Trigonometry : Trigonometric Functions and Graphs

Study concepts, example questions & explanations for Trigonometry

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Example Questions

Example Question #1 : Trigonometric Functions

Rt_triangle_letters

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

Possible Answers:

\(\displaystyle 41.8^\circ\)

\(\displaystyle \frac{2}{3}^\circ\)

\(\displaystyle 90^\circ\)

Undefined

\(\displaystyle 48.2^\circ\)

Correct answer:

\(\displaystyle 41.8^\circ\)

Explanation:

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

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

In this problem, that would be:

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

Plug in our given values:

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

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

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

\(\displaystyle a=41.8^\circ\)

Example Question #1 : Trigonometric Functions

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

 

 

Possible Answers:

\(\displaystyle cot\ x=-1\ or\ cot\ x=2\)

\(\displaystyle cot\ x=1\)

\(\displaystyle cot\ x=2\)

\(\displaystyle cot\ x=1\ or\ cot\ x=-2\)

\(\displaystyle cot\ x=1\ or\ cot\ x=2\)

Correct answer:

\(\displaystyle cot\ x=1\ or\ cot\ x=2\)

Explanation:

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

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

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

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

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

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

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

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

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

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

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

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

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

or

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

 

 

 

Example Question #3 : Trigonometric Functions

If \(\displaystyle sin\ x+\frac{1}{sin\ x}=2\), 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:

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

\(\displaystyle \Rightarrow sin^2x+1=2sin\ x\)

\(\displaystyle \Rightarrow sin^2x-2sin\ x+1\)

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

\(\displaystyle \Rightarrow sin\ x-1=0\)

\(\displaystyle \Rightarrow sin\ x=1\)

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

Example Question #1 : Trigonometric Functions And Graphs

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

Possible Answers:

\(\displaystyle 4+4\sqrt{3}\)

\(\displaystyle -8+4\sqrt{3}\)

\(\displaystyle 8-4\sqrt{3}\)

\(\displaystyle 8+4\sqrt{3}\)

\(\displaystyle 8+4\sqrt{2}\)

Correct answer:

\(\displaystyle 8+4\sqrt{3}\)

Explanation:

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

Example Question #5 : Trigonometric Functions

Find the value of the following expression:

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

Possible Answers:

\(\displaystyle 0\)

\(\displaystyle -1\)

\(\displaystyle 2\)

\(\displaystyle 1\)

\(\displaystyle -2\)

Correct answer:

\(\displaystyle 1\)

Explanation:

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

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

Example Question #1 : Trigonometric Functions

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

Possible Answers:

\(\displaystyle tan\ x=\pm 4\)

\(\displaystyle tan\ x=\pm 2\)

\(\displaystyle tan\ x=\pm 1\)

\(\displaystyle tan\ x=\pm 3\)

\(\displaystyle tan\ x=\pm \frac{1}{2}\)

Correct answer:

\(\displaystyle tan\ x=\pm \frac{1}{2}\)

Explanation:

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

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

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

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

Example Question #1 : Trigonometric Functions

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

Possible Answers:

\(\displaystyle 4\)

\(\displaystyle 1\)

\(\displaystyle 3\)

\(\displaystyle 2\)

\(\displaystyle 0\)

Correct answer:

\(\displaystyle 0\)

Explanation:

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

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

Example Question #81 : Trigonometry

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

Possible Answers:

\(\displaystyle -sin\ 25^{\circ}\)

\(\displaystyle -1\)

\(\displaystyle sin\ 25^{\circ}\)

\(\displaystyle 1\)

\(\displaystyle 0\)

Correct answer:

\(\displaystyle 1\)

Explanation:

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

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

Example Question #9 : Trigonometric Functions

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

Possible Answers:

\(\displaystyle 2A^2+1\)

\(\displaystyle 2A^2-1\)

\(\displaystyle A+1\)

\(\displaystyle A^2-1\)

\(\displaystyle A-1\)

Correct answer:

\(\displaystyle 2A^2-1\)

Explanation:

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

\(\displaystyle cos2x=2cos^2x-1=2A^2-1\)

Example Question #10 : Trigonometric Functions

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

Possible Answers:

\(\displaystyle 1\)

\(\displaystyle \frac{\sqrt{2}}{2}\)

\(\displaystyle 0\)

\(\displaystyle -1\)

\(\displaystyle -\frac{\sqrt{2}}{2}\)

Correct answer:

\(\displaystyle 0\)

Explanation:

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

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

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