# Thread: Having trouble with these 5 trig exercises

1. ## Having trouble with these 5 trig exercises

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.

2. Hello, wdma!

Here's #2 . . .

2) Find the exact value of: $\sin\left(-\frac{5\pi}{12}\right)$

$\sin\left(-\frac{5\pi}{12}\right) \;=\; -\sin\left(\frac{5\pi}{12}\right)$

. . . . . . . . $= \;-\sin\left(\frac{\pi}{6} + \frac{\pi}{4}\right)$ .**

. . . . . . . . $= \;-\bigg[\sin\frac{\pi}{6}\cos\frac{\pi}{4} + \sin\frac{\pi}{4}\cos\frac{\pi}{6}\bigg]$

. . . . . . . . $= \;-\bigg[\left(\frac{1}{2}\right)\left(\frac{1}{\sqrt{2}}\r ight) + \left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3 }}{2}\right)\bigg]$

. . . . . . . . $=\;-\bigg[\frac{1}{2\sqrt{2}} + \frac{\sqrt{3}}{2\sqrt{2}}\bigg]$

. . . . . . . . $= \;-\,\boxed{\frac{1+\sqrt{3}}{2\sqrt{2}}}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

** . $\frac{5\pi}{12} \;=\;\frac{2\pi}{12} + \frac{3\pi}{12} \;=\;\frac{\pi}{6} + \frac{\pi}{4}$ . . . see?

3. Originally Posted by wdma
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.

I prefer to give you hints and if you can't solve it ask for help

for the first question

$\cos^{-1}a=0 \Rightarrow \cos(\cos^{-1}a)=cos(0) \Rightarrow a=1$ since $\cos(\cos^{-1}a)=a$ the same thing for

$sin^{-1}(\frac{\pi}{6})=b$

$\frac{\pi}{6}=\sin(b)$
now you have a=1 and you have sin(b) now to find the value of

$\sin(a\pi+b)=sin(\pi+b)$ since a=1

use $\sin(x+y)=\sin(x)\cos(y)+\sin(y)\cos(x)$

the second question

$sin(\frac{-5\pi}{12})=??$

first thing you should know

$sin(-\theta)=-sin(\theta)$

you can write the angle

$\frac{5\pi}{12}=\frac{3\pi}{12}+\frac{2\pi}{12}$

then use the formula $\sin(x+y)=\sin(x)\cos(y)+\sin(y)\cos(x)$

4. Originally Posted by wdma
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.

1. $\cos{0}=1=a$ and $\sin^{-1}(\frac{\pi}{6})=.5507=b$. You do the rest.

5. $\sin{x}=\pm\frac{\sqrt{2}}{2}$. You do the rest.

5. Hi

Here is 4)

$tan(\beta)=\frac{12}{5}$

$tan(2\beta)=tan(\beta+\beta)=\frac{tan(\beta)+tan( \beta)}{1-tan(\beta)\cdot tan(\beta)}$

Now you know that $tan(\beta)=\frac{12}{5}$ , so this gives $\frac{\frac{24}{5}}{1-\left(\frac{12}{5}\right)^{2}} = -\frac{120}{119}$

6. Originally Posted by VonNemo19
1. $\cos{0}=1=a$ and $\sin^{-1}(\frac{\pi}{6})=.5507=b$. You do the rest.
Just a friendly reminder: in this case, leave $\sin^{-1}\left(\frac{\pi}{6}\right)$ as an exact value, because something will work out nicely in the simplification...

7. question three

$\frac{sin(2(x-y))}{2\cos(x)\cos(y)\cos(x-y)}$

use these formulas

$\sin(x-y)=\sin(x)\cos(y)-\sin(y)\cos(x)---------(1)$

$\sin2x=2sin(x)cos(x)------------------(2)$

$\cos(x-y)=\cos(x)\cos(y)+\sin(x)sin(y)--------(3)$

$\frac{2\sin(x-y)\cos(x-y)}{2\cos(x)\cos(y)\cos(x-y)}$

$\frac{2\sin(x-y)}{2\cos(x)cos(y)}$

$\frac{sin(x)cos(y)-sin(y)cos(x)}{cos(x)cos(y)}$

8. Hello again, wdma!

Here's my version of #3 . . .

3) Simplify as much as possible: . $\frac{\sin[2(x-y)]}{2\cos(x)\cos(y)\cos(x-y)}$
In the numerator, use: . $\sin2\theta \:=\:2\sin\theta\cos\theta$

We have: . $\frac{{\color{red}\rlap{/}}2\sin(x-y){\color{red}\rlap{/////////}}\cos(x-y)}{{\color{red}\rlap{/}}2\cos(x)\cos(y){\color{red}\rlap{/////////}}\cos(x-y)} \;=\;\frac{\sin(x-y)}{\cos(x)\cos(y)}$

. . $= \;\frac{\sin(x)\cos(y) - \cos(x)\sin(y)}{\cos(x)\cos(y)} \;=\;\frac{\sin(x){\color{red}\rlap{/////}}\cos(y)}{\cos(x){\color{red}\rlap{/////}}\cos(y)} - \frac{{\color{red}\rlap{/////}}\cos(x)\sin(y)}{{\color{red}\rlap{/////}}\cos(x)\cos(y)}$

. . $=\;\frac{\sin(x)}{\cos(x)} - \frac{\sin(y)}{\cos(y)} \;=\;\tan(x) - \tan(y)$

9. Originally Posted by Amer
I prefer to give you hints and if you can't solve it ask for help

for the first question

$\cos^{-1}a=0 \Rightarrow \cos(\cos^{-1}a)=cos(0) \Rightarrow a=1$ since $\cos(\cos^{-1}a)=a$ the same thing for

$sin^{-1}(\frac{\pi}{6})=b$

$\frac{\pi}{6}=\sin(b)$
now you have a=1 and you have sin(b) now to find the value of

$\sin(a\pi+b)=sin(\pi+b)$ since a=1

use $\sin(x+y)=\sin(x)\cos(y)+\sin(y)\cos(x)$

the second question

$sin(\frac{-5\pi}{12})=??$

first thing you should know

$sin(-\theta)=-sin(\theta)$

you can write the angle

$\frac{5\pi}{12}=\frac{3\pi}{12}+\frac{2\pi}{12}$

then use the formula $\sin(x+y)=\sin(x)\cos(y)+\sin(y)\cos(x)$

I'm still struggling with #1 and #5. Could anyone help me further with them?
Also, I would like to thank all of you so much, for being so kind and helpful!

10. Originally Posted by wdma
I'm still struggling with #1 and #5. Could anyone help me further with them?
Also, I would like to thank all of you so much, for being so kind and helpful!
5)
\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 $\pm\frac{\sqrt{2}}{2}$ (the alternate form is $\frac{1}{\sqrt{2}}$).

Ask yourself, the sine of what angles is $\frac{\sqrt{2}}{2}$? (There are two answers). Then ask yourself, the sine of what angles is $-\frac{\sqrt{2}}{2}$? (There are two more answers). If you have to, draw some triangles in the cartesian plane.

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