2 Marks Questions

Question 12 Marks

Prove that:
$\sqrt{2+\sqrt{2+2\cos4\text{x}}}=2\cos\text{x},0\cos\text{x},<\text{x}<\frac{\pi}{4}$

Answer

$\text{LHS}=\sqrt{2+\sqrt{2+2\cos4\text{x}}}$
$=\sqrt{2+\sqrt{2(1+\cos4\text{x})}}$
$=\sqrt{2+\sqrt{2.2\cos^22\text{x}}}$
$=\sqrt{2+2\cos2\text{x}}$
$=\sqrt{2(1+\cos2\text{x})}$
$=\sqrt{2.2\cos ^2\text{x}}$
$=2\cos\text{x}=\text{RHS}$

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Question 22 Marks

If $\cos\text{x}=\frac{4}{5}$ and x is acute, find $\tan 2\text{x}$

Answer

$\text{since}\ \theta\ \text{in acute},\text{so}\ 0\leq2\theta<\pi$
Now,
$\text{cos}\theta=\frac{4}{5}=\frac{\text{b}}{\text{h}}\Rightarrow\text{p}=3$
$\text{h}=5$
$\therefore\sin\theta=\frac{\text{p}}{\text{h}}=\frac{3}{5}$
$\tan\theta=\frac{\text{p}}{\text{b}}=\frac{3}{4}$
so,
$\tan2\theta=\frac{2\tan\theta}{1-\tan^2\theta}=\frac{2.\frac{3}{4}}{1-\Big(\frac{3}{4}\Big)^2}$
$=\frac{\frac{6}{4}}{\frac{7}{16}}=\frac{24}{7}$

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Question 32 Marks

Prove that:
$\frac{\cos2\text{x}}{1+\sin2\text{x}}=\tan(\frac{\pi}{4}-\text{x})$

Answer

$\text{LHS}=\frac{\cos2\text{x}}{1+\sin2\text{x}}$ $=\frac{\cos^2\text{x}-\sin^2\text{x}}{\sin^2+\cos^2\text{x}+2\sin\text{x}\cos\text{x}}$ $[\because\cos2\text{x}=\cos^2\text{x}-\sin^2\text{x}\ \&\sin^2\text{x}+\cos^2\text{x}=1]$ $=\frac{(\cos\text{x}-\sin\text{x})(\cos\text{x}+\sin\text{x})}{(\cos\text{x}+\sin\text{x})^2}$ $[\because\text{a}^2-\text{b}^2=(\text{a+b})(\text{a-b})]$ $=\frac{\cos\text{x}-\sin\text{x}}{\cos\text{x}+\sin\text{x}}$ Dividing num erator and demom enator by $\cos\text{x}$ $=\frac{1-\tan\text{x}}{1+\tan\text{x}}$ $=\tan(\frac{\pi}{4}-\text{x})=\frac{\tan\frac{\pi}{4}-\tan\text{x}}{1+\tan\frac{\pi}{4}\tan\text{x}}$ $=\frac{1-\tan\text{x}}{1+\tan\text{x}}\ \text{RHS}$Note: $\tan(\frac{\pi}{4}-\text{x})=\frac{\tan\frac{\pi}{4}-\tan\text{x}}{1+\tan\frac{\pi}{4}\tan\text{x}}$
$=\frac{1-\tan\text{x}}{1+\tan\text{x}}$

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Question 42 Marks

Prove that:
$\cos^2\frac{\pi}{8}+\cos^2\frac{3\pi}{8}+\cos^2\frac{5\pi}{8}+\cos^2\frac{7\pi}{8}=2$

