trigonometry help

[di95]

please can someone help me simplify this

 

[Dr.Peterson]

please can someone help me simplify this View attachment 27489

I'd expect you to be solving, not just simplifying; but it can benefit from the latter.

First, let's get rid of those decimals:
\(\sqrt{\sin^2\left(\frac{x}{2}\right)+2\sin\left(\frac{x}{2}\right)+1}-\sqrt{\left(4\sin\left(\frac{1}{2}\right)-6\right)^2}=-\frac{5}{2}\)

Next, observe that \(\sin\left(\frac{x}{2}\right)\) appears several times; make the substitution \(u=\sin\left(\frac{x}{2}\right)\). (I find myself wondering if there's supposed to be an x in the second radical.)

Then look for perfect squares.

Then follow our guidelines:

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That includes telling us the whole problem, including any instructions given with it, as well as showing some work so we can tell where you need help.

 

[di95]

hello, thank you for your answer. yes you are right, I need to solve this but because I wanted to try to solve alone first I asked help to simplify. I tried two method to simplify then try to solve it but I think I got it wrong
thank you for you help

 

[di95]

and second try to solve it. I'm not a student anymore so I forgot most it. Its for a work related improvement stage. thank you

 

[di95]

I'd expect you to be solving, not just simplifying; but it can benefit from the latter.

First, let's get rid of those decimals:
\(\sqrt{\sin^2\left(\frac{x}{2}\right)+2\sin\left(\frac{x}{2}\right)+1}-\sqrt{\left(4\sin\left(\frac{1}{2}\right)-6\right)^2}=-\frac{5}{2}\)

Next, observe that \(\sin\left(\frac{x}{2}\right)\) appears several times; make the substitution \(u=\sin\left(\frac{x}{2}\right)\). (I find myself wondering if there's supposed to be an x in the second radical.)

Then look for perfect squares.

Then follow our guidelines:

Posting Guidelines (Summary)

Welcome to our tutoring boards! :) This page summarizes the main points from our posting guidelines. As our name implies, we provide math help (primarily to students with homework). We do not generally post immediate answers or step-by-step solutions. We don't do your homework. We prefer to...

www.freemathhelp.com

That includes telling us the whole problem, including any instructions given with it, as well as showing some work so we can tell where you need help.

 

[lex]

Can you first check whether Dr Peterson's suggestion is the case, that the second root should be:
[MATH]\sqrt{(4\sin(0.5)\boldsymbol{x}-6)^2}[/MATH]

 

[di95]

Can you first check whether Dr Peterson's suggestion is the case, that the second root should be:
[MATH]\sqrt{(4\sin(0.5)\boldsymbol{x}-6)^2}[/MATH]

I checked twice and no it as i wrote √(4sin(0.5)−6)^2

 

[lex]

OK. (I still suspect they meant to write the other)!
[MATH][/MATH]However, in that case the second square root is just a number, which you can work out:
[MATH]\sqrt{(4\sin(0.5)-6)^2}=4.0823 \text{ (4d.p.)}[/MATH](This will avoid the kind of arithmetic slip you made with signs, up above).

Now you are solving:
[MATH]\sqrt{\left( \sin \tfrac{x}{2} +1 \right)^2}-4.0823=-2.5[/MATH]
i.e. [MATH]\sin \left(\tfrac{x}{2} \right) +1 -4.0823=-2.5 \hspace2ex[/MATH](since [MATH]\sin \left(\tfrac{x}{2} +1 \right)≥0[/MATH])

Hopefully you can solve that now.

 

[di95]

OK. (I still suspect they meant to write the other)!
[MATH][/MATH]However, in that case the second square root is just a number, which you can work out:
[MATH]\sqrt{(4\sin(0.5)-6)^2}=4.0823 \text{ (4d.p.)}[/MATH](This will avoid the kind of arithmetic slip you made with signs, up above).

Now you are solving:
[MATH]\sqrt{\left( \sin \tfrac{x}{2} +1 \right)^2}-4.0823=-2.5[/MATH]
i.e. [MATH]\sin \left(\tfrac{x}{2} \right) +1 -4.0823=-2.5 \hspace2ex[/MATH](since [MATH]\sin \left(\tfrac{x}{2} +1 \right)≥0[/MATH])

Hopefully you can solve that now.

thank you for your help,

is that right?

 

[lex]

Exactement!
All you need to do now is solve to get the values of x.

 

[lex]

@di95
(You still made two errors in the third line of the first page of your recent post: [MATH]\sqrt{\left(4\sin \left(\tfrac{1}{2}\right)-6\right)^2}=\left(6-4\sin \left(\tfrac{1}{2}\right)\right) \hspace1ex[/MATH] and then you should have used a bracket: .... [MATH]- \left(6-4\sin \left(\tfrac{1}{2}\right)\right)[/MATH]Thankfully that didn't stop you changing again to the correct values in the next line.)