Trigonometric Equations

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Math › Trigonometric Equations

Questions 1 - 10

In a right triangle, if

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

and

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

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

$\frac{4}{3}$

$\frac{5}{4}$

$\frac{3}{5}$

$\frac{3}{4}$

Explanation

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

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

In a right triangle, if

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

and

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

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

$\frac{4}{3}$

$\frac{5}{4}$

$\frac{3}{5}$

$\frac{3}{4}$

Explanation

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

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

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

-1

-2

Explanation

First we need to simplify each term:

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

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

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

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

So we will have:

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

Since we have the value of , we can divide the numerator and denominator by .

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

We can then substitute the value of in the obtained function.

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

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

Explanation

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

Plug in our given information:

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

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

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

$-\frac{11}{27}$

$\frac{13}{27}$

$-\frac{13}{27}$

$\frac{11}{27}$

Explanation

We can write:

$sinx-cosx=\frac{1}{3}$

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

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

$\Rightarrow 2sinxcosx=1-\frac{1}{9}$

$\Rightarrow sinxcosx=\frac{4}{9}$

We can also write:

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

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

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

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

-1

-2

Explanation

First we need to simplify each term:

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

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

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

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

So we will have:

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

Since we have the value of , we can divide the numerator and denominator by .

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

We can then substitute the value of in the obtained function.

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

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

Explanation

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

Plug in our given information:

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

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

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

Explanation

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

Plug in our given information:

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

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

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

$-\frac{11}{27}$

$\frac{13}{27}$

$-\frac{13}{27}$

$\frac{11}{27}$

Explanation

We can write:

$sinx-cosx=\frac{1}{3}$

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

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

$\Rightarrow 2sinxcosx=1-\frac{1}{9}$

$\Rightarrow sinxcosx=\frac{4}{9}$

We can also write:

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

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

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

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

Explanation

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

Plug in our given information:

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

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

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