Question

17. (1 point) Match each of the trigonometric expressions below with the equivalent non-trigonometric function from the following list. Enter the appropriate letter (A,B,C,D, or E) in cach blank. A. tan(sin '(x4)) B.cos(sin-'(x/4)) C. (1/2) sin(2 sin(x/4)) D.sin(tan (x/4)) E. costan (x/4)) 16 --- V16x² 2. 4 11111 3. 4. √16+x 4 16+ x2 √16- x² 16 X

Answer 1

For the letters A, B and C, let's use the following triangle:

Using the Pythagoras theorem, let's find the value of y:

`\begin{gathered} 4^2=x^2+y^2 \\ y^2=16-x^2 \\ y=\sqrt[]{16-x^2} \end{gathered}`

Now, let's find the equivalent for letter A:

`\begin{gathered} \tan (\sin ^(-1)((x)/(4))) \\ =\tan (k) \\ =(x)/(y) \\ =\frac{x}{\sqrt[]{16-x^2}} \end{gathered}`

So letter A is equivalent to number 1.

For letter B we have:

`\begin{gathered} \cos (\sin ^(-1)((x)/(4))) \\ =\cos (k) \\ =(y)/(4) \\ =\frac{\sqrt[]{16-x^2}}{4} \end{gathered}`

So letter B is equivalent to number 2.

For letter C we have:

`\begin{gathered} (1)/(2)\sin (2\sin ^(-1)((x)/(4))) \\ =(1)/(2)\sin (2k) \\ =(1)/(2)\cdot(2\sin (k)\cos (k) \\ =\sin (k)\cdot\cos (k)_{} \\ =(x)/(4)\cdot(y)/(4) \\ =(x)/(16)\cdot\sqrt[]{16-x^2} \end{gathered}`

So letter C is equivalent to number 5.

Now, for letters D and E, let's use this triangle:

Finding y, we have that:

`\begin{gathered} y^2=x^2+4^2 \\ y=\sqrt[]{16+x^2^{}} \end{gathered}`

So for letter D we have:

`\begin{gathered} \sin (\tan ^(-1)((x)/(4))) \\ =\sin (k) \\ =(x)/(y) \\ =\frac{x}{\sqrt[]{16+x^2}} \end{gathered}`

So letter D is equivalent to number 3.

For letter E we have:

`\begin{gathered} \cos (\tan ^(-1)((x)/(4))) \\ =\cos (k) \\ =(4)/(y) \\ =\frac{4}{\sqrt[]{16+x^2}} \end{gathered}`

So letter E is equivalent to number 4.