High School Math : Solving Trigonometric Equations

Study concepts, example questions & explanations for High School Math

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Example Questions

Example Question #1 : Trigonometric Equations

If \(\displaystyle tanx=\frac{1}{3}\), find \(\displaystyle f(x)=\frac{sin(3\pi +x)+sin(x-\frac{\pi }{2})}{cos(x-\pi )+cos(\frac{3\pi }{2}+x)}\).

Possible Answers:

0

-2

-1

1

2

Correct answer:

2

Explanation:

First we need to simplify each term:

 

\(\displaystyle sin(3\pi +x)= -sinx\)

\(\displaystyle sin(x-\frac{\pi }{2})=cosx\)

\(\displaystyle cos(x-\pi )=- cosx\)

\(\displaystyle cos(\frac{3\pi }{2}+x)=sinx\)

 

So we will have:

 

\(\displaystyle f(x)=\frac{-sinx-cosx}{-cosx+sinx}\)

Since we have the value of \(\displaystyle tanx\), we can divide the numerator and denominator by \(\displaystyle cosx\).

 

\(\displaystyle f(x)=\frac{-\frac{sinx}{cosx}-\frac{cosx}{cosx}}{\frac{-cosx}{cosx}+\frac{sinx}{cosx}}=\frac{-tanx-1}{-1+tanx} = \frac{1+tanx}{1-tanx}\)

We can then substitute the value of \(\displaystyle tanx\) in the obtained function.

 

\(\displaystyle f(x)=\frac{1+\frac{1}{3}}{1-\frac{1}{3}}=\frac{\frac{4}{3}}{\frac{2}{3}}=2\)

Example Question #2 : Trigonometric Equations

A right triangle has an angle of \(\displaystyle 41^\circ\). If the side opposite it has a length of \(\displaystyle 8\), what is the hypotenuse?

Possible Answers:

\(\displaystyle 10.6\)

\(\displaystyle 9.2\)

\(\displaystyle 5.25\)

\(\displaystyle 6.04\)

\(\displaystyle 12.19\)

Correct answer:

\(\displaystyle 12.19\)

Explanation:

For this problem we will use the sine function as \(\displaystyle \sin=\frac{O}{H}\).

Plug in our given information:

\(\displaystyle \sin(41^\circ)=\frac{8}{H}\)

\(\displaystyle H=\frac{8}{\sin(41^\circ)}\)

\(\displaystyle H=12.19\)

Example Question #3 : Trigonometric Equations

A right triangle has an angle of \(\displaystyle 72^\circ\). If the side adjacent to it has a length of \(\displaystyle 11\), what is the hypotenuse?

Possible Answers:

\(\displaystyle 3.4\)

\(\displaystyle 3.57\)

\(\displaystyle 11.56\)

\(\displaystyle 35.6\)

\(\displaystyle 10.46\)

Correct answer:

\(\displaystyle 35.6\)

Explanation:

For this problem we will use the cosine function as \(\displaystyle \cos=\frac{A}{H}\).

Plug in our given information:

\(\displaystyle \cos(72^\circ)=\frac{11}{H}\)

\(\displaystyle H=\frac{11}{\cos(72^\circ)}\)

\(\displaystyle H=35.6\)

Example Question #4 : Trigonometric Equations

In a right triangle, if 

\(\displaystyle cos(x)=\frac{3}{5}\)

and 

\(\displaystyle cos(y)=\frac{4}{5}\)

then what does \(\displaystyle \tan(x)\) equal?

Possible Answers:

\(\displaystyle \frac{3}{5}\)

\(\displaystyle \frac{3}{4}\)

\(\displaystyle \frac{4}{3}\)

\(\displaystyle \frac{5}{4}\)

\(\displaystyle 1\)

Correct answer:

\(\displaystyle \frac{4}{3}\)

Explanation:

One can draw a right triangle with acute angles \(\displaystyle \angle x\) and \(\displaystyle \angle y\). The side adjacent to \(\displaystyle \angle y\) is 4, and the side adjacent to \(\displaystyle \angle x\)  is 3.  

\(\displaystyle tan(x)=\frac{opposite}{adjacent}\) 

Example Question #13 : Functions

If \(\displaystyle sinx-cosx=\frac{1}{3}\) , find the value of the function \(\displaystyle f(x)=sin^{3}x+cos^{3}x\).

Possible Answers:

\(\displaystyle \frac{13}{27}\)

\(\displaystyle -\frac{13}{27}\)

\(\displaystyle \frac{11}{27}\)

1

\(\displaystyle -\frac{11}{27}\)

Correct answer:

\(\displaystyle \frac{13}{27}\)

Explanation:

We can write:

\(\displaystyle sinx-cosx=\frac{1}{3}\)

\(\displaystyle \Rightarrow (sinx-cosx)^{2}=\frac{1}{9}\)

\(\displaystyle \Rightarrow sin^2x+cos^2x-2sinxcosx=\frac{1}{9}\)

\(\displaystyle \Rightarrow 2sinxcosx=1-\frac{1}{9}\)

\(\displaystyle \Rightarrow sinxcosx=\frac{4}{9}\)

We can also write:

 

\(\displaystyle (sinx-cosx)^3=sin^3x+cos^3x+3sinxcos^2x-3cosxsin^2x\)

\(\displaystyle \Rightarrow sin^3x+cos^3x=(sinx-cosx)^3+3sinxcosx(sinx-cosx)\)

 

Now we can substitute the values of (sinx - cosx) and (sinx * cosx) in the obtained function:

 

\(\displaystyle \Rightarrow sin^3x+cos^3x=(\frac{1}{3})^3+3*\frac{4}{9}*\frac{1}{3}=\frac{1}{27}+\frac{4}{9}=\frac{13}{27}\)

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