# Make the internet a better place to learn

2

## 5sin(x)−2=2sin(x)+4 within [0, 2π] has solutions. True or False? if true explain why and what those solutions are.

Tidsresa ∞
Featured yesterday

5sin(x)−2=2sin(x)+4

$\implies 3 \sin x = 6$

$\implies \sin x = 2$

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

But this equation doesn't have any solutions because $\sin \theta$ is defined only within $\left[- 1 , 1\right]$ hence, $\theta$ can have a range [-π/2,π/2]

So, ${\sin}^{- 1} 2$ is undefined.

Hope this helps.:)

1

## How do you find arc length of an arc that subtends a central angle of 60 degrees in a circle with radius 25m?

Jim G.
Featured yesterday

$\text{length "~~26.18" m to 2 dec. places}$

#### Explanation:

$\text{the length of an arc given it subtends a known angle at}$
$\text{the centre is}$

$\text{arc length (l) "="circumference "xx"fraction of circle}$

$\Rightarrow l = 2 \pi r \times \frac{60}{360}$

$\textcolor{w h i t e}{l} = 2 \pi \times 25 \times \frac{1}{6} = \frac{25 \pi}{3} \approx 26.18 \text{ m}$

2

## Given the right triangle below, find cos theta, sin theta, tan theta, sec theta, csc theta, and cot theta. Do not approximate, find exact answers and show all work?

Jim G.
Featured yesterday

$\text{see explanation}$

#### Explanation:

$\text{to find some of the ratios we will require the third side of}$
$\text{the triangle}$

$\text{since it is right use "color(blue)"Pythagoras' theorem}$

$\text{third side } = \sqrt{{11}^{2} - {6}^{2}} = \sqrt{85}$

•color(white)(x)costheta="adjacent"/"hypotenuse"=6/11

$\Rightarrow \sec \theta = \frac{1}{\cos} \theta = \frac{1}{\frac{6}{11}} = \frac{11}{6}$

•color(white)(x)sintheta="opposite"/"hypotenuse"=sqrt85/11

$\Rightarrow \csc \theta = \frac{1}{\sin} \theta = \frac{1}{\frac{\sqrt{85}}{11}} = \frac{11}{\sqrt{85}}$

•color(white)(x)tantheta=sintheta/costheta

$\textcolor{w h i t e}{\times \times \times} = \frac{\sqrt{85}}{11} \times \frac{11}{6} = \frac{\sqrt{85}}{6}$

$\Rightarrow \cot \theta = \frac{1}{\tan} \theta = \frac{1}{\frac{\sqrt{85}}{6}} = \frac{6}{\sqrt{85}}$

2

## How do you answer this sin^2x+cos^2x+cot^2x/1+tan^2x=cot^2x ?

Tidsresa ∞
Featured yesterday

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

$\text{LHS}$:

$\frac{\textcolor{red}{{\sin}^{2} x + {\cos}^{2} x} + {\cot}^{2} x}{\textcolor{t e a l}{1 + {\tan}^{2} x}}$

We know the $\boldsymbol{\text{Pythagorean Identities:}}$

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

• $\textcolor{t e a l}{1 + {\tan}^{2} x = {\sec}^{2} x}$

• $\textcolor{\in \mathrm{di} g o}{1 + {\cot}^{2} x = {\csc}^{2} x}$

$\boldsymbol{\text{Reciprocal Identities:}}$

• $\cos x = \frac{1}{\sec x}$

• $\sin x = \frac{1}{\csc x}$

Employing all of them,

=(color(red)1+cot^2x)/(color(teal)(sec^2x)

$= \frac{\textcolor{\in \mathrm{di} g o}{1 + {\cot}^{2} x}}{{\sec}^{2} x}$

$= \frac{\textcolor{\in \mathrm{di} g o}{{\csc}^{2} x}}{{\sec}^{2} x}$

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

=(1/sinx)^2×(cosx/1)^2

$= {\cot}^{2} x$ $\left[\because \frac{1}{\tan x} = \cot x = \frac{\cos x}{\sin x}\right]$

$\implies \text{RHS}$

Hence, proved! :)

Hope this helps. :)

2

## Prove the identity using the following rules, Algebra, reciprocal, quotient, Pythagorean, and odd/even?

Yahia M.
Featured yesterday

check the explanation below

#### Explanation:

$\csc \frac{- x}{\sec} \left(- x\right) = \frac{\frac{1}{\sin} \left(- x\right)}{\frac{1}{\cos} \left(- x\right)} = \cos \frac{- x}{\sin} \left(- x\right)$ color(green)(rarr "reciprocal"

now substitute

$\cos \frac{- x}{\sin} \left(- x\right) + \cos \frac{- x}{\sin} \left(- x\right) = 2 \cos \frac{- x}{\sin} \left(- x\right)$

color(blue)(sinx " is odd so "sin(-x)=-sinx"

color(blue)( cosx " is even so "cos(-x)=cosx

Substitute

$= 2 \cos \frac{x}{-} \sin x$color(green)(rarreven and odd " functions"

