2 Marks Questions

Question 12 Marks

Evaluate the following:
If A = 45°, verify that:
$\cos2\text{A}=2\cos^2\text{A}-1=1-2\sin^2\text{A}$

Answer

$\text{A}=45^\circ$
$\Rightarrow2\text{A}=2\times45^\circ=90^\circ$
$\cos2\text{A}=2\cos90^\circ=0$
$2\cos^2\text{A}-1=2\cos^245-1$
$=2\times\Big(\frac{1}{\sqrt{2}}\Big)^2-1$
$=2\times\frac12-1=1-1=0$
Now, $1-2\sin^2\text{A}=1-2\times\Big(\frac{1}{\sqrt{2}}\Big)^2$
$=1-2\times\frac12=1-1=0$
$\therefore\ \cos2\text{A}=2\cos^2\text{A}-1=1-2\sin^2\text{A}$

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Question 22 Marks

Evaluate the following:
Verify the following:
$\cos60^\circ\cos30^\circ+\sin60^\circ\sin30^\circ=\cos30^\circ$

Answer

$\cos60^\circ\cos30^\circ+\sin60^\circ\sin30^\circ$
$=\Big(\frac12\Big)\times\Big(\frac{\sqrt{3}}{2}\Big)+\Big(\frac{\sqrt{3}}{2}\Big)\times\Big(\frac12\Big)$
$=\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4}=\frac{\sqrt{3}}{2}$
Also, $\cos30^\circ=\frac{\sqrt{3}}{2}$
$\therefore\ \cos60^\circ\cos30^\circ+\sin60^\circ\sin30^\circ=\cos30^\circ$

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Question 32 Marks

Evaluate the following:
$\frac{4}{\cot^230^\circ}+\frac{1}{\sin^230^\circ}-2\cos^245^\circ-\sin^20^\circ$

Answer

On substituting values of various T-ratious, we get:
$\frac{4}{\cot^230^\circ}+\frac{1}{\sin^230^\circ}-2\cos^245^\circ-\sin^20^\circ$
$=\frac{4}{\big(\sqrt{3}\big)^2}+\frac{1}{\big(\frac12\big)^2}-2\times\Big(\frac{1}{\sqrt{2}}\Big)^2-(0)^2$
$=\frac43+\frac11-2\times\frac12-0$
$=\frac43+4-1$
$=\frac43+3=\frac{4+9}{3}=\frac{13}{3}$

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Question 42 Marks

Evaluate the following:
$\frac{5\cos^260^\circ+4\sec^230^\circ-\tan^245^\circ}{\sin^230^\circ+\cos^230^\circ}$

Answer

$\frac{5\cos^260^\circ+4\sec^230^\circ-\tan^245^\circ}{\sin^230^\circ+\cos^230^\circ}$
$=\frac{5\big(\frac12\big)^2+4\big(\frac{2}{\sqrt{3}}\big)^2-(1)^2}{\big(\frac12\big)^2+\Big(\frac{\sqrt{3}}{2}\Big)^2}$
$=\frac{\Big(\frac54+\frac{4+4}{3}-1\Big)}{\Big(\frac14+\frac34\Big)}$
$=\frac{\Big(\frac{5}{4}+\frac{16}{3}-\frac{1}{1}\Big)}{\big(\frac44\big)}$
$=\frac{\big(\frac{15+64-12}{12}\big)}{\big(\frac44\big)}$
$=\frac{\big(\frac{67}{12}\big)}{(1)}$
$=\frac{67}{12}$

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Question 52 Marks

Evaluate the following:
Show that:
$\frac{\cos30^\circ+\sin60^\circ}{1+\sin30^\circ+\cos60^\circ}=\cos30^\circ$

Answer

$\text{L.H.S.}=\frac{\cos30^\circ+\sin60^\circ}{1+\sin30^\circ+\cos60^\circ}=\frac{\Big(\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}\Big)}{\Big(1+\frac{1}{2}+\frac12\big)}$
$=\frac{\frac{\sqrt{3}+\sqrt{3}}{2}}{\frac{2+1+1}{2}}=\frac{\sqrt{3}}{2}$
Also,
$\text{R.H.S.}=\cos30^\circ=\frac{\sqrt{3}}{2}$
Hence, L.H.S. = R.H.S.

