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Trigonometry : Trigonometric Functions and Graphs

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Example Questions

Example Question #1 : Trigonometric Functions

Rt_triangle_letters

In this figure, if angle $c=90^\circ$ $\displaystyle c=90^\circ$, side X=12 $\displaystyle X=12$, and side Z=8 $\displaystyle Z=8$, what is the measure of angle $\displaystyle a$?

Possible Answers:

$41.8^\circ$ $\displaystyle 41.8^\circ$

$\frac{2}{3}^\circ$ $\displaystyle \frac{2}{3}^\circ$

$90^\circ$ $\displaystyle 90^\circ$

Undefined

$48.2^\circ$ $\displaystyle 48.2^\circ$

Correct answer:

$41.8^\circ$ $\displaystyle 41.8^\circ$

Explanation:

Since $c=90^\circ$ $\displaystyle c=90^\circ$, we know we are working with a right triangle.

That means that $\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}$ $\displaystyle \sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}$.

In this problem, that would be:

$\sin(a)=\frac{\text{Z}}{\text{X}}$ $\displaystyle \sin(a)=\frac{\text{Z}}{\text{X}}$

Plug in our given values:

$\sin(a)=\frac{8}{12}$ $\displaystyle \sin(a)=\frac{8}{12}$

$\sin(a)=\frac{2}{3}$ $\displaystyle \sin(a)=\frac{2}{3}$

$a=\sin^{-1}(\frac{2}{3})$ $\displaystyle a=\sin^{-1}(\frac{2}{3})$

$a=41.8^\circ$ $\displaystyle a=41.8^\circ$

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Example Question #1 : Trigonometric Functions

If $sin^2x+2cos^2x+3(sin\ x)(cos\ x)=3$ $\displaystyle sin^2x+2cos^2x+3(sin\ x)(cos\ x)=3$, give the value of $cot\ x$ $\displaystyle cot\ x$.

Possible Answers:

$cot\ x=-1\ or\ cot\ x=2$ $\displaystyle cot\ x=-1\ or\ cot\ x=2$

$cot\ x=1$ $\displaystyle cot\ x=1$

$cot\ x=2$ $\displaystyle cot\ x=2$

$cot\ x=1\ or\ cot\ x=-2$ $\displaystyle cot\ x=1\ or\ cot\ x=-2$

$cot\ x=1\ or\ cot\ x=2$ $\displaystyle cot\ x=1\ or\ cot\ x=2$

Correct answer:

$cot\ x=1\ or\ cot\ x=2$ $\displaystyle cot\ x=1\ or\ cot\ x=2$

Explanation:

$sin^2x+2cos^2x+3(sin\ x)(cos\ x)=3$ $\displaystyle sin^2x+2cos^2x+3(sin\ x)(cos\ x)=3$

$\Rightarrow sin^2x+(cos^2x+cos^2x)+3(sin\ x)(cos\ x)=3$ $\displaystyle \Rightarrow sin^2x+(cos^2x+cos^2x)+3(sin\ x)(cos\ x)=3$

Use the identity $\displaystyle sin^2x+cos^2x=1$ .

$\Rightarrow 1+cos^2x+3(sin\ x)(cos\ x)=3\Rightarrow cos^2x+3(sin\ x)(cos\ x)=3-1=2$ $\displaystyle \Rightarrow 1+cos^2x+3(sin\ x)(cos\ x)=3\Rightarrow cos^2x+3(sin\ x)(cos\ x)=3-1=2$

Now we should divide both sides by sin^2x $\displaystyle sin^2x$:

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

We can use the identity $\frac{1}{sin^2x}=1+cot^2x$ $\displaystyle \frac{1}{sin^2x}=1+cot^2x$.

$\Rightarrow cot^2x+3cot\ x=2(1+cot^2x)$ $\displaystyle \Rightarrow cot^2x+3cot\ x=2(1+cot^2x)$

$\Rightarrow cot^2x+3cot\ x=2+2cot^2x$ $\displaystyle \Rightarrow cot^2x+3cot\ x=2+2cot^2x$

$\Rightarrow 2cot^2x-cot^2x-3cot\ x+2=0$ $\displaystyle \Rightarrow 2cot^2x-cot^2x-3cot\ x+2=0$

$\Rightarrow cot^2x-3cot\ x+2=0$ $\displaystyle \Rightarrow cot^2x-3cot\ x+2=0$

$\Rightarrow (cot\ x-1)(cot\ x-2)=0$ $\displaystyle \Rightarrow (cot\ x-1)(cot\ x-2)=0$

$cot\ x-1=0\Rightarrow cot\ x=1$ $\displaystyle cot\ x-1=0\Rightarrow cot\ x=1$

$cot\ x-2=0\Rightarrow cot\ x=2$ $\displaystyle cot\ x-2=0\Rightarrow cot\ x=2$

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Example Question #3 : Trigonometric Functions

If $sin\ x+\frac{1}{sin\ x}=2$ $\displaystyle sin\ x+\frac{1}{sin\ x}=2$, find the value of $\displaystyle sin^5x+sin^7x$.

