Westonci.ca is the premier destination for reliable answers to your questions, brought to you by a community of experts. Experience the convenience of getting reliable answers to your questions from a vast network of knowledgeable experts. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.

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

17 1 Point Match Each Of The Trigonometric Expressions Below With The Equivalent Nontrigonometric Function From The Following List Enter The Appropriate Letter class=

Sagot :

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

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

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

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

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

So letter A is equivalent to number 1.

For letter B we have:

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

So letter B is equivalent to number 2.

For letter C we have:

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

So letter C is equivalent to number 5.

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

Finding y, we have that:

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

So for letter D we have:

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

So letter D is equivalent to number 3.

For letter E we have:

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

So letter E is equivalent to number 4.

View image RickaylaN746747
View image RickaylaN746747