# Make the internet a better place to learn

1

## How do you graph y=-sec2x?

sankarankalyanam
Featured 1 month ago

As below.

#### Explanation:

Standard form of Secant function is color(violet)(y = A sec (Bx - C) + D

$\text{Given } y = - \sec \left(2 x\right)$

$A = - 1 , B = 2 , C = 0 , D = 0$

$A m p l i t u \mathrm{de} = \text{None for secant function}$

$\text{Period } = \frac{2 \pi}{|} B | = \frac{2 \pi}{2} = \pi$

$\text{Phase Shift } = - \frac{C}{B} = 0$

$\text{Vertical Shift } = D = 0$

graph{-sec (2x) [-10, 10, -5, 5]}

2

## How do I solve (2sqrt2-2)cosalpha=3sin^2alpha+sqrt2-4?

Ratnaker Mehta
Featured 1 month ago

$\therefore \alpha = 2 k \pi \pm \arccos \left[\frac{- \left(\sqrt{2} - 1\right) \pm \sqrt[4]{2}}{3}\right] , k \in \mathbb{Z}$.

#### Explanation:

Replacing ${\sin}^{2} \alpha$ in the right member by $\left(1 - {\cos}^{2} \alpha\right)$, we have,

$\left(2 \sqrt{2} - 2\right) \cos \alpha = 3 - 3 {\cos}^{2} \alpha + \sqrt{2} - 4$.

$\therefore 3 {\cos}^{2} \alpha + 2 \left(\sqrt{2} - 1\right) \cos \alpha - \left(\sqrt{2} - 1\right) = 0$.

This is a quadr. eqn. in $\cos \alpha$, and, its discriminant

$\Delta = {\left(2 \sqrt{2} - 2\right)}^{2} + 4 \cdot 3 \cdot \left(\sqrt{2} - 1\right)$,

$= \left(8 - 8 \sqrt{2} + 4\right) + 12 \sqrt{2} - 12$,

$\therefore \Delta = 4 \sqrt{2}$.

Therefore, using the quadr. formula, we have,

$\cos \alpha = \frac{- 2 \left(\sqrt{2} - 1\right) \pm \sqrt{4 \sqrt{2}}}{2 \cdot 3}$,

$= \frac{- 2 \left(\sqrt{2} - 1\right) \pm 2 \sqrt[4]{2}}{2 \cdot 3}$,

$\therefore \cos \alpha = \frac{- \left(\sqrt{2} - 1\right) \pm \sqrt[4]{2}}{3}$.

$\therefore \alpha = 2 k \pi \pm \arccos \left[\frac{- \left(\sqrt{2} - 1\right) \pm \sqrt[4]{2}}{3}\right] , k \in \mathbb{Z}$.

1

## Prove that:Sectheta -1/sectheta+1=sin^2 theta/(1+costheta)^2 ?

Ratnaker Mehta
Featured 1 month ago

Kindly see a Proof in the Explanation.

#### Explanation:

$\frac{\sec \theta - 1}{\sec \theta + 1}$,

$= \left(\frac{1}{\cos} \theta - 1\right) \div \left(\frac{1}{\cos} \theta + 1\right)$,

$= \frac{1 - \cos \theta}{\cos} \theta \div \frac{1 + \cos \theta}{\cos} \theta$,

$= \frac{1 - \cos \theta}{1 + \cos \theta}$,

$= \frac{1 - \cos \theta}{1 + \cos \theta} \times \frac{1 + \cos \theta}{1 + \cos \theta}$,

$= \frac{1 - {\cos}^{2} \theta}{1 + \cos \theta} ^ 2$,

$= {\sin}^{2} \frac{\theta}{1 + \cos \theta} ^ 2$, as desired!

1

## How do you solve 2cosx+1=0 for 0°<=x<=360°?

maganbhai P.
Featured 1 month ago

$x = {120}^{\circ} \mathmr{and} {240}^{\circ}$

#### Explanation:

Here,

$2 \cos x + 1 = 0 , w h e r e , {0}^{\circ} \le x \le {360}^{\circ}$

$\implies 2 \cos x = - 1$

$\implies \cos x = - \frac{1}{2} < 0 \implies I {I}^{n d} Q u a \mathrm{dr} a n t \mathmr{and} I I {I}^{r d} Q u a \mathrm{dr} a n t$

$\left(i\right) {90}^{\circ} < x < {180}^{\circ} \implies x = {120}^{\circ} \to \left(I {I}^{n d} Q u a \mathrm{dr} a n t\right)$

$\left(i i\right) {180}^{\circ} < x < {270}^{\circ} \implies x = {240}^{\circ} \to \left(I I {I}^{r d} Q u a \mathrm{dr} a n t\right)$

Hence,

$x = {120}^{\circ} \mathmr{and} {240}^{\circ}$

1

## Prove that sin5π/18+cos4π/9=cosπ/9?

maganbhai P.
Featured 1 month ago

We know that,
color(blue)(cosC+cosD=2cos((C+D)/2)cos((C-D)/2)to(1)

