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Trigonometry : Trigonometry

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Example Questions

Example Question #2 : Trigonometric Functions

If $x=\frac{\pi}{6}$ , find the value of $\frac{1+cos\ x}{sin^3x}$ .

Possible Answers:

$8+4\sqrt{2}$

$8-4\sqrt{3}$

$4+4\sqrt{3}$

$-8+4\sqrt{3}$

$8+4\sqrt{3}$

Correct answer:

$8+4\sqrt{3}$

Explanation:

$x=\frac{\pi}{6}\Rightarrow \frac{{1+cos\ x}}{}{sin^3x}=\frac{{1+cos\ (\pi/6)}}{sin^3(\pi/6)}=\frac{1+\frac{\sqrt{3}}{2}}{(\frac{1}{2})^3}=4(2+\sqrt{3})=8+4\sqrt{3}$

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Example Question #2 : Trigonometric Functions

Find the value of the following expression:

$sin^2 (\frac{\pi}{15})+\frac{1}{sec^2(\frac{\pi}{15})}$

Possible Answers:

Correct answer:

Explanation:

We know that $sec\ x= \frac{1}{cos\ x}$ .

$sin^2 (\frac{\pi}{15})+\frac{1}{sec^2(\frac{\pi}{15})}=sin^2 (\frac{\pi}{15})+\frac{1}{\frac{1}{cos^2(\frac{\pi}{15})}}=sin^2 (\frac{\pi}{15})+cos^2 (\frac{\pi}{15})=1$

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Example Question #2 : Trigonometric Functions

If $sec\ x=\frac{\sqrt{5}}{2}$ , give $tan\ x$ :

Possible Answers:

$tan\ x=\pm 1$

$tan\ x=\pm \frac{1}{2}$

$tan\ x=\pm 3$

$tan\ x=\pm 4$

$tan\ x=\pm 2$

Correct answer:

$tan\ x=\pm \frac{1}{2}$

Explanation:

We need to use the identity sec^2x=1+tan^2x :

$sec^2x=1+tan^2x\Rightarrow (\frac{\sqrt{5}}2)^2=1+tan^2x$

$\Rightarrow tan^2x=\frac{5}{4}-1\Rightarrow tan^2x=\frac{1}{4}$

$\Rightarrow tan\ x=\pm \frac{1}{2}$

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Example Question #3 : Trigonometric Functions

Give the value of $sec\ 26^{\circ} - csc\ 64^{\circ}$ .

Possible Answers:

Correct answer:

Explanation:

We need to use the identity $csc(90^{\circ}-x)=sec\ x$ :

$sec\ 26^{\circ} - csc\ 64^{\circ}=sec\ 26^{\circ} - csc (90-26)^{\circ}=sec\ 26^{\circ} -sec\ 26^{\circ}=0$

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Example Question #81 : Trigonometry

Find $\frac{cos\ 65^{\circ}}{sin\ 25^{\circ}}$ .

Possible Answers:

$sin\ 25^{\circ}$

$-sin\ 25^{\circ}$

Correct answer:

Explanation:

We need to use the identity $cos(90^{\circ}-x)=sin\ x$ :

$\frac{cos\ 65^{\circ}}{sin\ 25^{\circ}}=\frac{cos (90-25)^{\circ}}{sin\ 25^{\circ}}=\frac{sin\ 25^{\circ}}{sin\ 25^{\circ}}=1$

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Example Question #82 : Trigonometry

If $cos\ x=A$ , give $cos\ 2x$ in terms of .

Possible Answers:

2A^2-1

A^2-1

2A^2+1

A+1

A-1

Correct answer:

2A^2-1

Explanation:

We need to use the identity cos2x=2cos^2x-1 .

cos2x=2cos^2x-1=2A^2-1

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Example Question #83 : Trigonometry

If $cos\ x=\frac{\sqrt{2}}{2}$ , find $cos\ 2x$ .