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Question 62 Marks

Evaluate the following:
$\frac{\sin30^\circ}{\cos45^\circ}+\frac{\cot45^\circ}{\sec60^\circ}-\frac{\sin60^\circ}{\tan45^\circ}+\frac{\cos30^\circ}{\sin90^\circ}$

Answer

$\frac{\sin30^\circ}{\cos45^\circ}+\frac{\cot45^\circ}{\sec60^\circ}-\frac{\sin60^\circ}{\tan45^\circ}+\frac{\cos30^\circ}{\sin90^\circ}$
$=\frac{\big(\frac14\big)}{\big(\frac{1}{\sqrt{2}}\big)}+\frac12-\frac{\Big(\frac{\sqrt{3}}{2}\Big)}{1}+\frac{\Big(\frac{\sqrt{3}}{2}\Big)}{1}$
$=\frac{\sqrt{2}}{2}+\frac12-\frac{{\sqrt{3}}}{2}+\frac{\sqrt{3}}{2}$
$=\frac{\sqrt{2}+1}{2}$

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Question 72 Marks

Evaluate the following:
Verify the following:
$2\sin30^\circ\cos30^\circ=\sin60^\circ$

Answer

$2\sin30^\circ\cos30^\circ=\sin60^\circ$
$=2\times\frac{1}{2}\times\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2}$
Also, $\sin60^\circ=\frac{\sqrt{3}}{2}$
$\therefore\ 2\sin30^\circ\cos30^\circ=\sin60^\circ$

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Question 82 Marks

Evaluate the following:
Verify the following:
$\sin60^\circ\cos30^\circ-\cos60^\circ\sin30^\circ=\sin30^\circ$

Answer

$\sin60^\circ\cos30^\circ-\cos60^\circ\sin30^\circ=\sin30^\circ$
$=\Big(\frac{\sqrt{3}}{2}\Big)\times\Big(\frac{\sqrt{3}}{2}\Big)-\Big(\frac12\Big)\times\Big(\frac12\Big)$
$=\frac34-\frac14=\frac24=\frac12$
Also, $\sin30^\circ=\frac{1}{2}$
$\therefore\ \sin60^\circ\cos30^\circ-\cos60^\circ\sin30^\circ$

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Question 92 Marks

Evaluate the following:
$\cot^230^\circ-2\cos^230^\circ-\frac34\sec^245^\circ+\frac14\text{cosec}^230^\circ$

Answer

On substituting values of various T-ratious, we get:
$\cot^230^\circ-2\cos^230^\circ-\frac34\sec^245^\circ+\frac14\text{cosec}^230^\circ$
$=\big(\sqrt{3}\big)^2-2\times\Big(\frac{\sqrt{3}}{2}\Big)^2-\frac34\times\big(\sqrt{2}\big)^2+\frac14\times(2)^2$
$=3-2\times\frac34-\frac34\times2+\frac14\times4$
$=3-\frac32-\frac32+1$
$=4-\Big(\frac{3}{2}+\frac{3}{2}\Big)$
$=4-3$
$=1$

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Question 102 Marks

Evaluate the following:
$2\cos^260^\circ+3\sin^245^\circ-3\sin^230^\circ+2\cos^290^\circ$

Answer

On substituting values of various T-ratious, we get:
$2\cos^260^\circ+3\sin^245^\circ-3\sin^230^\circ+2\cos^290^\circ$
$=2\times\Big(\frac12\Big)^2+3\times\Big(\frac{1}{\sqrt{2}}\Big)^2-3\times\Big(\frac12\Big)^2+2\times(0)^2$
$=2\times\frac14+3\times\frac12-3\times\frac14+0$
$=\Big(\frac12+\frac32-\frac34\Big)=\Big(\frac{2+6-3}{4}\Big)=\frac54$

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Question 112 Marks

Evaluate the following:
If A = 45°, verify that:
$\sin2\text{A}=2\sin\text{A}\cos\text{A}$

Answer

$\text{A}=45^\circ$
$\Rightarrow2\text{A}=2\times45^\circ=90^\circ$
$\sin2\text{A}=\sin90^\circ=1$
$2\sin\text{A}\cos\text{A}=2\sin45^\circ\cos45^\circ$
$=2\times\frac{1}{\sqrt{2}}\times\frac{1}{\sqrt{2}}=1$
$\therefore\ \sin2\text{A}=2\sin\text{A}\cos\text{A}$

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MATHS 2 Marks Questions

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