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## How do you simplify sin(arccos(x))?

George C.
Featured 3 months ago

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

#### Explanation:

From Pythagoras, we have:

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

If $x \in \left[- 1 , 1\right]$ and $\theta = \arccos \left(x\right)$ then:

$\theta \in \left[0 , \pi\right]$

$\sin \left(\theta\right) \ge 0$

Hence:

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

Note we can use the non-negative square root since we have already established that $\sin \left(\arccos \left(x\right)\right) \ge 0$

## How would you use the sum and difference identities to prove: 2sin(a+b)sin(a-b) = cos(2b) - cos(2a) ?

dk_ch
Featured 3 months ago

Let $a + b = x \mathmr{and} a - b = y$

So $2 a = x + y \mathmr{and} 2 b = x - y$

Now $R H S = \cos 2 b - \cos 2 a$

$= \cos \left(x - y\right) - \cos \left(x + y\right)$

$= \cos x \cos y + \sin x \sin y - \cos x \cos y + \sin x \sin y$

$= 2 \sin x \sin y$

$= 2 \sin \left(a + b\right) \sin \left(a - b\right) = L H S$

## HS trigonometry question? arc cot(x)+2arc sin (sqrt(3)/2)= pi The rad is only over three. Can someone help me?

sente
Featured 3 months ago

$x = \frac{\sqrt{3}}{3}$

#### Explanation:

$\text{arccot} \left(x\right) + 2 \arcsin \left(\frac{\sqrt{3}}{2}\right) = \pi$

The inverse sine function $\arcsin \left(x\right)$ is defined as the unique value in the interval $\left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$ such that $\sin \left(\arcsin \left(x\right)\right) = x$. On that interval, we have $\sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$ as a well known angle. Thus $\arcsin \left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}$

$\implies \text{arccot} \left(x\right) + \frac{2 \pi}{3} = \pi$

$\implies \text{arccot} \left(x\right) = \frac{\pi}{3}$

$\implies \cot \left(\text{arccot} \left(x\right)\right) = \cot \left(\frac{\pi}{3}\right)$

$\implies x = \cot \left(\frac{\pi}{3}\right)$

$= \cos \frac{\frac{\pi}{3}}{\sin} \left(\frac{\pi}{3}\right)$

$= \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}$

$= \frac{1}{\sqrt{3}}$

$= \frac{\sqrt{3}}{3}$

## How do you simplify the expression : arcsin((x+1)/sqrt(2*(x²+1))) ?

dk_ch
Featured 3 months ago

Given expression =arcsin((x+1)/sqrt(2*(x²+1)))
Let $x = \tan \theta$

So $\theta = {\tan}^{-} 1 x$

Inserting $x = \tan \theta$ the given expression becomes

$= \arcsin \left(\frac{\tan \theta + 1}{\sqrt{2 \cdot \left({\tan}^{2} \theta + 1\right)}}\right)$

$= \arcsin \left(\frac{\sin \frac{\theta}{\cos} \theta + 1}{\sqrt{2 \cdot \left({\sec}^{2} \theta\right)}}\right)$

$= \arcsin \left(\frac{1}{\sqrt{2}} \left(\frac{\sin \theta \sec \theta + 1}{\sec \theta}\right)\right)$

$= \arcsin \left(\frac{1}{\sqrt{2}} \left(\frac{\sin \theta \cancel{\sec} \theta}{\cancel{\sec}} \theta + \frac{1}{\sec \theta}\right)\right)$

$= \arcsin \left(\frac{1}{\sqrt{2}} \left(\sin \theta + \cos \theta\right)\right)$

$= \arcsin \left(\frac{1}{\sqrt{2}} \sin \theta + \frac{1}{\sqrt{2}} \cos \theta\right)$

$= \arcsin \left(\cos \left(\frac{\pi}{4}\right) \sin \theta + \sin \left(\frac{\pi}{4}\right) \cos \theta\right)$

$= \arcsin \left(\sin \left(\theta + \frac{\pi}{4}\right)\right)$

$= \theta + \frac{\pi}{4}$

$= {\tan}^{-} 1 x + \frac{\pi}{4}$

## How do you use the sum and difference formula to simplify cos((17pi)/12)?

dk_ch
Featured 3 months ago

$\cos \left(\frac{17 \pi}{12}\right)$

$\cos \left(\frac{24 \pi - 7 \pi}{12}\right)$

$= \cos \left(2 \pi - \frac{7 \pi}{12}\right)$

$= \cos \left(\frac{7 \pi}{12}\right)$

$= \cos \left(\frac{4 \pi + 3 \pi}{12}\right)$

$= \cos \left(\frac{4 \pi}{12} + \frac{3 \pi}{12}\right)$

$= \cos \left(\frac{\pi}{3} + \frac{\pi}{4}\right)$

$= \cos \left(\frac{\pi}{3}\right) \cos \left(\frac{\pi}{4}\right) - \sin \left(\frac{\pi}{3}\right) \sin \left(\frac{\pi}{4}\right)$

$= \frac{1}{2} \cdot \frac{1}{\sqrt{2}} - \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}}$

$= \frac{1 - \sqrt{3}}{2 \sqrt{2}}$

$= \frac{1}{4} \left(\sqrt{2} - \sqrt{6}\right)$

## What is the definition of the Inverse Sine Function?

Ratnaker Mehta
Featured 3 months ago

The Inverse Sine Function is denoted by, $a r c \sin$ or, ${\sin}^{-} 1$ and

is defined by,

$a r c \sin x = \theta , x \in \left[- 1 , 1\right] \iff \sin \theta = x , \theta \in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$

#### Explanation:

The Inverse Sine Function is denoted by, $a r c \sin$ or, ${\sin}^{-} 1$ and

is defined by,

$a r c \sin x = \theta , x \in \left[- 1 , 1\right] \iff \sin \theta = x , \theta \in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$

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