3 years ago

Use unit circle to determine sine, cosine and tangent functions of the angle 5pi/4

2 answers:

5 0

5pi/4 is located in quadrant III.
In Q III, sin and cos are negative, tan is positive.
T-bar for 5pi/4 is pi/4.
sin and cos are - (sqrt 2)/2
Tan is 1

Kobotan [32]3 years ago

5 0

Sin(5π/4)=o/h
cos(5π/4)=a/h
tan(5π/4)=o/a

You might be interested in

What is the y intercept of the line

Juliette [100K]

Because a function must pass the vertical line test , a function can have at most one y -intercept .

The y -intercept is often referred to with just the y -value.

4 0

3 years ago

Read 2 more answers

Write as an expression and simplify: the tripled product of 6a and b^2

klasskru [66]

Answer:

(6ab^2)^3

Step-by-step explanation:

3 0

4 years ago

Can anyone give me some points

rosijanka [135]

For what???????????????

3 0

3 years ago

Read 2 more answers

Solve the equation 2cosA = 3tanA

zavuch27 [327]

$\bf sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta)
\\\\\\
tan(\theta)=\cfrac{sin(\theta)}{cos(\theta)}\\\\
-----------------------------\\\\

2cos(A)=3tan(A)\implies 2cos(A)=3\cfrac{sin(A)}{cos(A)}
\\\\\\
2cos^2(A)=3sin(A)\implies 2[1-sin^2(A)]=3sin(A)
\\\\\\
2-2sin^2(A)=3sin(A)\implies 2sin^2(A)+3sin(A)-2$

$\bf \\\\\\
0=[2sin(A)-1][sin(A)+2]\implies 
\begin{cases}
0=2sin(A)-1\\
1=2sin(A)\\
\frac{1}{2}=sin(A)\\\\
sin^{-1}\left( \frac{1}{2} \right)=\measuredangle A\\\\
\frac{\pi }{6},\frac{5\pi }{6}\\
----------\\
0=sin(A)+2\\
-2=sin(A)
\end{cases}$

now, as far as the second case....well, sine of anything is within the range of -1 or 1, so -1 < sin(A) < 1

now, we have -2 = sin(A), which simply is out of range for a valid sine, so there's no angle with such sine

so, only the first case are the valid angles for A

8 0

3 years ago

the boiling of jet fuel is 390f. rounded t the nearest degree, what is the temperature in degrees celsius? where c represents de

e-lub [12.9K]

390f = 198.889c
Hope this helps!

7 0

3 years ago