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## What is the amplitude, period and the phase shift of #y = 2 sin (1/4 x)#?

Featured 2 weeks ago

The amplitude is $= 2$. The period is $= 8 \pi$ and the phase shift is $= 0$

#### Explanation:

We need

$\sin \left(a + b\right) = \sin a \cos b + \sin b \cos a$

The period of a periodic function is $T$ iif

$f \left(t\right) = f \left(t + T\right)$

Here,

$f \left(x\right) = 2 \sin \left(\frac{1}{4} x\right)$

Therefore,

$f \left(x + T\right) = 2 \sin \left(\frac{1}{4} \left(x + T\right)\right)$

where the period is $= T$

So,

$\sin \left(\frac{1}{4} x\right) = \sin \left(\frac{1}{4} \left(x + T\right)\right)$

$\sin \left(\frac{1}{4} x\right) = \sin \left(\frac{1}{4} x + \frac{1}{4} T\right)$

$\sin \left(\frac{1}{4} x\right) = \sin \left(\frac{1}{4} x\right) \cos \left(\frac{1}{4} T\right) + \cos \left(\frac{1}{4} x\right) \sin \left(\frac{1}{4} T\right)$

Then,

$\left\{\begin{matrix}\cos \left(\frac{1}{4} T\right) = 1 \\ \sin \left(\frac{1}{4} T\right) = 0\end{matrix}\right.$

$\iff$, $\frac{1}{4} T = 2 \pi$

$\iff$, $T = 8 \pi$

As

$- 1 \le \sin t \le 1$

Therefore,

$- 1 \le \sin \left(\frac{1}{4} x\right) \le 1$

$- 2 \le 2 \sin \left(\frac{1}{4} x\right) \le 2$

The amplitude is $= 2$

The phase shift is $= 0$ as when $x = 0$

$y = 0$

graph{2sin(1/4x) [-6.42, 44.9, -11.46, 14.2]}

## How to prove that #tan112 1/2=-sqrt2-1#?

Featured 2 weeks ago

#### Explanation:

We need

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

Therefore,

$\cos \left(\frac{x}{2}\right) = \sqrt{\frac{1 + \cos x}{2}}$

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

$\tan \left(\frac{x}{2}\right) = \sin \frac{\frac{x}{2}}{\cos} \left(\frac{x}{2}\right) = \sqrt{\frac{1 - \cos x}{1 + \cos x}}$

$= \sqrt{\frac{\left(1 - \cos x\right) \left(1 - \cos x\right)}{\left(1 + \cos x\right) \left(1 - \cos x\right)}}$

$= \sqrt{{\left(1 - \cos x\right)}^{2} / \left(1 - {\cos}^{2} x\right)}$

$= \sqrt{{\left(1 - \cos x\right)}^{2} / \left({\sin}^{2} x\right)}$

And finally,

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

Here,

$x = 225$

$\cos \left(225\right) = - \frac{1}{\sqrt{2}}$

$\sin \left(225\right) = - \frac{1}{\sqrt{2}}$

$\tan \left(\frac{225}{2}\right) = \frac{1 - \left(- \frac{1}{\sqrt{2}}\right)}{- \frac{1}{\sqrt{2}}} = - \left(\sqrt{2} + 1\right)$

## This is the question ?

Abhishek K.
Featured 2 weeks ago

Given,

$\rightarrow \left(\frac{x}{a}\right) \cos \theta + \left(\frac{y}{b}\right) \sin \theta = 1. \ldots . \left(1\right)$

$\rightarrow \left(\frac{x}{a}\right) \sin \theta - \left(\frac{y}{b}\right) \cos \theta = 1. \ldots . . \left(2\right)$

Squaring and adding equation $\left(1\right)$ and $\left(2\right)$,we get

$\rightarrow \frac{{x}^{2}}{{a}^{2}} {\cos}^{2} \theta \cancel{+ 2 \left(\frac{x y}{a b}\right) \cos \theta \cdot \sin \theta} + \frac{{y}^{2}}{{b}^{2}} {\sin}^{2} \theta + \frac{{x}^{2}}{{a}^{2}} {\sin}^{2} \theta \cancel{- 2 \left(\frac{x y}{a b}\right) \sin \theta \cdot \cos \theta} + \frac{{y}^{2}}{{b}^{2}} {\cos}^{2} \theta = {1}^{2} + {1}^{2}$

$\rightarrow \frac{{x}^{2}}{{a}^{2}} \left({\cos}^{2} \theta + {\sin}^{2} \theta\right) + \frac{{y}^{2}}{{b}^{2}} \left({\sin}^{2} \theta + {\cos}^{2} \theta\right) = 2$

$\therefore \frac{{x}^{2}}{{a}^{2}} + \frac{{y}^{2}}{{b}^{2}} = 2$

## How do you graph #y = sin(x - π/4) + 2#?

sankarankalyanam
Featured 1 week ago

As below.

#### Explanation:

$y = \sin \left(x - \left(\frac{\pi}{4}\right)\right) + 2$

Standard Form of a sine function is $y = A \sin \left(B x - C\right) + D$

$\therefore A = 1 , B = 1 , C = \frac{\pi}{4} , D = 2$

$A m p l i t u \mathrm{de} = | A | = 1$

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

Phase Shift $= - \frac{C}{B} = \frac{\pi}{4}$, $\frac{\pi}{4}$ to the right#

Vertical Shift $= D = 2$

graph{sin(x-(pi/4)) + 2 [-10, 10, -5, 5]}

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

Ratnaker Mehta
Featured 4 days 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!

## 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 4 days 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:

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

$a + b = 2.5396 + 12.6541$

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

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