All categories

1 year ago

Write down the indicated trigonometric value and write as decimal and round to thousandths place

Mathematics

1 answer:

mestny [16]1 year ago

8 0

The trigonometric values for a right triangle can be obtained through the measurements of the different sides,

$\begin{gathered} sin(\theta)=\frac{op}{hyp} \\ cos(\theta)=\frac{ad}{hyp} \\ tan(\theta)=\frac{op}{ad} \end{gathered}$

this can be found in a triangle like this,

then, identify the measurements in the given triangle

then,

$\begin{gathered} sin(A)=\frac{10}{26}=\frac{5}{13} \\ cos(A)=\frac{24}{26}=\frac{12}{13} \\ tan(A)=\frac{10}{24}=\frac{5}{12} \end{gathered}$

You might be interested in

2 (3)^3 + 5 And explain how you got it

ser-zykov [4K]

Answer:

Step-by-step explanation:

3^3=27

2x27=54

27+54+5=59

8 0

3 years ago

Write the given expression in terms of x and y only. sin(sin−1(x) + cos−1(y))

yan [13]

If you're using the app, try seeing this answer through your browser: brainly.com/question/2308127

_______________

Write the expression below in terms of x and y only:

(I'm going to call it "E")

$\mathsf{E=sin\!\left[sin^{-1}(x)+cos^{-1}(y)\right]\qquad\quad(i)}$

Let

$\begin{array}{lcl} \mathsf{\alpha=sin^{-1}(x)}&\qquad&\mathsf{then~-\,\dfrac{\pi}{2}\le \alpha\le \dfrac{\pi}{2}}\\\\\\ \mathsf{\beta=cos^{-1}(x)}&\qquad&\mathsf{then~0\le \beta\le \pi.} \end{array}$

so the expression becomes

$\mathsf{E=sin(\alpha+\beta)}\\\\ \mathsf{E=sin\,\alpha\,cos\,\beta+sin\,\beta\,cos\,\alpha\qquad\quad(ii)}$

• Finding $\mathsf{sin\,\alpha:}$

$\mathsf{sin\,\alpha=sin\!\left[sin^{-1}(x)\right]}\\\\ \mathsf{sin\,\alpha=x\qquad\quad\checkmark}$

• Finding $\mathsf{cos\,\alpha:}$

$\mathsf{sin^2\,\alpha=x^2}\\\\ \mathsf{1-cos^2\,\alpha=x^2}\\\\ \mathsf{cos^2\,\alpha=1-x^2}\\\\ \mathsf{cos\,\alpha=\sqrt{1-x^2}\qquad\quad\checkmark}$

because $\mathsf{cos\,\alpha}$ is positive for $\mathsf{\alpha\in \left[-\frac{\pi}{2},\,\frac{\pi}{2}\right].}$

• Finding $\mathsf{cos\,\beta:}$

$\mathsf{cos\,\beta=cos\!\left[cos^{-1}(y)\right]}\\\\ \mathsf{cos\,\beta=y\qquad\quad\checkmark}$

• Finding $\mathsf{sin\,\beta:}$

$\mathsf{cos^2\,\alpha=y^2}\\\\
 \mathsf{1-sin^2\,\beta=y^2}\\\\ \mathsf{sin^2\,\beta=1-y^2}\\\\ 
\mathsf{sin\,\beta=\sqrt{1-y^2}\qquad\quad\checkmark}$

because $\mathsf{sin\,\beta}$ is positive for $\mathsf{\beta\in [0,\,\pi].}$

Finally, you get

$\mathsf{E=x\cdot y +\sqrt{1-y^2}\cdot \sqrt{1-x^2}}\\\\\\ \therefore~~\mathsf{sin\!\left[sin^{-1}(x)+cos^{-1}(y)\right]=x\cdot y +\sqrt{1-y^2}\cdot \sqrt{1-x^2}\qquad\quad\checkmark}$

I hope this helps. =)

Tags: inverse trigonometric trig function sine cosine sin cos arcsin arccos sum angles trigonometry

6 0

4 years ago

I’ll give brainliest to first correct answer.

kirza4 [7]

Ok but u forgot to add the picture

4 0

3 years ago

Which statement is true about the underlined digit in the number 65,888?

Ghella [55]

Sixty five thousand eight hundred and eighty eight

8 0

4 years ago