpha}{2}}{1+\tan^2\frac{\alpha}{2}}=\frac{1-\text{k}^2}{1+\text{k}^2}\ .....(\text{A})$
Also,
$\frac{3+5\cos\beta}{5+3\cos\beta}=\frac{3+5\Bigg(\frac{1-\tan^2\frac{\beta}{2}}{1+\tan^2\frac{\beta}{2}}\Bigg)}{5+3\Bigg(\frac{1-\tan^2\frac{\beta}{2}}{1+\tan^2\frac{\beta}{2}}\Bigg)}$
$=\frac{3+5\Big(\frac{1-4\text{k}^2}{1+4\text{k}^2}\Big)}{5+3\Big(\frac{1-4\text{k}^2}{1+4\text{k}^2}\Big)}$
$=\frac{8-8\text{k}^2}{8+8\text{k}^2}=\frac{1-\text{k}^2}{1+\text{k}^2}\ .....(\text{B})$
From(A) & (B)
$\cos\alpha=\frac{3+5\cos\beta}{5+3\cos\beta}$
View full question & answer→Question 165 Marks
If $2\tan\alpha=3\tan\beta,$ prove that $\tan(\alpha-\beta)=\frac{\sin2\beta}{5-\cos2\beta}$
AnswerWe have,
$2\tan\alpha=3\tan\beta$
$\Rightarrow\frac{\tan\alpha}{\tan\beta}=\frac{3}{2}$
Let $\tan\alpha=3\text{k}$ and $\tan\beta=2\text{k}$
Also,
$\frac{\sin2\beta}{5-\cos2\beta}=\frac{\frac{2\tan\beta}{1+\tan^2\beta}}{5-\Big(\frac{1-\tan^\beta}{1+\tan^2\beta}\Big)}$
$=\frac{\frac{2.2\text{k}}{1+4\text{k}^2}}{5-\Big(\frac{1-4\text{k}^2}{1+4\text{k}^2}\Big)}$
$=\frac{4\text{k}}{5+20\text{k}^2-1+4\text{k}^2}$
$=\frac{4\text{k}}{4+24\text{K}^2}=\frac{\text{K}}{1+6\text{k}^2}\ ...\text{(B)}$
From (A) & (B)
$\tan(\alpha-\beta)=\frac{\sin2\beta}{5-\cos2\beta}$
View full question & answer→Question 175 Marks
If $\sin\alpha+\sin\beta=\text{a}$ and $\cos\alpha+\cos\beta=\text{b}$ prove that
$\sin(\alpha+\beta)=\frac{2\text{ab}}{\text{a}^2+\text{b}^2}$
Answerwe have,
$\sin\alpha+\sin\beta=\text{a}\ \&\ \cos\alpha+\cos\beta=\text{b}\ .....(\text{A})$
squaring and adding, we get
$\sin^2\alpha+\sin^2\beta+2\sin\alpha\sin\beta+\cos^2\alpha+\cos^2\beta+2\cos\alpha\cos\beta=\text{a}^2+\text{b}^2$
$\Rightarrow1+1+2(\sin\alpha\sin\beta+\cos\alpha\cos\beta)=\text{a}^2+\text{b}^2$
$\Rightarrow2(\sin\alpha\sin\beta+\cos\alpha\cos\beta)=\text{a}^2+\text{b}^2-2$
$\therefore2\cos(\alpha+\beta)=\text{a}^2+\text{b}^2-2$
Thus,
$\cos(\alpha-\beta)=\frac{\text{a}^2+\text{b}^2-2}{2}$
Again,
$\sin\alpha+\sin\beta=\text{a}\Rightarrow2\sin\frac{\alpha+\beta}{2}.\cos\frac{\alpha-\beta}{2}=\text{a}$
$\cos\alpha+\cos\beta=\text{b}\Rightarrow2\cos\frac{\alpha+\beta}{2}.\cos\frac{\alpha-\beta}{2}=\text{b}$
$\Rightarrow\tan\frac{\alpha+\beta}{2}=\frac{\text{a}}{\text{b}}\ .....\text{(B)}$
Now,
$\sin(\alpha+\beta)=\frac{1\tan\frac{\alpha+\beta}{2}}{1}+\tan^2\Big(\frac{\alpha+\beta}{2}\Big)$
thus,
$\sin(\alpha+\beta)=\frac{2\text{ab}}{\text{a}^2+\text{b}^2}$
View full question & answer→Question 185 Marks
Prove that:
