Trigonometric Equations - Math

Card 0 of 20

In a right triangle, if

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

and

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

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

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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 $sinx-cosx=\frac{1}{3}$$ , find the value of the function .

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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)}$ .

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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?

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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?

Tap to see back →

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)}$$

In a right triangle, if

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

and

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

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

Tap to see back →

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 $sinx-cosx=\frac{1}{3}$$ , find the value of the function .

Tap to see back →

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)}$ .

Tap to see back →

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?

Tap to see back →

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?

Tap to see back →

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)}$$

In a right triangle, if

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

and

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

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

Tap to see back →

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 $sinx-cosx=\frac{1}{3}$$ , find the value of the function .

Tap to see back →

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)}$ .

Tap to see back →

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?

Tap to see back →

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?

Tap to see back →

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)}$$

In a right triangle, if

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

and

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

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

Tap to see back →

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 $sinx-cosx=\frac{1}{3}$$ , find the value of the function .

Tap to see back →

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)}$ .

Tap to see back →

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?

Tap to see back →

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?

Tap to see back →

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)}$$