Answer

$\text{LHS}=\cos^2\frac{\pi}{8}+\cos^2\frac{3\pi}{8}+\cos^2\frac{5\pi}{8}+\cos^2\frac{7\pi}{8}$
$\cos^2\frac{\pi}{8}+\cos^2\frac{3\pi}{8}+\cos^2\big(\pi-\frac{3\pi}{8}\big)+\cos^2\big(\pi-\frac{\pi}{8}\big)$
$\cos^2\frac{\pi}{8}+\cos^2\frac{3\pi}{8}+\cos^2\frac{3\pi}{8}+\cos^2\frac{\pi}{8}$
$=2\big(\cos^2\frac{\pi}{8}+\cos^2\frac{3\pi}{8}\big)$
$=2\big(\cos^2\frac{\pi}{8}+\cos^2\big(\frac{\pi}{2}-\frac{\pi}{8}\big)\big)$
$=2\big(\cos^2\frac{\pi}{8}+\sin^2\frac{\pi}{8}$
$=2\ \text{RHS}$

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Question 52 Marks

If $\text{a}\cos2\text{x}+\text{b}\sin2\text{x}=\text{c}$ has $\alpha$ and $\beta$ as its roots, then prove that,
$\tan(\alpha+\beta)=\frac{\text{b}}{\text{a}}$

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
$\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}=\frac{2\text{b}}{\text{c+a=c+a}}=\frac{\text{b}}{\text{a}}$

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Question 62 Marks

Prove that:
$1+\cos^22\text{x}=2(\cos^4\text{x}+\sin^4\text{x})$

Answer

$\text{LHS}=1+\cos^22\text{x}$
$=1+(\cos^2\text{x}-\sin^2\text{x})$ $[\because\cos2\text{x}=\cos^2\text{x}-\sin^2\text{x}]$
$=1+\cos^4\text{x}+\sin^4\text{x}-2\sin^2\text{x}.\cos^2\text{x}$
$=(\sin^2\text{x}+\cos^2\text{x})^2+\cos^4\text{x}+\sin^4\text{x}-2\sin\text{x},\cos^2\text{x}$ $[\because\sin^2\text{x}+\cos^2\text{x}=1]$
$=\sin^4\text{x}+\cos^4\text{x}+2\sin^2\text{x}\cos^2\text{x}+\cos^4\text{x}+\sin^4\text{x}-2\sin^2\text{x}.\cos^2\text{x}$
$=2\big(\cos^4\text{x}+\sin^4\text{x}\big)\ \text{RHS}$

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Question 72 Marks

Prove that:
$\sin^2\frac{\pi}{8}+\sin^2\frac{3\pi}{8}+\sin^2\frac{5\pi}{8}+\sin^2\frac{7\pi}{8}=2$

Answer

$\text{LHS}=\sin^2\frac{\pi}{8}+\sin^2\frac{3\pi}{8}+\sin^2\frac{5\pi}{8}+\sin^2\frac{7\pi}{8}$
$=\sin^2\big(\frac{\pi}{2}-\frac{3\pi}{8}\big)+\sin^2\big(\frac{\pi}{2}-\frac{\pi}{8}\big)+\sin^2\frac{5\pi}{8}+\sin^2\frac{7\pi}{8}$
$=\cos^2\frac{3\pi}{8}+\sin^2\frac{\pi}{8}+\sin^2\big(\pi-\frac{3\pi}{8}\big)+\sin^2\big(\pi-\frac{\pi}{8}\big)$
$=\cos^2\frac{3\pi}{8}+\sin^2\frac{\pi}{8}+\sin^2\frac{3\pi}{8}+\cos^2\frac{\pi}{8}$
$=\big(\cos^2\frac{\pi}{8}+\sin^2\frac{\pi}{8}\big)+\big(\cos^2\frac{3\pi}{8}+\sin^2\frac{3\pi}{8}\big)$
$=1+1=2=\text{RHS}$
Hence proved.