$\cos^32\text{x}+3\cos2\text{x}=4(\cos^6\text{x}-\sin^6\text{x})$
Answer$\cos^32\text{x}+3\cos2\text{x}=4(\cos^6\text{x}-\sin^6\text{x})$
$\text{RHS}=4\Big[(\cos^2\theta)^3-(\sin^2\theta)^3\Big]$
$=4(\cos^2\theta-\sin^2\theta)\Big[\cos^4\theta+\sin^4\theta+\sin^2\theta+\cos^2\theta\Big]$
$=4\cos^2\theta\Big[(\cos^2\theta-\sin^2\theta)+2\sin^2\theta+\sin^2\theta+\sin^2\theta+\cos^2\theta\Big]$
$=4\cos^2\theta[\cos^22\theta+3\sin^2\theta\cos^2\theta]$
$=4\cos^2\theta\Big[\cos^22\theta+3\Big(\frac{1-\cos^2\theta}{2}\Big)\Big(\frac{1+\cos^2\theta}{2}\Big)\Big]$
$=4\cos^2\theta\Big(\cos^22\theta+\frac{3}{4}(1-\cos^22\theta)\Big)$
$=\cos^2\theta\big[4\cos^22\theta+3-3\cos^22\theta\big]$
$=\cos^2\theta[\cos^22\theta+3]$
$=\cos^32\theta+3\cos^2\theta\ \text{LHS}$
$\text{LHS=RHS}$
View full question & answer→Question 195 Marks
Show thet:
$3(\sin\text{x}-\cos\text{x})+6(\sin\text{x}+\cos)^2+4(\sin^6\text{x}+\cos^6\text{x})=13$
Answer$\text{LHS}=3(\sin\text{x}-\cos\text{x})+6(\sin\text{x}+\cos)^2+4(\sin^6\text{x}+\cos^6\text{x})$
$=3[\sin^4\text{x}-4\sin^3\text{x}\cos\text{x}+6\sin^2\text{x}\cos^2\text{x}-4\sin\text{x}\cos^3\text{x}+\cos^4\text{x}]$
$+6[\sin^2\text{x}+2\sin\text{x}\cos\text{x}+\cos^2\text{x}]+4(\sin^6\text{x}+\cos^6\text{x})$
$\big[\because(\text{a}-\text{b})^4=\text{a}^4-4\text{a}^3\text{b}+6\text{a}^2\text{b}^2-4\text{ab}^3+\text{b}^4$ by binomial expainsion$\big]$
$=3\big[\sin^4\text{x}+\cos^4\text{x}-4\sin\text{x}\cos\text{x}(\sin^2\text{x}+\cos^2\text{x})+6\sin^2\text{x}\cos^2\text{x}\big]$
$+6[1+2\sin\text{x}\cos\text{x}]+4\big[(\cos^2\text{x}+\sin^2\text{x})(\cos^4\text{x}-\cos^2\text{x}+\sin^4\text{x})\big]$
$\big[\because\text{a}^3+\text{b}^3=(\text{a+b})(\text{a}^2-\text{ab+b}^2)$
$=7[\sin^4\text{x}+\cos^4\text{x}]+18\sin^2\text{x}\cos^4\text{x}-4\sin^2\text{x}\cos^2\text{x}+6$
$=7[\sin^4\text{x}+\cos^4\text{x}2\sin^2\cos^2]+6$
$=7[\sin^2\text{x}+\cos^2\text{x}+2\sin^2\text{x}\cos^2\text{x}]+6$
$=7+6$
$=13=\text{RHS}$
View full question & answer→Question 205 Marks
If $\sin\alpha=\frac{4}{5}$ and $\cos\beta=\frac{5}{13},$ prove that $\cos\frac{\alpha-\beta}{2}=\frac{8}{\sqrt{65}}$
AnswerWe have,
$\sin\alpha=\frac{4}{5}\ \&\cos\beta=\frac{4}{5}\Rightarrow\cos\alpha=\frac{3}{5}\ \&\cos\alpha=\frac{3}{5}\ \&\sin\beta=\frac{12}{13}$
$\therefore\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha.\sin\beta$
$=\frac{3}{5}.\frac{5}{13}+\frac{4}{5}.\frac{12}{13}$
$=\frac{15}{65}+\frac{48}{65}=\frac{63}{65}$
Now,