Possible Answers:

$\displaystyle 0$

$\displaystyle -2$

$\displaystyle 2$

$\displaystyle -1$

$\displaystyle 1$

Correct answer:

$\displaystyle 2$

Explanation:

$sin\ x+\frac{1}{sin\ x}=2\Rightarrow \frac{sin^2x+1}{sin\ x}=2$ $\displaystyle sin\ x+\frac{1}{sin\ x}=2\Rightarrow \frac{sin^2x+1}{sin\ x}=2$

$\Rightarrow sin^2x+1=2sin\ x$ $\displaystyle \Rightarrow sin^2x+1=2sin\ x$

$\Rightarrow sin^2x-2sin\ x+1$ $\displaystyle \Rightarrow sin^2x-2sin\ x+1$

$\Rightarrow (sinx-1)^2=0$ $\displaystyle \Rightarrow (sinx-1)^2=0$

$\Rightarrow sin\ x-1=0$ $\displaystyle \Rightarrow sin\ x-1=0$

$\Rightarrow sin\ x=1$ $\displaystyle \Rightarrow sin\ x=1$

Therefore, $\displaystyle sin^5x+sin^7x=1^5+1^7=1+1=2$.

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Example Question #1 : Trigonometric Functions And Graphs

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

Possible Answers:

$4+4\sqrt{3}$ $\displaystyle 4+4\sqrt{3}$

$-8+4\sqrt{3}$ $\displaystyle -8+4\sqrt{3}$

$8-4\sqrt{3}$ $\displaystyle 8-4\sqrt{3}$

$8+4\sqrt{3}$ $\displaystyle 8+4\sqrt{3}$

$8+4\sqrt{2}$ $\displaystyle 8+4\sqrt{2}$

Correct answer:

$8+4\sqrt{3}$ $\displaystyle 8+4\sqrt{3}$

Explanation:

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

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Example Question #5 : Trigonometric Functions

Find the value of the following expression:

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

Possible Answers:

$\displaystyle 0$

$\displaystyle -1$

$\displaystyle 2$

$\displaystyle 1$

$\displaystyle -2$

Correct answer:

$\displaystyle 1$

Explanation:

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

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

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Example Question #1 : Trigonometric Functions

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

Possible Answers:

$tan\ x=\pm 4$ $\displaystyle tan\ x=\pm 4$

$tan\ x=\pm 2$ $\displaystyle tan\ x=\pm 2$

$tan\ x=\pm 1$ $\displaystyle tan\ x=\pm 1$

$tan\ x=\pm 3$ $\displaystyle tan\ x=\pm 3$

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

Correct answer:

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

Explanation:

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

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

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

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

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Example Question #1 : Trigonometric Functions

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

Possible Answers:

$\displaystyle 4$

$\displaystyle 1$

$\displaystyle 3$

$\displaystyle 2$

$\displaystyle 0$

Correct answer:

$\displaystyle 0$

Explanation:

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

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

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Example Question #81 : Trigonometry

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

Possible Answers:

$-sin\ 25^{\circ}$ $\displaystyle -sin\ 25^{\circ}$

$\displaystyle -1$

$sin\ 25^{\circ}$ $\displaystyle sin\ 25^{\circ}$

$\displaystyle 1$

$\displaystyle 0$

Correct answer:

$\displaystyle 1$

Explanation:

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

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

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Example Question #9 : Trigonometric Functions

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

Possible Answers:

2A^2+1 $\displaystyle 2A^2+1$

2A^2-1 $\displaystyle 2A^2-1$

A+1 $\displaystyle A+1$

A^2-1 $\displaystyle A^2-1$

A-1 $\displaystyle A-1$

Correct answer:

2A^2-1 $\displaystyle 2A^2-1$

Explanation:

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

$\displaystyle cos2x=2cos^2x-1=2A^2-1$

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Example Question #10 : Trigonometric Functions

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

Possible Answers:

$\displaystyle 1$

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

$\displaystyle 0$

$\displaystyle -1$

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

Correct answer:

$\displaystyle 0$

Explanation:

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

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

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Trigonometry : Trigonometric Functions and Graphs

Study concepts, example questions & explanations for Trigonometry

All Trigonometry Resources

Example Questions

Example Question #1 : Trigonometric Functions

Example Question #1 : Trigonometric Functions

Example Question #3 : Trigonometric Functions

Example Question #1 : Trigonometric Functions And Graphs

Example Question #5 : Trigonometric Functions

Example Question #1 : Trigonometric Functions

Example Question #1 : Trigonometric Functions

Example Question #81 : Trigonometry

Example Question #9 : Trigonometric Functions

Example Question #10 : Trigonometric Functions

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