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## Sin^4x=1/8(3-4cos2x+cos4x) How to prove?

Abhishek K.
Featured 1 week ago

$L H S = {\sin}^{4} x$

$= {\left[\frac{2 {\sin}^{2} x}{2}\right]}^{2}$

$= \frac{1}{4} \left[{\left(1 - \cos 2 x\right)}^{2}\right]$

$= \frac{1}{4} \left[1 - 2 \cos 2 x + {\cos}^{2} \left(2 x\right)\right]$

$= \frac{2}{4 \cdot 2} \left[1 - 2 \cos 2 x + {\cos}^{2} \left(2 x\right)\right]$

$= \frac{1}{8} \left[2 - 4 \cos 2 x + 2 {\cos}^{2} \left(2 x\right)\right]$

$= \frac{1}{8} \left[2 - 4 \cos 2 x + 1 + \cos 4 x\right]$

$= \frac{1}{8} \left[3 - 4 \cos 2 x + \cos 4 x\right] = R H S$

## How do you find the sum of all solutions of cosxcos(x+pi/3)cos(pi/3-x)=1/4. if x is in [0, 6pi]?

Shwetank Mauria
Featured 1 week ago

Sum of all solutions is $30 \pi$.

#### Explanation:

$\cos x \cos \left(x + \frac{\pi}{3}\right) \cos \left(\frac{\pi}{3} - x\right) = \frac{1}{4}$ can be written as

$4 \cos x \cos \left(x + \frac{\pi}{3}\right) \cos \left(\frac{\pi}{3} - x\right) = 1$

or $2 \cos x \left[2 \cos \left(x + \frac{\pi}{3}\right) \cos \left(\frac{\pi}{3} - x\right)\right] = 1$

or $2 \cos x \left[\cos \left(x + \frac{\pi}{3} + \frac{\pi}{3} - x\right) + \cos \left(x + \frac{\pi}{3} - \frac{\pi}{3} + x\right)\right] = 1$

or $2 \cos x \left[\cos \left(\frac{2 \pi}{3}\right) + \cos 2 x\right] = 1$

or $2 \cos x \left[- \frac{1}{2} + 2 {\cos}^{2} x - 1\right] = 1$

or $4 {\cos}^{3} x - 3 \cos x = 1$

or $\cos 3 x = 1 = \cos 0$

i.e. $3 x = 2 n \pi$, where $n$ is an integer.

or $x = \frac{2 n \pi}{3}$, where $n$ is an integer.

and in $\left[0 , 6 \pi\right]$ we will have

$x = \left[0 , \textcolor{red}{\frac{2 \pi}{3} , \frac{4 \pi}{3}} , 2 \pi , \textcolor{red}{\frac{8 \pi}{3} , \frac{10 \pi}{3}} , 4 \pi , \textcolor{red}{\frac{14 \pi}{3} , \frac{16 \pi}{3}} , 6 \pi\right]$

and their sum is $\textcolor{red}{2 \pi} + 2 \pi + \textcolor{red}{6 \pi} + 4 \pi + \textcolor{red}{10 \pi} + 6 \pi = 30 \pi$

## Sin (x+y) + sin (x-y) / sin (x+y) - sin (x-y) = tan x cot y ?

Ratnaker Mehta
Featured 5 days ago

Kindly go through the Explanation.

#### Explanation:

$\sin \left(x + y\right) + \sin \left(x - y\right)$,

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

$\therefore \sin \left(x + y\right) + \sin \left(x - y\right) = 2 \sin x \cos y \ldots \ldots \left({\ast}^{1}\right)$.

Otherwise,

$\sin \left(x + y\right) + \sin \left(x - y\right)$,

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

$= 2 \sin x \cos y$.

Similarly, $\sin \left(x + y\right) - \sin \left(x - y\right) = 2 \cos x \sin y \ldots \ldots \left({\ast}^{2}\right)$.

From $\left({\ast}^{1}\right) \mathmr{and} \left({\ast}^{2}\right)$, we have,

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

$= \left(\sin \frac{x}{\cos} y\right) \cdot \left(\cos \frac{y}{\sin} y\right)$,

$= \tan x \cot y ,$ as desired!

## How do you use sum and difference identities to simplify csc(pi/2 +x) - sec(pi - x)?

Hoat V.
Featured 5 days ago

0

#### Explanation:

$\csc \left(\theta\right) = \frac{1}{\sin} \left(\theta\right) , \sec \left(\theta\right) = \frac{1}{\cos} \left(\theta\right)$
$\frac{1}{\sin \left(\frac{\pi}{2} + x\right)} - \frac{1}{\cos \left(\pi - x\right)}$

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

$\frac{1}{\sin \left(\frac{\pi}{2}\right) \cdot \cos x + \cos \left(\frac{\pi}{2}\right) \cdot \sin x} - \frac{1}{\cos \pi \cdot \cos x + \sin \pi \cdot \sin x}$
Using either calculator or unit circle,
$\cos \left(\frac{\pi}{2}\right) = 0 , \sin \left(\pi\right) = 0 , \cos \left(\pi\right) = 1 , \sin \left(\frac{\pi}{2}\right) = 1$

$\frac{1}{\cos x} - \frac{1}{\cos x} = 0$

## If cos θ=4/5 on the interval (3pi/2, 2pi) find the exact value of tan 2θ?

dk_ch
Featured 5 days ago

Given $\cos \theta = \frac{4}{5} \mathmr{and} \theta \in \left(\frac{3 \pi}{2} , 2 \pi\right) i . e . {Q}_{4}$

So $\cos \theta \mathmr{and} S e c \theta \to + v e$

But $\tan \theta \to - v e$

Now $\tan \theta = - \sqrt{{\sec}^{2} \theta - 1} = - \sqrt{{\left(\frac{5}{4}\right)}^{2} - 1} = - \frac{3}{4}$

Now

$\tan 2 \theta = \frac{2 \tan \theta}{1 - {\tan}^{2} \theta} = \frac{2 \times \left(- \frac{3}{4}\right)}{1 - {\left(- \frac{3}{4}\right)}^{2}}$

$= - 2 \times \frac{3}{4} \times \frac{16}{7} = - \frac{24}{7}$

## How do you use a half-angle formula to find the exact value of cos((3pi)/8)?

Jim G.
Featured 2 days ago

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

#### Explanation:

$\text{using the "color(blue)"half angle formula}$

•color(white)(x)cos(x/2)=+-sqrt((1+cosx)/2)

$\left(\frac{x}{2}\right) = \frac{3 \pi}{8} \Rightarrow x = \frac{3 \pi}{4}$

$\Rightarrow \cos \left(\frac{3 \pi}{8}\right) = \pm \sqrt{\frac{1 + \cos \left(\frac{3 \pi}{4}\right)}{2}}$

$\frac{3 \pi}{8} \text{ is in first quadrant }$

$= + \sqrt{\frac{1 - \cos \left(\frac{\pi}{4}\right)}{2}}$

$= + \sqrt{\frac{1 - \frac{1}{\sqrt{2}}}{2}} = + \sqrt{\frac{1 - \frac{\sqrt{2}}{2}}{2}} = \sqrt{\frac{2 - \sqrt{2}}{4}}$

$\Rightarrow \cos \left(\frac{3 \pi}{8}\right) = \frac{1}{2} \sqrt{2 - \sqrt{2}}$

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