$\cos\Big(\frac{\alpha-\beta}{2}\Big)=\sqrt{\frac{1+\cos(\alpha-\beta)}{2}}$
$=\sqrt{\frac{1+\frac{63}{65}}{2}}$
$=\sqrt{\frac{128}{65\times2}}=\sqrt{\frac{64}{65}}$
$=\pm\frac{8}{\sqrt{65}}$
$\therefore\cos\Big(\frac{\alpha-\beta}{2}\Big)=\frac{8}{\sqrt{65}}$
View full question & answer→Question 215 Marks
Prove that:
$\sin5\text{x}=5\sin\text{x}-20\sin^3\text{x}+16\sin^5\text{x}$
Answer$\text{LHS}=\sin5\text{x}=\sin(3\text{x}+2\text{x})$
$=\sin3\text{x}\cos2\text{x}+\cos3\text{x}.\sin2\text{x}$
$=(3\sin\text{x}-4\sin^3\text{x})(1-2\sin^3\text{x})\\+(s\cos^3\text{x}=3\cos\text{x})2\sin\text{x}\cos\text{x}.$
$=3\sin\text{x}-4\sin^3\text{x}-6\sin^3\text{x}\\+8\sin^5\text{x}(8\cos^4\text{x}-6\cos^2\text{x})\sin\text{x}$
$=3\sin\text{x}-10\sin^3\text{x}+8\sin^5\text{x}\\+8\sin\text{x}\big((1-\sin^2\text{x})^2-6\sin\text{x}(1-\sin^2\text{x})\big)$
$=13\sin\text{x}-10\sin^3\text{x}+8\sin^5\text{x}+8\sin\text{x}\\-16\sin^3\text{x}8\sin^5\text{x}-6\sin\text{x}+6\sin^3\text{x}$
$=5\sin\text{x}-20\sin^3\text{x}+16\sin^5\text{x}=\text{RHS}$
View full question & answer→Question 225 Marks
If $\cos\text{x}=\frac{\cos\alpha+\cos\beta}{1+\cos\alpha\cos\beta}$ prove that $\tan\frac{\text{x}}{2}=\pm\tan\frac{\alpha}{2}\tan\frac{\beta}{2}$
AnswerWe have,
$\cos\theta=\frac{\cos\alpha+\cos\beta}{1+\cos\alpha.\cos\beta}$
Now,
$\cos\theta=\frac{1-\tan^2\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}$
$=\frac{1-\tan^2\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}=\frac{\cos\alpha+\cos\beta}{1+\cos\alpha\cos\beta}$
By componende and dividendo, we get
$\frac{\big(1-\tan^2\frac{\theta}{2}\big)+\big(1+\tan^2\frac{\theta}{2}\big)}{\big(1+\tan^2\frac{\theta}{2}\big)-\big(1-\tan^2\frac{\theta}{2}\big)}=\frac{1+\cos\alpha\cos\beta+\cos\alpha\cos\beta}{-(1+\cos\alpha\cos\beta-\cos\alpha\cos\beta)}$
$\frac{2}{2\tan^2\frac{\theta}{2}}=\frac{(1+\cos\alpha)(1+\cos\beta)}{(1-\cos\alpha)(1-\cos\alpha)}$
$\tan^2\frac{\theta}{2}=\frac{(1-\cos\alpha)(1-\cos\beta)}{(1+\cos\alpha)(1+\cos\alpha)}$
$=\frac{2\sin^2\frac{\alpha}{2}.1\sin^2\frac{\beta}{2}}{2\cos^2\frac{\alpha}{2}.1\cos^2\frac{\beta}{2}}$
$\tan\frac{\theta}{2}=\pm\tan\frac{\alpha}{2}.\tan\frac{\beta}{2}$
View full question & answer→Question 235 Marks
If $\cos\text{x}=-\frac{3}{5}$ and x lies in the IIIrd quadrant, find the values of $\cos\frac{\text{x}}{2},\sin\frac{\text{x}}{2},\sin2\text{x}.$
Answer$\text{since}\cos\text{x}=-\frac{3}{5}=\frac{\text{b}}{\text{h}}$
$\Rightarrow\text{b}=3,\text{h}=5$
$\Rightarrow\text{p}=4$
Now, x loes on third quad.
