# Math Help - Some problems...

1. ## Some problems...

I've been having some trouble with some trig problems. It would be great if someone could help me to do them. I'm not very sure how to put in symbols such as pie, etc. so I apologize in advance for having to write it out.

*Using a sum/difference formula, find the exact value of tan pie/12.

*Simplify/evaluate to find the exact value of 1-2sinsquared15 degrees.

*Verify (sinx-cosx)squared= 1-sin2x.

*Find solutions for 2cos squared x+sin squared x=2cosx if X is between 0 degrees and 360 degrees.

* Find solutions for 2 sin squared x+5sinx=-3 if x is between 0 degrees and 360 degrees.

Thank you!

I've been having some trouble with some trig problems. It would be great if someone could help me to do them. I'm not very sure how to put in symbols such as pie, etc. so I apologize in advance for having to write it out.

*Using a sum/difference formula, find the exact value of tan pie/12.
Note that,
$\tan^2(x/2)=\frac{1-\cos x}{1+\cos x}$
Therefore if $x=\pi/6$ then you have,
$\tan^2(\pi/12)=\frac{1-\cos (\pi/6)}{1+\cos (\pi/6)}$ but $\cos (\pi/6)=\sqrt{3}/2$
Thus,
$\frac{1-\frac{\sqrt{3}}{2}}{1+\frac{\sqrt{3}}{2}}$
Simplifies to,
$\frac{2-\sqrt{3}}{2+\sqrt{3}}$
Rationalize,
${7-4\sqrt{3}}$
Therefore,
$\tan=\sqrt{7-4\sqrt{3}}=2-\sqrt{3}$

*Simplify/evaluate to find the exact value of 1-2sinsquared15 degrees.
Because you know that,
$\cos 2x=1-2\sin^2x$
Overhere $x=15^o$ thus, $\cos 2x=\cos 30^o=\frac{\sqrt{3}}{2}$

*Verify (sinx-cosx)squared= 1-sin2x.
Open parantheses,
$(\sin x-\cos x)(\sin x-\cos x)=\sin^2x-2\sin x\cos x+\cos 2x$
Use the fact that, $\sin^2x+\cos^2x=1$
and that, $2\sin x\cos x=\sin x$ thus we have,
$1-\sin 2x$ and proof is complete.

Using a sum/difference formula, find the exact value of $\tan \frac{\pi}{12}$
The formula is: $\tan(A \pm B)\;=\;\frac{\tan A \pm \tan B}{1 \mp \tan A\cdot\tan B}$

We note that: $\frac{\pi}{12}\:=\:\frac{\pi}{3} - \frac{\pi}{4}$

Hence: $\tan \frac{\pi}{12}\;=\;\tan\left(\frac{\pi}{3} - \frac{\pi}{4}\right) \;=\;\frac{\tan\frac{\pi}{3} - \tan\frac{\pi}{4}}{1 + \tan\frac{\pi}{3}\cdot\tan\frac{\pi}{4}} \;= \;\frac{\sqrt{3} - 1}{1 + \sqrt{3}}$

Rationalize: $\frac{\sqrt{3} - 1}{\sqrt{3} + 1}\cdot\frac{\sqrt{3} - 1}{\sqrt{3} - 1} \;= \;\frac{3 - 2\sqrt{3} + 1}{3 - 1}\;=\;\frac{4 - 2\sqrt{3}}{2}\;=\;2 - \sqrt{3}

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