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## Question 6c606

Sunayan S
Featured 2 weeks ago

Please give a look here...

#### Explanation:

color(red)((1+sinA)/cosA+cosB/(1-sinB)

=color(green)({cos (A/2)+sin(A/2)}^2/{cos^2(A/2)-sin^2(A/2)}+{cos^2(B/2)-sin^2(B/2)}/{cos(B/2)-sin(B/2)}^2

=color(red)((cos(A/2)+sin(A/2))/{cos(A/2)-sin(A/2)}+(cos(B/2)+sin(B/2))/(cos(B/2)-sin(B/2))

=color(green)([{(cos(A/2)+sin(A/2))cdot(cos(B/2)-sin(B/2))}+{(cos(B/2)+sin(B/2))cdot(cos(A/2)-sin(A/2))}]/{(cos(A/2)-sin(A/2))cdot(cos(B/2)-sin(B/2))}

=color(red)({cos(A/2) cdot cos(B/2)-cos(A/2) cdot sin(B/2)+cos(B/2) cdot sin(A/2)-sin(A/2) cdot sin(B/2)+cos(B/2)cos(A/2)-cos(B/2) cdot sin(A/2)+sin(B/2) cdot cos(A/2)-sin(B/2) cdot sin(A/2)}/{cos(A/2)cos(B/2)-cos(A/2) cdot sin(B/2)-sin(A/2) cdot cos(B/2)+sin(A/2) cdot sin(B/2)}

=color(green)((2{cos(A/2)cdot cos(B/2)-sin(A/2) cdot sin(B/2)})/({cos(A/2)cdot cos(B/2)+sin(A/2) cdot sin(B/2)}-{sin(A/2)cos(B/2)+cos(A/2)sin(B/2)}

=color(red)((2 cos((A+B)/2))/{cos((A-B)/2)-sin((A+B)/2)}

=color(green)((2 cos((A+B)/2)cdot2sin((A-B)/2))/{2 cdot cos((A-B)/2) cdot sin((A-B)/2)-2 cdot sin((A+B)/2)cdot sin((A-B)/2)

=color(red)((2(sinA-sinB))/(sin(A-B)+cosA-cosB)

Hope it helps...
Thank you...

## Write each expression as a single angle and evaluate if possible? 2Sin30°Cos30°

Øko
Featured 2 weeks ago

$2 \sin \left({30}^{\circ}\right) \cos \left({30}^{\circ}\right) = \sin \left({60}^{\circ}\right) = \frac{\sqrt{3}}{2}$

#### Explanation:

Use the identity

• $\sin \left(2 x\right) = 2 \sin \left(x\right) \cos \left(x\right)$

Applying this identity to your example

$2 \sin \left({30}^{\circ}\right) \cos \left({30}^{\circ}\right) = \sin \left(2 \cdot {30}^{\circ}\right) = \sin \left({60}^{\circ}\right) = \frac{\sqrt{3}}{2}$

## A triangle has sides A, B, and C. Sides A and B are of lengths 1 and 2, respectively, and the angle between A and B is (2pi)/3 . What is the length of side C?

sankarankalyanam
Featured 6 days ago

color(brown)(c = sqrt 7

#### Explanation:

$a = 1 , b = 2 , \hat{C} = \frac{2 \pi}{3}$

Applying Law of Cosine,

${c}^{2} = {a}^{2} + {b}^{2} - 2 a b \sin C$

${c}^{2} = {1}^{2} + {2}^{2} - \left(2 \cdot 1 \cdot 2 \cdot \cos \left(\frac{2 \pi}{3}\right)\right)$

${c}^{2} = 5 - 4 \cos \left(\frac{2 \pi}{3}\right) = 5 + 4 \cos \left(\frac{\pi}{3}\right)$ as $\cos \left(\frac{2 \pi}{3}\right) = - \cos \left(\frac{\pi}{3}\right)$ and cos is negative in II quadrant.

${c}^{2} = 5 + \left(4 \cdot \left(\frac{1}{2}\right)\right) = 5 + 2 = 7$

color(brown)(c = sqrt 7

## How do you convert radian measures to degrees?

F. Javier B.
Featured 4 days ago

Throug the equivalence pi=180º and de direct proportionality of arcs and angles. See examples

#### Explanation:

If we have 45º degrees convert to radians is easy

$\frac{\pi}{x} = \frac{180}{45}$. Then $x = \frac{45 \pi}{180} = \frac{\pi}{4}$. So 45º=pi/4 rad

By the other hand, if we have $\frac{\pi}{3}$ rads, to convert to degrees, we use the same equivalence

$\frac{\pi}{\frac{\pi}{3}} = \frac{180}{y}$. Trasposing terms y=(180pi)/(3pi)=180/3=60º#

## Solve for all values of X ? Sinx= cosx+1

VNVDVI
Featured yesterday

$x = \frac{\pi}{2} + n \pi$ where $n$ is any integer.

#### Explanation:

We first want to get this equation in terms of only one trigonometric function. Let's say we want to put everything in terms of cosine, as that will work best in this case.

Let's square both sides:

${\sin}^{2} x = {\left(\cos x + 1\right)}^{2}$

${\sin}^{2} x = {\cos}^{2} x + 2 \cos x + 1$

Recall the identity

${\sin}^{2} x + {\cos}^{2} x = 1$

Solving for ${\sin}^{2} x ,$ we get

${\sin}^{2} x = 1 - {\cos}^{2} x$

Thus,

$1 - {\cos}^{2} x = {\cos}^{2} x + 2 \cos x + 1$

And everything is in terms of cosine. Let's move everything to one side:

${\cos}^{2} x + 3 \cos x = 0$ (Our $1 ' s$ cancel out)

We can factor out an instance of $\cos x :$

$\cos x \left(\cos x + 3\right) = 0$

We now solve the following equations:

$\cos x = 0$

$\cos x + 3 = 0$

For $\cos x = 0 , x = \frac{\pi}{2} + n \pi$ where $n$ is an integer because $\cos x = 0$ for $\frac{\pi}{2} , 3 \frac{\pi}{2} , 5 \frac{\pi}{2} , 7 \frac{\pi}{2} , \ldots$, meaning values of $x$ which cause $\cos x = 0$ are infinitely many and repeat every $\pi$ units (so we add $n \pi$).

$F \mathmr{and} \cos x + 3 = 0$, let's solve for $\cos x :$

$\cos x = - 3$

We already see no values of $x$ solve this. $- 1 \le \cos x \le 1 ,$ always. $\cos x$ cannot be $- 3.$

## A 24 foot tree casts a shadow that is 9 feet long . At the same time a nearby building cats a shadow 45 feet long . how tall is the building?

Gió
Featured 6 hours ago

I got $120 \text{ft}$

#### Explanation:

Consider the diagram:

The angles $\alpha$ are the same so we can write for the two triangles (tree and building):
$\frac{h}{l} = \frac{H}{L}$
in numbers:
$\frac{24}{9} = \frac{H}{45}$
so that:
$H = \frac{24}{9} \cdot 45 = 120 \text{ft}$

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