$\therefore\sin2\text{x}=2\sin\text{x}.\cos\text{x}$
$=2\Big(\frac{-4}{5}\Big).\Big(\frac{-3}{5}\Big)=\frac{24}{25}$
$\because\pi<\text{x}<\frac{3\pi}{2}\Rightarrow\frac{\pi}{2}<\frac{\text{x}}{2}<\frac{3\pi}{4}$
Which means $\frac{\text{x}}{2}$ lies in secound quadrant
so, $\cos\frac{\text{x}}{2}=\sqrt{\frac{1+\cos\text{x}}{2}}$ $[\because1+\cos2\theta=2\cos^2\theta]$
$=\sqrt{\frac{1-\frac{3}{5}}{2}}=\frac{-1}{\sqrt{5}}$ (-ve sigh because od second quad. where cos D is -ve)
Also,
$\sin\frac{\text{x}}{2}=\frac{\sin\text{x}}{1\cos\frac{\text{x}}{2}}$ $[\because\sin2\text{A}=2\sin\text{A}\cos\text{A}]$
$=\Bigg(\frac{\frac{-4}{5}}{2\Big(\frac{-1}{\sqrt{5}}\Big)}\Bigg)$
$=\frac{2}{\sqrt{5}}$
View full question & answer→Question 245 Marks
$\sin5\text{x}=5\cos^4\text{x}\sin\text{x}-10\cos^2\text{x}\sin^3\text{x}+\sin^5\text{x}$
Answer$\text{LHS}=\sin5\text{x}$
$=\sin\text{x}(3\text{x}+2\text{x})$
$=\sin3\text{x}\times\cos2\text{x}+\cos3\text{x}\times\sin2\text{x}$
$=(3\sin\text{x}-4\sin^3\text{x})(2\cos^2-1)\\+(4\cos^32\cos^2-3\cos\text{x})\times2\sin\text{x}\cos\text{x}$
$=-3\sin\text{x}-4\sin^3\text{x}+6\sin\text{x}\cos^2\text{x}\\-8\sin^3\text{x}\cos^2\text{x}+8\sin\text{x}\cos^4\text{x}-6\sin\text{x}\cos^2\text{x}$
$=8\sin\text{x}\cos^4\text{x}-8\sin^3\text{x}\cos^2\text{x}-3\sin\text{x}+4\sin^3\text{x}$
$=5\sin\text{x}\cos^4\text{x}-10\sin^3\text{x}\cos^2\text{x}-3\sin\text{x}\\+3\sin\text{x}\cos^4\text{x}+4\sin^3\text{x}+2\sin^3\text{x}\cos^2\text{x}$
$=5\sin\text{x}\cos^4\text{x}-10\sin^3\text{x}\cos^2\text{x}\\-3\sin\text{x}(1-\cos^4\text{x})+2\sin^3\text{x}(2+\cos^2\text{x})$
$=5\sin\text{x}\cos^4\text{x}-10\sin^3\text{x}\cos^2\text{x}\\-3\sin\text{x}(1-\cos^2\text{x})(1+\cos^2\text{x})+2\sin^3\text{x}(2+\cos^2\text{x})$
$=5\sin\text{x}\cos^4\text{x}-10\sin^3\text{x}\cos^2\text{x}\\-3\sin^3\text{x}(1+\cos^2\text{x})+2\sin^3\text{x}(2+\cos^2\text{x})$
$=5\sin\text{x}\cos^4\text{x}-10\sin^3\text{x}\cos^2\text{x}\\-3\sin^3\text{x}[3(1+\cos^2\text{x})-2(2+\cos^2\text{x})]$