#### Explanation:

We have to prove ,

$\sin \left(\frac{5 \pi}{18}\right) + \cos \left(\frac{4 \pi}{9}\right) = \cos \left(\frac{\pi}{9}\right)$

We take,

$L H S = \sin \left(\frac{5 \pi}{18}\right) + \cos \left(\frac{4 \pi}{9}\right)$

$\textcolor{w h i t e}{L H S} = \cos \left(\frac{\pi}{2} - \frac{5 \pi}{18}\right) + \cos \left(\frac{4 \pi}{9}\right)$

$\textcolor{w h i t e}{L H S} = \cos \left(\frac{9 \pi - 5 \pi}{18}\right) + \cos \left(\frac{4 \pi}{9}\right)$

color(white)(LHS)=color(blue)(cos((2pi)/9)+cos((4pi)/9)...toApply(1)

color(white)(LHS)=color(blue)(2cos(((2pi)/9+(4pi)/9)/2)cos(((2pi)/9-(4pi)/9)/2)

$\textcolor{w h i t e}{L H S} = 2 \cos \left(\frac{6 \pi}{18}\right) \cos \left(\frac{- 2 \pi}{18}\right)$

$\textcolor{w h i t e}{L H S} = 2 \cos \left(\frac{\pi}{3}\right) \cos \left(- \frac{\pi}{9}\right)$

color(white)(LHS)=2(1/2)cos(pi/9)...to[becausecos(- theta)=costheta ]

$\textcolor{w h i t e}{L H S} = \cos \left(\frac{\pi}{9}\right)$

$L H S = R H S$

2

## Boris is sitting in a movie theatre, 11 meters from the screen. The angle of elevation from his line of sight to the top of the screen is 13degrees, and the angle of depression from his line of sight to the bottom of the screen is 49degrees?

Anjali G
Featured 1 month ago

$\text{Screen height " = 15.2 " meters}$

#### Explanation:

Using the variable definitions of $a$ and $b$ as shown in the diagram above, we can use trigonometry to solve for the height of the screen, $a + b$.

$\tan \left({13}^{\circ}\right) = \frac{a}{11}$

$a = 11 \tan \left({13}^{\circ}\right)$

$\tan \left({49}^{\circ}\right) = \frac{b}{11}$

$b = 11 \tan \left({49}^{\circ}\right)$

Therefore, the height of the screen $a + b$ is:

$a + b = 11 \tan \left({13}^{\circ}\right) + 11 \tan \left({49}^{\circ}\right)$

$a + b = 2.5396 + 12.6541$

$a + b = 15.1936 \text{ meters}$

1

## How do you divide (i+2) / (9i+4) in trigonometric form?

sankarankalyanam
Featured 1 month ago

color(blue)((2 + i) / (4 + i9) = 0.175 - i 0.144

#### Explanation:

To divide $\frac{2 + i}{4 + i 9}$ using trigonometric form.

${z}_{1} = \left(2 + i\right) , {z}_{2} = \left(4 + i 9\right)$

${r}_{1} = \sqrt{{2}^{2} + {1}^{2}} = \sqrt{5}$

${r}_{2} = \sqrt{{9}^{2} + {4}^{2}} = \sqrt{97}$

${\theta}_{1} = \arctan \left(\frac{1}{2}\right) = {26.57}^{\circ}$

${\Theta}_{2} = \arctan \left(\frac{9}{4}\right) = {66.06}^{\circ}$

${r}_{1} / {r}_{2} = \sqrt{5} \frac{\cdot}{\sqrt{97}} \approx 0.227$

${z}_{1} / {z}_{2} = \left({r}_{1} / {r}_{2}\right) \cdot \left(\cos \left({\theta}_{1} - {\theta}_{2}\right) + i \sin \left({\theta}_{1} - {\theta}_{2}\right)\right)$

${z}_{1} / {z}_{2} = \left(0.227\right) \cdot \left(\cos \left(26.57 - 66.06\right) + i \sin \left(26.57 - 66.06\right)\right)$

${z}_{1} / {z}_{2} = 0.227 \cdot \left(\cos \left(- 39.49\right) + i \sin \left(- 39.49\right)\right) = 0.227 \left(0.772 - i 0.636\right)$

color(blue)((2 + i) / (4 + i9) = 0.175 - i 0.144

1

## What is the product of 5 (cos frac{3\pi}{4}+isinfrac{3\pi}{4} ) \cdot \sqrt{3}(cosfrac{\pi}{2}+i sinfrac{\pi}{2})?

sankarankalyanam
Featured 1 month ago

color(maroon)(=> -5sqrt(3/2) + i 5sqrt(3/2) or $- 5 \sqrt{\frac{3}{2}} \left(1 - i\right)$