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Question 82 Marks

Prove that:
$\frac{1-\cos2\text{x}+\sin2\text{x}}{1+\cos2\text{x}+\sin2\text{x}}=\tan\text{x}$

Answer

$\text{LHS}=\frac{1-\cos2\text{x}+\sin2\text{x}}{1+\cos2\text{x}+\sin2\text{x}}$
$=\frac{2\sin^2\text{x}+2\sin\text{x},\cos\text{x}}{2\cos^2+2\sin\text{x},\cos\text{x}}$
$=\frac{2\sin\text{x}(\sin\text{x}+\cos\text{x})}{2\cos\text{x}(\cos\text{x}+\sin\text{x})}$
$=\frac{\sin\text{x}}{\cos\text{x}}$
$=\tan\text{x}=\text{RHS}$

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Question 92 Marks

Prove that:
$(\sin^3\text{x}+\sin\text{x})\sin\text{x}+(\cos3\text{x}-\cos\text{x})\cos\text{x}=0$

Answer

$\text{LHS}=(\sin3\text{A}+\sin\text{A})\sin\text{A}(\cos3\text{A}-\cos\text{A})\cos\text{A}$
$\Rightarrow2\sin2\text{A}.\cos\text{A}.\sin\text{A}+(-2\sin2\text{A}.\sin\text{A}\cos\text{A})$ $\begin{bmatrix}\because\sin\text{C}+\sin\text{D}=2\sin\frac{\text{C+D}}{2}.\cos\frac{\text{C}-\text{D}}{2}\\\cos\text{C}-\cos\text{D}=-2\sin\frac{\text{C+D}}{2}.\sin\frac{\text{C+D}}{2}\end{bmatrix}$
$\Rightarrow2\sin2\text{A}.\cos\text{A}.\sin\text{A}-2\sin2\text{A}\cos\text{A}.\sin\text{A}$
$\Rightarrow0=\text{RHS}$

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Question 102 Marks

Prove that:
$\frac{\cos\text{x}}{1-\sin\text{x}}=\tan(\frac{\pi}{4}+\frac{\pi}{2})$

Answer

$\text{LHS}=\frac{\cos\text{x}}{1-\sin\text{x}}$
$\frac{\cos^2\frac{\text{x}}{2}-\sin^2\frac{\text{x}}{2}}{\sin^2\frac{\text{x}}{2}+\cos^2\frac{\text{x}}{2}-2\sin\frac{\text{x}}{2},\frac{\cos\text{x}}{2}}$ $[\because\cos2\text{x}=\cos^2\text{x}-\sin2\text{x}\ \&\sin^2\text{x}+\cos^2\text{x}]=1$
$\frac{(\cos\frac{\text{x}}{2}-\sin\frac{\text{x}}{2})(\cos\frac{\text{x}}{2}+\sin\frac{\text{x}}{2})}{(\cos\frac{\text{x}}{2}-\sin\frac{\text{x}}{2})}$
$=\frac{\cos\frac{\text{x}}{2}+\sin\frac{\text{x}}{2}}{\cos\frac{\text{x}}{2}-\sin\frac{\text{x}}{2}}$
Dividing numeretor and denominator by $\cos\frac {\text{x}}{2}$
$=\frac{1+\tan\frac{\text{x}}{2}}{1=\tan\frac{\text{x}}{2}}$
$\tan(\frac{\pi}{4}+\frac{\text{x}}{2})=\text{RHS}$

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Question 112 Marks

If $\cos\alpha+\cos\beta=0\sin\alpha+\sin\beta,$ then prove that $\cos2\alpha+\cos2\beta=-2\cos(\alpha+\beta).$

Answer

$\cos\alpha+\cos\beta=0\sin\alpha+\sin\beta$
Squaring on both sides gives
$\cos^2\alpha+\cos^2\beta+2\cos\alpha\cos\beta=\sin^2\alpha+\sin^2\beta+2\sin\alpha\sin\beta$
Bring square trems on one side, we get
$\cos^2\alpha+\cos^2\beta-2(-2\sin\alpha\sin\beta=\cos^2\alpha+\cos^2\beta)-2(\alpha+\beta)$