Possible Answers:

$-\frac{\sqrt{2}}{2}$

$\frac{\sqrt{2}}{2}$

Correct answer:

Explanation:

We need to use the identity cos2x=2cos^2x-1 .

$cos2x=2cos^2x-1=2(\frac{\sqrt{2}}{2})^2-1=2(\frac{1}{2})-1=1-1=0$

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Example Question #11 : Trigonometric Functions

If $sin\ x=A$ , express $sin\ 2x$ in terms of . ( $0<x<30^{\circ}$ )

Possible Answers:

$2A\sqrt{1+A^2}$

$2\sqrt{1-A^2}$

$A\sqrt{1+A^2}$

$2A\sqrt{1-A^2}$

$A\sqrt{1-A^2}$

Correct answer:

$2A\sqrt{1-A^2}$

Explanation:

We need to use the double angle formula:

$sin\ 2x=2(sin\ x)(cos\ x)$

$sin\ x$ is known, but we need to find $cos \ x$ :

$sin^2x+cos^2x=1\Rightarrow cos\ x=\pm \sqrt{1-sin^2x}$

In this problem is a first quadrant angle ( $0<x<30^{\circ}$ ), so we can only use the positive value for $cos\ x$ .

$cos\ x=\sqrt{1-sin^2x}=\sqrt{1-A^2}$

$sin\ 2x=2(sin\ x)(cos\ x)=2A\sqrt{1-A^2}$

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Example Question #11 : Trigonometric Functions

If $tan\ x=\sqrt{3}$ , what is the value of $cos\ x$ ? ( $0^{\circ}<x<90^{\circ}$ )

Possible Answers:

0.5

0.75

0.25

$\frac{\sqrt{3}}2{}$

Correct answer:

0.5

Explanation:

We need to use a Pythagorean identity:

$sec^2x=1+tan^2x=1+(\sqrt{3})^2=1+3=4\Rightarrow sec\ x=\pm 2$

Since $0^{\circ}<x<90^{\circ}$ , we can only use the positive value of $sec\ x$ . That means $sec\ x=2$ .

$cos\ x=\frac{1}{sec\ x}=\frac{1}{2}=0.5$

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Example Question #85 : Trigonometry

If $cot\ x=\sqrt{3}$ , what is the value of $cos\ x$ ? ( $0^{\circ}<x<90^{\circ}$ )

Possible Answers:

$\frac{1}{2}$

$\frac{\sqrt{3}}{2}$

$\frac{\sqrt{2}}{2}$

Correct answer:

$\frac{\sqrt{3}}{2}$

Explanation:

We need to use a Pythagorean identity:

$csc^2x=1+cot^2x=1+(\sqrt{3})^2=1+3=4\Rightarrow csc\ x=\pm 2$

Since $0^{\circ}<x<90^{\circ}$ , we can only use the positive value for $csc\ x$ . That means $csc\ x=2$ .

$sin\ x=\frac{1}{csc\ x}=\frac{1}{2}$

Based on another Pythagorean identity we have:

$cos^2x+sin^2x=1\Rightarrow cos^2x=1-sin^2x=1-(\frac{1}{2})^2=1-\frac{1}{4}=\frac{3}{4}$

$\Rightarrow cos\ x=\pm \sqrt{(\frac{3}{4})}=\pm \frac{\sqrt{3}}{2}$

Since $0^{\circ}<x<90^{\circ}$ , we can again only use the positive value for $cos\ x$ .

$cos\ x=\frac{\sqrt{3}}{2}$

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Trigonometry : Trigonometry

Study concepts, example questions & explanations for Trigonometry

All Trigonometry Resources

Example Questions

Example Question #2 : Trigonometric Functions

Example Question #2 : Trigonometric Functions

Example Question #2 : Trigonometric Functions

Example Question #3 : Trigonometric Functions

Example Question #81 : Trigonometry

Example Question #82 : Trigonometry

Example Question #83 : Trigonometry

Example Question #11 : Trigonometric Functions

Example Question #11 : Trigonometric Functions

Example Question #85 : Trigonometry

All Trigonometry Resources

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