#### Explanation:

${Z}_{1} \cdot {Z}_{2} = {r}_{1} \cdot {r}_{2} \left(\cos \left({\theta}_{1} + {\theta}_{2}\right) + i \sin \left({\theta}_{1} + {\theta}_{2}\right)\right)$

Z_1 = 5 (cos ((3pi)/4) + i sin ((3pi)/4)

${Z}_{2} = \sqrt{3} \left(\cos \left(\frac{\pi}{2}\right) + i \sin \left(\frac{\pi}{2}\right)\right)$

${r}_{1} = 5 , {r}_{2} = \sqrt{3} , {\theta}_{1} = \frac{3 \pi}{4} , {\theta}_{2} = \frac{\pi}{2}$

Z_1 * Z_2 = 5 sqrt3 ((cos ((3pi)/4 + pi/2) + i sin ((3pi)/4 + pi/2))

$\implies 5 \sqrt{3} \cdot \left(- \left(\frac{1}{\sqrt{2}}\right) + i \left(\frac{1}{\sqrt{2}}\right)\right)$

color(maroon)(=> -5sqrt(3/2) + i 5sqrt(3/2) or $- 5 \sqrt{\frac{3}{2}} \left(1 - i\right)$

1

## If a=5 b=12 c=13 in a trangle ABC than what is the value of tan (B/4)=?

maganbhai P.
Featured 4 weeks ago

$\tan \left(\frac{B}{4}\right) = \frac{\sqrt{13} - 3}{2}$

#### Explanation:

Here,

$a = 5 , b = 12 \mathmr{and} c = 13$

$\implies B C = 5 , C A = 12 , A B = 13$

$\implies \overline{A B}$ , is the hypotenuse of $\triangle A B C$

So, $m \angle C = {90}^{\circ} \implies \angle A \mathmr{and} \angle B$ are acute angles.

Now, color(red)(cosB=("side adjacent to B")/("hypotenus")=5/13

Using half angle formula :

$\cos \left(\frac{B}{2}\right) = \frac{1 + \cos B}{2} = \frac{1 + \frac{5}{13}}{2} = \frac{9}{13} = {\left(\frac{3}{\sqrt{13}}\right)}^{2}$

=>color(red)(cos(B/2)=+3/sqrt13...to[because mangleB < 90^circ]

Also ,we know the half angle formula:

color(blue)(tan^2(theta/2)=(1-costheta)/(1+costheta)

Let us take, $\theta = \left(\frac{B}{2}\right)$

:.color(blue)(tan^2((B/2)/2)=(1-cos(B/2))/(1+cos(B/2))

=>tan^2(B/4)=(1-3/sqrt13)/(1+3/sqrt13)=(sqrt13-3)/(sqrt13+3)=(sqrt13-3)/(sqrt13+3)xxcolor(green)((1)

=>tan^2(B/4)=(sqrt13-3)/(sqrt13+3)xxcolor(green)((sqrt13-3)/(sqrt13-3)

$\implies {\tan}^{2} \left(\frac{B}{4}\right) = {\left(\sqrt{13} - 3\right)}^{2} / \left({\left(\sqrt{13}\right)}^{2} - {\left(3\right)}^{2}\right)$

$\implies {\tan}^{2} \left(\frac{B}{4}\right) = {\left(\sqrt{13} - 3\right)}^{2} / 4$

$\implies {\left(\tan \left(\frac{B}{4}\right)\right)}^{2} = {\left(\frac{\sqrt{13} - 3}{2}\right)}^{2.} . . \to m \angle B < {90}^{\circ}$

$\implies \tan \left(\frac{B}{4}\right) = \frac{\sqrt{13} - 3}{2}$

2

## If tanx/(1-cotx) + cotx/(1-tanx) = -1, what is the value of tanx?

dk_ch
Featured 4 weeks ago

$\tan \frac{x}{1 - \cot x} + \cot \frac{x}{1 - \tan x} = - 1$

$\implies - \tan \frac{x}{1 - \frac{1}{\tan} x} - \frac{\frac{1}{\tan} x}{1 - \tan x} = 1$

$\implies - {\tan}^{2} \frac{x}{\tan x - 1} - \frac{1}{\tan x \left(1 - \tan x\right)} = 1$

$\implies {\tan}^{2} \frac{x}{1 - \tan x} - \frac{1}{\tan x \left(1 - \tan x\right)} = 1$

$\implies \frac{{\tan}^{3} x - 1}{\tan x \left(1 - \tan x\right)} = 1$

$\implies \frac{\left(\tan x - 1\right) \left({\tan}^{2} x + \tan x + 1\right)}{\tan x \left(1 - \tan x\right)} = 1$

$\tan x = 1$ will make LHS undefined so neglected.
Hence

$\implies - \frac{{\tan}^{2} x + \tan x + 1}{\tan} x = 1$

$\implies {\tan}^{2} x + 2 \tan x + 1 = 0$

$\implies {\left(\tan x + 1\right)}^{2} = 0$

$\implies \tan x = - 1$