$=5\sin\text{x}\cos^4\text{x}-10\sin^3\text{x}\cos^2\text{x}\\-3\sin^3\text{x}[3+3\cos^2\text{x}-4-2\cos^2\text{x}]$
$=5\sin\text{x}\cos^4\text{x}-10\sin^3\text{x}\cos^2\text{x}-3\sin^3\text{x}[\cos^2\text{x-1}]$
$=5\sin\text{x}\cos^4\text{x}-10\sin^3\text{x}\cos^2\text{x}-3\sin^3\text{x}\times(-\sin^2\text{x})$
$=5\sin\text{x}\cos^4\text{x}-10\sin^3\text{x}\cos^2\text{x}+\sin^5\text{x}$
$=5\cos^4\text{x}\sin\text{x}-10\cos^2\text{x}\sin^3\text{x}+\sin^5\text{x}$
$=\text{RHS}$
View full question & answer→Question 255 Marks
Prove that:
$\cos\frac{\pi}{15}\cos\frac{2\pi}{15}\cos\frac{3\pi}{15}\cos\frac{4\pi}{15}\cos\frac{5\pi}{15}\cos\frac{6\pi}{15}\cos\frac{7\pi}{15}=\cos\frac{1}{128}$
Answer$\text{LHS}=\cos\frac{\pi}{15}\cos\frac{2\pi}{15}\cos\frac{3\pi}{15}\cos\frac{4\pi}{15}\cos\frac{5\pi}{15}\cos\frac{6\pi}{15}\cos\frac{7\pi}{15}\cos\frac{1}{128}$
$=\cos\frac{\pi}{15}\cos\frac{2\pi}{15}\cos\frac{4\pi}{15}\big(\cos\frac{3\pi}{15}\cos\frac{6\pi}{15}\big)\times\big(-\cos\frac{8\pi}{15}\big)$
$=-\frac{1}{2}\big[\cos\frac{\pi}{15}\cos\frac{2\pi}{15}\cos\frac{4\pi}{15}\cos\frac{8\pi}{15}\big]\times\frac{1}{2}\times\big(\cos\frac{3\pi}{15}\cos\frac{6\pi}{15}\big)$
$=-\frac{1}{2}\times\frac{2^3}{2^4\sin\frac{\pi}{15}}[2\sin\frac{\pi}{15}\cos\frac{\pi}{15}\cos\frac{2\pi}{15}\cos\frac{4\pi}{15}\cos\frac{8\pi}{15}]\times\frac{2}{2^2\times\sin\frac{3\pi}{15}}(2\sin\frac{3\pi}{15}\cos\frac{3\pi}{15}\cos\frac{6\pi}{15})$
$=-\frac{2^3}{132\sin\frac{\pi}{15}}[\sin\frac{2\pi}{15}\cos\frac{2\pi}{15}\cos\frac{4\pi}{15}\cos\frac{8\pi}{15}]\times\frac{2}{4\sin\frac{3\pi}{15}}(\sin\frac{6\pi}{15}\cos\frac{6\pi}{15})$
$=-\frac{2^3}{132\sin\frac{\pi}{15}}[2\sin\frac{2\pi}{15}\cos\frac{2\pi}{15}\cos\frac{4\pi}{15}\cos\frac{8\pi}{15}]\times\frac{1}{4\sin\frac{3\pi}{15}}(2\sin\frac{6\pi}{15}\cos\frac{6\pi}{15})$
$=-\frac{2}{32\sin\frac{\pi}{15}}[\sin\frac{16\pi}{15}]\times\frac{\sin\frac{12\pi}{15}}{4\sin\frac{3\pi}{15}}$
$=-\frac{\sin\big(\pi+\frac{\pi}{15}\big)}{128\sin\frac{\pi}{15}}\times\frac{\sin\big(\pi-\frac{3\pi}{15}\big)}{\sin\frac{3\pi}{15}}$