$= - 2 \cot \left(x\right)$color(green)(rarrquoitent

1

## Angle x is in the second quadrant and cos x = -4/5. what is sin x to the nearest tenth?

Jim G.
Featured yesterday

$\sin x = \frac{3}{5}$

#### Explanation:

$\text{using the "color(blue)"trigonometric identity}$

•color(white)(x)sin^2x+cos^2x=1

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

$\text{x is in second quadrant where } \sin x > 0$

$\Rightarrow \sin x = + \sqrt{1 - {\left(- \frac{4}{5}\right)}^{2}}$

$\textcolor{w h i t e}{\Rightarrow \sin x} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5} = 0.6$

0

## If cosA=4/5. Find cos2A?

maganbhai P.
Featured yesterday

$\cos 2 A = \frac{7}{25}$

#### Explanation:

Here,

$\cos A = \frac{4}{5} > 0 \implies {I}^{s t} Q u a \mathrm{dr} a n t \mathmr{and} I {V}^{t h} Q u a \mathrm{dr} a n t$

We know that,

$\cos 2 A = 2 {\cos}^{2} A - 1 = 2 {\left(\frac{4}{5}\right)}^{2} - 1$

$\therefore \cos 2 A = 2 \times \frac{16}{25} - 1 = \frac{32}{25} - 1$

$\implies \cos 2 A = \frac{32 - 25}{25} = \frac{7}{25} > 0$

$\implies \cos 2 A = \frac{7}{25}$

2

## How do you prove  cos^4(x) - sin^4(x) = cos(2x)?

Tidsresa ∞
Featured yesterday

${\cos}^{4} x - {\sin}^{4} x$

$= \left(\textcolor{red}{{\cos}^{2} x + {\sin}^{2} x}\right) \left(\textcolor{b l u e}{{\cos}^{2} x - {\sin}^{2} x}\right)$ $\left[\because {a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)\right]$

We know the Pythagorean Identity,

$\textcolor{red}{{\sin}^{2} x + {\cos}^{2} x = 1} \implies \textcolor{b l u e}{{\sin}^{2} x = 1 - {\cos}^{2} x}$

So,

=color(red)1×(cos^2 x- sin^2 x)

$= {\cos}^{2} x - \textcolor{b l u e}{\left(1 - {\cos}^{2} x\right)}$

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

$= \cos 2 x \implies \text{RHS}$ $\left[\because \text{Double Angle Formula:} \cos 2 x = 2 {\cos}^{2} x - 1 = 1 - 2 {\sin}^{2} x = {\cos}^{2} x - {\sin}^{2} x\right]$

Hope this helps. :)

1

## How do you convert 570^o to radians?

Jim G.
Featured 15 hours ago

$\frac{19 \pi}{6} \text{ radians}$

#### Explanation:

$\text{to convert from "color(blue)"degrees to radians}$

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{\text{radian measure "="degree measure } \times \frac{\pi}{180}} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\Rightarrow {\cancel{570}}^{19} / 1 \times \frac{\pi}{\cancel{180}} ^ 6 = \frac{19 \pi}{6} \leftarrow \textcolor{red}{\text{exact value}}$

1

## How do I solve given 0<x<2pi 2sinxcosx=sqrt2cosx?

Dean R.
Featured 15 hours ago

$x \in \left\{\frac{\pi}{4} , \frac{\pi}{2} , \frac{3 \pi}{4} , \frac{3 \pi}{2}\right\}$

#### Explanation:

$2 \sin x \cos x = \sqrt{2} \cos x$

$2 \sin x \cos x - \sqrt{2} \cos x = 0$

$\cos x \left(2 \sin x - \sqrt{2}\right) = 0$

$\cos x = 0$ or $\sin x = \setminus \frac{\sqrt{2}}{2}$

$x = {90}^{\circ} + {180}^{\circ} k \quad$ for integer $k$ or

$x = {45}^{\circ} + {360}^{\circ} k$ or

$x = {135}^{\circ} + {360}^{\circ} k$

In radians from $0$ to $2 \pi$ that's

$x = \left\{\frac{\pi}{4} , \frac{\pi}{2} , \frac{3 \pi}{4} , \frac{3 \pi}{2}\right\}$

Check:

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

$x = \frac{\pi}{4} , \quad \sin \left(2 x\right) = \sin \left(\frac{\pi}{2}\right) = 1$, sqrt 2 cos (pi/4) = 1 quad sqrt

$x = \frac{\pi}{2} , \quad \sin \left(2 x\right) = 0$, sqrt 2 cos (pi/2) = 0 quad sqrt

I'll leave the other to you to check.