All categories

2 years ago

If (cos)x= 1/2 what is sin(x) and tan(x)? Explain your steps in complete sentences.

2 answers:

6 0

The correct answers are:
1) sin(x) = $\frac{ \sqrt{3} }{2}$
2) tan(x) = $\sqrt{3}$

Explanation:
Given:
$cos(x) = \frac{1}{2}$

Step 1:
Since, according to the Trigonometric identity:
$sin^2(x) + cos^2(x) = 1$ -- (1)

Step 2:
Plug in the value of cos(x) in equation (1):
$sin^2(x) + ( \frac{1}{2} )^2 = 1 \\ sin^2(x) + \frac{1}{4} = 1 \\ sin^2(x) = \frac{3}{4}$

Step 3:
Take square-root on both sides:
$\sqrt{sin^2(x)} = \sqrt{\frac{3}{4}}$

sin(x) = $\frac{ \sqrt{3} }{2}$

Now to find the tan(x), we would use the following formula:

tan(x) = $\frac{sin(x)}{cos(x)}$ --- (2)

Plug in the values of sin(x) and cos(x) in equation (2):
tan(x) = $\frac{ \frac{ \sqrt{3} }{2} }{ \frac{1}{2} }$

Hence tan(x) = $\sqrt{3}$

Wewaii [24]2 years ago

3 0

Answer:

The value of $\sin x=\frac{\sqrt{3}}{2}$

The value of $\tan x=\sqrt{3}$

Step-by-step explanation:

Given = $\cos x=\frac{1}{2}$

To find : $\sin x=?$ and $\tan x=?$

Solution:

We know that, if the cosine function of an angle is $\frac{1}{2}$ then the angle is equal to the 60°.

$\cos x=\frac{1}{2}$

x = 60°

The value of the :

$\sin x=\sin 60^o=\frac{\sqrt{3}}{2}$

We know that ratio of sine function to the cosine function is equal tto the tangent

$\tan x=\frac{\sin x}{\cos x}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}$

The value of $\sin x=\frac{\sqrt{3}}{2}$

The value of $\tan x=\sqrt{3}$

You might be interested in

A total of $228,000 will be evenly spent to renovate 24 apartments. If the first 10 apartments have been completed and paid for,

tigry1 [53]

Step-by-step explanation:

Step 1: Calculate per apartment;

⟼ 228,000 ÷ 24

⟼ 9,500

Step 2: Multiply the price per apartment by 10;

⟼ 9,500 × 10

⟼ 95,000

Step 3: Substract the amount of 10 apartments from 24 apartments.

⟼ 228,000 - 95,000

⟼ 133,000

Therefore, $133,000 is still available for the remaining apartments.

5 0

2 years ago

Which is the graph of the inverse secant function?

Vilka [71]

The graph of the inverse secant function would possibly be the third one I think.

7 0

2 years ago

Read 2 more answers

Assuming that the equation defines x and y implicitly as differentiable functions xequalsf(t), yequalsg(t), find the slope of

Doss [256]

Answer:

$\dfrac{dx}{dt} = -8,\dfrac{dy}{dt} = 1/8\\$

Hence, the slope , $\dfrac{dy}{dx} = \dfrac{-1}{64}$

Step-by-step explanation:

We need to find the slope, i.e. $\dfrac{dy}{dx}$ .

and all the functions are in terms of $t$ .

So this looks like a job for the 'chain rule', we can write:

$\dfrac{dy}{dx} = \dfrac{dy}{dt} .\dfrac{dt}{dx} -Eq(A)$

Given the functions

$x = f(t)\\y = g(t)\\$

and

$x^3 +4t^2 = 37 -Eq(B)\\2y^3 - 2t^2 = 110 - Eq(C)$

we can differentiate them both w.r.t to $t$

first we'll derivate Eq(B) to find dx/dt

$x^3 +4t^2 = 37\\3x^2\frac{dx}{dt} + 8t = 0\\\dfrac{dx}{dt} = \dfrac{-8t}{3x^2}\\$

we can also rearrange Eq(B) to find x in terms of t , $x = (37 - 4t^2)^{1/3}$ . This is done so that $\frac{dx}{dt}$ is only in terms of t.

$\dfrac{dx}{dt} = \dfrac{-8t}{3(37 - 4t^2)^{2/3}}\\$

we can find the value of this derivative using t = 3, and plug that value in Eq(A).

$\dfrac{dx}{dt} = \dfrac{-8t}{3(37 - 4t^2)^{2/3}}\\\dfrac{dx}{dt} = \dfrac{-8(3)}{3(37 - 4(3)^2)^{2/3}}\\\dfrac{dx}{dt} = -8$

now let's differentiate Eq(C) to find dy/dt

$2y^3 - 2t^2 = 110\\6y^2\frac{dy}{dt} -4t = 0\\\dfrac{dy}{dt} = \dfrac{4t}{6y^2}$

rearrange Eq(C), to find y in terms of t, that is $y = \left(\dfrac{110 + 2t^2}{2}\right)^{1/3}$ . This is done so that we can replace y in $\frac{dy}{dt}$ to make only in terms of t

$\dfrac{dy}{dt} = \dfrac{4t}{6y^2}\\\dfrac{dy}{dt}=\dfrac{4t}{6\left(\dfrac{110 + 2t^2}{2}\right)^{2/3}}\\$

we can find the value of this derivative using t = 3, and plug that value in Eq(A).

$\dfrac{dy}{dt} = \dfrac{4(3)}{6\left(\dfrac{110 + 2(3)^2}{2}\right)^{2/3}}\\\dfrac{dy}{dt} = \dfrac{1}{8}$

Finally we can plug all of our values in Eq(A)

but remember when plugging in the values that $\frac{dy}{dt}$ is being multiplied with $\frac{dt}{dx}$ and NOT $\frac{dx}{dt}$ , so we have to use the reciprocal!