$=-\frac{-\sin\frac{\pi}{15}}{128\sin\frac{\pi}{15}}\times\frac{\sin\frac{3\pi}{15}}{\sin\frac{3\pi}{15}}$
$=\frac{1}{128}$
$=\text{RHS}$
View full question & answer→Question 265 Marks
$\tan\text{x}\tan(\text{x}+\frac{\pi}{3})+\tan\text{x}(\frac{\pi}{3}-\text{x})\\+\tan(\text{x}+\frac{\pi}{3})\tan(\text{x}-\frac{\pi}{3})=-3$
AnswerWe have to prove that
$\sin5\text{x}=5\cos^4\text{x}\sin\text{x}-10\cos^2\text{x}\sin^3\text{x}+\sin^5\text{x}$
$\text{LHS}=\sin\text{x}=\sin(3\text{x}+2\text{x})$
$=\sin3\text{x}\cos2\text{x}+\cos3\text{x}.\sin2\text{x}$
$=(3\sin\text{x}-4\sin^3\text{x})(2\cos^2\text{x}-1)\\+(4\cos^3\text{x}-3\cos\text{x})2\sin\text{x}\cos\text{x}.$
$=-3\sin\text{x}+4\sin^3\text{x}+6\sin\text{x}\cos^2\text{x}-8\sin^3\text{x}\cos^2\text{x}\\+8\cos^4\text{x}\sin\text{x}-6\cos^2\text{x}\sin\text{x}$
$=8\cos^4\text{x}\sin\text{x}-8\sin^3\text{x}\cos^2\text{x}-3\sin\text{x}+4\sin^3\text{x}$
$=5\cos^4\text{x}\sin\text{x}-10\sin^3\text{x}\cos^2\text{x}-3\sin\text{x}\\+3\cos^4\text{x}\sin\text{x}+4\sin^3\text{x}+2\sin^3\text{x}\cos^2\text{x}$
$=5\cos^4\text{x}\sin\text{x}-10\sin^3\text{x}\cos^2\text{x}\\-3\sin\text{x}(1-\cos^4\text{x})+2\sin^3\text{x}(2+\cos^2\text{x})$
$=5\cos^4\text{x}\sin\text{x}-10\sin^3\text{x}\cos^2\text{x}\\-3\sin\text{x}(1-\cos^2\text{x})(1+\cos^2\text{x})+2\sin^3\text{x}(2+\cos^2\text{x})$
$=5\cos^4\text{x}\sin\text{x}-10\sin^3\text{x}\cos^2\text{x}\\-3\sin^3\text{x}(1+\cos^2)+2\sin^3(2+\cos^2\text{x})$
$=5\cos^4\text{x}\sin\text{x}-10\sin^3\text{x}\cos^2\text{x}\\-\sin^3\text{x}\big[3(1+\cos^2\text{x})-2(2+\cos^2\text{x})\big]$
$=5\cos^4\text{x}\sin\text{x}-10\sin^3\text{x}\cos^2\text{x}\\-\sin^3\text{x}\big[3+3\cos^2\text{x}-4-2\cos^2\text{x}\big]$
$=5\cos^4\text{x}\sin\text{x}-10\sin^3\text{x}\cos^2\text{x}-\sin^3\text{x}\big[\cos^2\text{x}-1\big]$
$=5\cos^4\text{x}\sin\text{x}-10\sin^3\text{x}\cos^2\text{x}+\sin^5\text{x}$
$=\text{RHS}$
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