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Question 122 Marks

Prove that:
$\cos78^\circ\cos42^\circ\cos36^\circ=\frac{1}{8}$

Answer

$\text{LHS}=\cos78^\circ.\cos42^\circ.\cos36^\circ$
$=\frac{(2\cos78^\circ.\cos42^\circ)}{2}.\cos36^\circ$
$=\frac{1}{2}(\cos120^\circ+\cos36^\circ).\cos36^\circ$
$=\frac{1}{2}\Big(\frac{-1}{2}+\frac{\sqrt{5}+1}{4}\Big)\frac{\sqrt{5}+1}{4}$
$=\frac{1}{8}\frac{\big[-2\big(\sqrt{5}+1\big)+5+1+2\sqrt{5}\big]}{4}$
$=\frac{1}{4}\big[\frac{4}{4}\big]$
$=\frac{1}{8}$
$=\text{RHS}$

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Question 132 Marks

Prove that:
$\cos^2(\frac{\pi}{4}-\text{x})-\sin^2(\frac{\pi}{4}-\text{x})=\sin2\text{x}$

Answer

$\text{LHS}=\cos^2\Big(\frac{\pi}{4}-\text{x}\Big)-\sin^2\Big(\frac{\pi}{4}-\text{x}\Big)$
$=\cos2\Big(\frac{\pi}{4}-\text{x}\Big)$ $\Big[\because\cos2\text{x}=\cos^2\theta-\sin^2\text{x}\Big]$
$=\cos(\frac{\pi}{2}-2\text{x})$ $\Big[\because\cos\Big(\frac{\pi}{2}-\text{x}\Big)=\sin\text{x}\Big]$
$=\sin2\text{x}=\text{RHS}$

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Question 142 Marks

Prove that:
$\sin^224^\circ-\sin^26^\circ=\frac{\sqrt{5}-1}{8}$

Answer

$\text{LHS}=\sin^224^\circ-\sin^26^\circ$
$=\sin(24+6)\sin(24-6)\\ [\because\sin(\text{A+B})\sin(\text{A}-\text{B}=\sin^2\text{A}-\sin^2\text{B})]$
$=\sin30^\circ\sin18^\circ$
$=\frac{1}{2}.\frac{\sqrt{5}-1}{4}\Big[\because\sin18^\circ=\frac{\sqrt{5}-1}{4}\Big]$
$=\frac{\sqrt{5}-1}{8}$
$=\text{RHS}$

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Question 152 Marks

Prove that:
$\frac{\sin\text{x}+\sin2\text{x}}{1+\cos\text{x}+\cos2\text{x}}=\tan\text{x}$

Answer

$\text{LHS}=\frac{\sin\text{x}+\sin2\text{x}}{1+\cos\text{x}+\cos2\text{x}}$
$=\frac{\sin\text{x}+2\sin\text{x},\cos\text{x}}{\cos\text{x}+(1+\cos2\text{x})}$
$=\frac{\sin\text{x}(1+2\cos\text{x})}{\cos\text{x}+2\cos\text{x}}$
$=\frac{\sin\text{x}(1+2\cos\text{x})}{\cos\text{x}(1+2\cos\text{x})}$
$=\frac{\sin\text{x}}{\cos\text{x}}$
$=\tan\text{x}=\text{RHS}$

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Question 162 Marks

Prove that:
$\sin^2\frac{2\pi}{5}=\sin^2\frac{\pi}{3}=\frac{\sqrt{5}-1}{8}$

Answer

We have,
$\sin^272^\circ-\sin^260^\circ,$
$=\sin^2(90^\circ-18^\circ)-\Big(\frac{\sqrt{3}}{2}\Big)^2$
$=\cos^218^\circ-\frac{3}{4}$
$\Bigg(\frac{\sqrt{10+2+\sqrt{5}}}{4}\Bigg)^2-\frac{3}{4}\Bigg[\because\cos18^\circ=\frac{\sqrt{10+2\sqrt{5}}}{4}\Bigg]$
$=\frac{10+2\sqrt{5}}{16}-\frac{3}{4}$
$=\frac{10+2\sqrt{5}-12}{16}$
$=\frac{2\sqrt{5}-2}{16}$
$=\frac{\sqrt{5}-1}{8}$

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