Hello everybody. I was solving my set of exercises, but got stuck with these 5 exercises... Could anyone help me with it? Even if you can help with only one of them, I would be really glad.. Thank you all so much.
Hello everybody. I was solving my set of exercises, but got stuck with these 5 exercises... Could anyone help me with it? Even if you can help with only one of them, I would be really glad.. Thank you all so much.
Hello, wdma!
Here's #2 . . .
2) Find the exact value of: $\displaystyle \sin\left(-\frac{5\pi}{12}\right)$
$\displaystyle \sin\left(-\frac{5\pi}{12}\right) \;=\; -\sin\left(\frac{5\pi}{12}\right)$
. . . . . . . .$\displaystyle = \;-\sin\left(\frac{\pi}{6} + \frac{\pi}{4}\right)$ .**
. . . . . . . .$\displaystyle = \;-\bigg[\sin\frac{\pi}{6}\cos\frac{\pi}{4} + \sin\frac{\pi}{4}\cos\frac{\pi}{6}\bigg] $
. . . . . . . .$\displaystyle = \;-\bigg[\left(\frac{1}{2}\right)\left(\frac{1}{\sqrt{2}}\r ight) + \left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3 }}{2}\right)\bigg] $
. . . . . . . .$\displaystyle =\;-\bigg[\frac{1}{2\sqrt{2}} + \frac{\sqrt{3}}{2\sqrt{2}}\bigg]$
. . . . . . . .$\displaystyle = \;-\,\boxed{\frac{1+\sqrt{3}}{2\sqrt{2}}}$
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
** . $\displaystyle \frac{5\pi}{12} \;=\;\frac{2\pi}{12} + \frac{3\pi}{12} \;=\;\frac{\pi}{6} + \frac{\pi}{4}$ . . . see?
I prefer to give you hints and if you can't solve it ask for help
for the first question
$\displaystyle \cos^{-1}a=0 \Rightarrow \cos(\cos^{-1}a)=cos(0) \Rightarrow a=1 $ since $\displaystyle \cos(\cos^{-1}a)=a $ the same thing for
$\displaystyle sin^{-1}(\frac{\pi}{6})=b $
$\displaystyle \frac{\pi}{6}=\sin(b) $
now you have a=1 and you have sin(b) now to find the value of
$\displaystyle \sin(a\pi+b)=sin(\pi+b)$ since a=1
use $\displaystyle \sin(x+y)=\sin(x)\cos(y)+\sin(y)\cos(x)$
the second question
$\displaystyle sin(\frac{-5\pi}{12})=??$
first thing you should know
$\displaystyle sin(-\theta)=-sin(\theta)$
you can write the angle
$\displaystyle \frac{5\pi}{12}=\frac{3\pi}{12}+\frac{2\pi}{12}$
then use the formula $\displaystyle \sin(x+y)=\sin(x)\cos(y)+\sin(y)\cos(x)$
Hi
Here is 4)
$\displaystyle tan(\beta)=\frac{12}{5} $
$\displaystyle tan(2\beta)=tan(\beta+\beta)=\frac{tan(\beta)+tan( \beta)}{1-tan(\beta)\cdot tan(\beta)} $
Now you know that $\displaystyle tan(\beta)=\frac{12}{5} $ , so this gives $\displaystyle \frac{\frac{24}{5}}{1-\left(\frac{12}{5}\right)^{2}} = -\frac{120}{119}$
question three
$\displaystyle \frac{sin(2(x-y))}{2\cos(x)\cos(y)\cos(x-y)}$
use these formulas
$\displaystyle \sin(x-y)=\sin(x)\cos(y)-\sin(y)\cos(x)---------(1)$
$\displaystyle \sin2x=2sin(x)cos(x)------------------(2)$
$\displaystyle \cos(x-y)=\cos(x)\cos(y)+\sin(x)sin(y)--------(3)$
$\displaystyle \frac{2\sin(x-y)\cos(x-y)}{2\cos(x)\cos(y)\cos(x-y)}$
$\displaystyle \frac{2\sin(x-y)}{2\cos(x)cos(y)}$
$\displaystyle \frac{sin(x)cos(y)-sin(y)cos(x)}{cos(x)cos(y)}$
Hello again, wdma!
Here's my version of #3 . . .
In the numerator, use: .$\displaystyle \sin2\theta \:=\:2\sin\theta\cos\theta$3) Simplify as much as possible: .$\displaystyle \frac{\sin[2(x-y)]}{2\cos(x)\cos(y)\cos(x-y)} $
We have: .$\displaystyle \frac{{\color{red}\rlap{/}}2\sin(x-y){\color{red}\rlap{/////////}}\cos(x-y)}{{\color{red}\rlap{/}}2\cos(x)\cos(y){\color{red}\rlap{/////////}}\cos(x-y)} \;=\;\frac{\sin(x-y)}{\cos(x)\cos(y)}$
. . $\displaystyle = \;\frac{\sin(x)\cos(y) - \cos(x)\sin(y)}{\cos(x)\cos(y)} \;=\;\frac{\sin(x){\color{red}\rlap{/////}}\cos(y)}{\cos(x){\color{red}\rlap{/////}}\cos(y)} - \frac{{\color{red}\rlap{/////}}\cos(x)\sin(y)}{{\color{red}\rlap{/////}}\cos(x)\cos(y)} $
. . $\displaystyle =\;\frac{\sin(x)}{\cos(x)} - \frac{\sin(y)}{\cos(y)} \;=\;\tan(x) - \tan(y) $
5)
$\displaystyle \begin{aligned}
2\sin^2 x &= 1 \\
\sin^2 x &= \frac{1}{2} \\
\sin x &= \pm \frac{\sqrt{2}}{2}
\end{aligned}$
You should have recognized the ratios $\displaystyle \pm\frac{\sqrt{2}}{2}$ (the alternate form is $\displaystyle \frac{1}{\sqrt{2}}$).
Ask yourself, the sine of what angles is $\displaystyle \frac{\sqrt{2}}{2}$? (There are two answers). Then ask yourself, the sine of what angles is $\displaystyle -\frac{\sqrt{2}}{2}$? (There are two more answers). If you have to, draw some triangles in the cartesian plane.
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