# Math Help - sinus, cosinus expression

1. ## sinus, cosinus expression

sinx = -2/(5^0.5)

3pi/2 < x < 2pi

the question that i don´t understand is that i shuold use this values and decide the exacly value of this expression: cos(2x-3pi/4)

so i did this way

arcsin(-2/(5^0.5) = x = -63.4

cos(2*(-63.4)-3pi/4)

are i doing right ?

2. Originally Posted by paulaa
sinx = -2/(5^0.5)

3pi/2 < x < 2pi

the question that i don´t understand is that i shuold use this values and decide the exacly value of this expression: cos(2x-3pi/4)

so i did this way

arcsin(-2/(5^0.5) = x = -63.4

cos(2*(-63.4)-3pi/4)

are i doing right ?
But... 3pi/2 < x < 2pi (270* < x < 360*)

3. sinx = -2/(5^0.5)
3pi/2 < x < 2pi

x = 2*pi*n-sin^(-1)(2/sqrt(5)), där n = Z

n=1.

x = 2*pi*1-sin^(-1)(2/sqrt(5))= ~ 5.17604

is this the answer for this question ?

4. $\displaystyle \cos{\left(2x - \frac{3\pi}{4}\right)} = \cos{2x}\cos{\frac{3\pi}{4}} + \sin{2x}\sin{\frac{3\pi}{4}}$

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

$\displaystyle = -\cos{\frac{\pi}{4}}\cos{2x} + \sin{\frac{\pi}{4}}\sin{2x}$

$\displaystyle = -\frac{\sqrt{2}}{2}\cos{2x} + \frac{\sqrt{2}}{2}\sin{2x}$

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

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

$\displaystyle = -\frac{\sqrt{2}}{2}\left[1 - 2\left(-\frac{2}{\sqrt{5}}\right)^2\right] + \sqrt{2}\left(-\frac{2}{\sqrt{5}}\right)\sqrt{1 - \left(-\frac{2}{\sqrt{5}}\right)^2}$

$\displaystyle = -\frac{\sqrt{2}}{2}\left[1 - 2\left(\frac{4}{5}\right)\right] - \frac{2\sqrt{2}}{\sqrt{5}}\sqrt{1 - \frac{4}{5}}$

$\displaystyle = -\frac{\sqrt{2}}{2}\left(-\frac{3}{5}\right) - \frac{2\sqrt{2}}{\sqrt{5}}\left(\frac{1}{\sqrt{5}} \right)$

$\displaystyle = \frac{3\sqrt{2}}{10} - \frac{2\sqrt{2}}{5}$

$\displaystyle = \frac{3\sqrt{2}}{10} - \frac{4\sqrt{2}}{10}$

$\displaystyle = -\frac{\sqrt{2}}{10}$.

5. Originally Posted by paulaa
sinx = -2/(5^0.5)

3pi/2 < x < 2pi

the question that i don´t understand is that i should use this value and decide the exact value of this expression: cos(2x-3pi/4)

so i did this way

arcsin(-2/(5^0.5) = x = -63.4

cos(2*(-63.4)-3pi/4)

are i doing right ?
Not quite!

You are being asked to calculate an exact value, so you need to proceed
similarly to Prove It's method.

$\displaystyle\ Sinx=-\frac{2}{\sqrt{5}}$

You could reference the Unit Circle also and write

$Cos({\pi}-A)=-CosA,\;\;\;Sin({\pi}-A)=SinA$

Also, the angle x is in the 4th quadrant from the range of x given.

You could sketch a right-angled triangle and use Pythagoras' theorem to find $Cosx$

$\displaystyle\ 2^2+1^2=5\Rightarrow\ Cosx=\frac{1}{\sqrt{5}}$

Therefore....

$\displaystyle\ Cos\left(2x-\frac{3{\pi}}{4}\right)=Cos(2x)Cos\left(\frac{3{\p i}}{4}\right)+Sin(2x)Sin\left(\frac{3{\pi}}{4}\rig ht)$

from $Cos(A-B)=CosACosB+SinASinB$

and since

$\displaystyle\ Cos\left(\frac{3{\pi}}{4}\right)=-Cos\left(\frac{\pi}{4}\right)=-\frac{1}{\sqrt{2}}$

while

$\displaystyle\ Sin\left(\frac{3{\pi}}{4}\right)=Sin\left(\frac{\p i}{4}\right)=\frac{1}{\sqrt{2}}$

then

$\displaystyle\ Cos\left(2x-\frac{3{\pi}}{4}\right)=-\frac{Cos(2x)}{\sqrt{2}}+\frac{Sin(2x)}{\sqrt{2}}$

Using

$\displaystyle\ Sin^2x=\frac{1}{2}\left[1-Cos(2x)\right]\Rightarrow\ Cos(2x)=1-2Sin^2x=1-2\left[\frac{4}{5}\right]=-\frac{3}{5}$

and

$\displaystyle\ Sin(2x)=2SinxCosx=-\frac{4}{\sqrt{5}}\;\left[\frac{1}{\sqrt{5}}\right]=-\frac{4}{5}$

$\displaystyle\Rightarrow\ Cos\left(2x-\frac{3{\pi}}{4}\right)=\frac{3}{5\sqrt{2}}-\frac{4}{5\sqrt{2}}=-\frac{1}{5\sqrt{2}}$

6. Originally Posted by Archie Meade
Not quite!

You are being asked to calculate an exact value, so you need to proceed
similarly to Prove It's method.

$\displaystyle\ Sinx=-\frac{2}{\sqrt{5}}$

You could reference the Unit Circle also and write

$Cos({\pi}-A)=-CosA,\;\;\;Sin({\pi}-A)=SinA$

Also, the angle x is in the 4th quadrant from the range of x given.

You could sketch a right-angled triangle and use Pythagoras' theorem to find $Cosx$

$\displaystyle\ 2^2+1^2=5\Rightarrow\ Cosx=\frac{1}{\sqrt{5}}$

Therefore....

$\displaystyle\ Cos\left(2x-\frac{3{\pi}}{4}\right)=Cos(2x)Cos\left(\frac{3{\p i}}{4}\right)+Sin(2x)Sin\left(\frac{3{\pi}}{4}\rig ht)$

from $Cos(A-B)=CosACosB+SinASinB$

and since

$\displaystyle\ Cos\left(\frac{3{\pi}}{4}\right)=-Cos\left(\frac{\pi}{4}\right)=-\frac{1}{\sqrt{2}}$

while

$\displaystyle\ Sin\left(\frac{3{\pi}}{4}\right)=Sin\left(\frac{\p i}{4}\right)=\frac{1}{\sqrt{2}}$

then

$\displaystyle\ Cos\left(2x-\frac{3{\pi}}{4}\right)=-\frac{Cos(2x)}{\sqrt{2}}+\frac{Sin(2x)}{\sqrt{2}}$

Using

$\displaystyle\ Sin^2x=\frac{1}{2}\left[1-Cos(2x)\right]\Rightarrow\ Cos(2x)=1-2Sin^2x=1-2\left[\frac{4}{5}\right]=-\frac{3}{5}$

and

$\displaystyle\ Sin(2x)=2SinxCosx=-\frac{4}{\sqrt{5}}\;\left[\frac{1}{\sqrt{5}}\right]=-\frac{4}{5}$

$\displaystyle\Rightarrow\ Cos\left(2x-\frac{3{\pi}}{4}\right)=\frac{3}{5\sqrt{2}}-\frac{4}{5\sqrt{2}}=-\frac{1}{5\sqrt{2}}$
$\displaystyle = -\frac{\sqrt{2}}{10}$ :P