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what is tan7π/12 using the tam sum or difference formulas? 

What is sin105 using the sin sum or difference formulas?

 

please help I don't understand trigonometry 

Guest May 15, 2018
 #1
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\(\tan\big(\frac{7\pi}{12}\big)=\tan\big( \frac{\pi}{3}+\frac{\pi}{4} \big)\)

 

And the sum of two angles formula for tan is:

 

\(\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}\)         so......

 

\(\tan\big( \frac{\pi}{3}+\frac{\pi}{4} \big)=\frac{\tan\frac{\pi}{3}+\tan\frac{\pi}{4}}{1-\tan\frac{\pi}{3}\tan\frac{\pi}{4}}\)

                                                        And we know   \(\tan\frac\pi3=\sqrt3\)   and   \(\tan\frac\pi4=1\)    so...

\(\tan\big( \frac{\pi}{3}+\frac{\pi}{4} \big)=\frac{ \sqrt3 +1}{1-(\sqrt3)(1)}\\~\\ \tan\big( \frac{\pi}{3}+\frac{\pi}{4} \big)=\frac{ \sqrt3 +1}{1-\sqrt3}\)

                                                        Multiply numerator and denominator by  \((1+\sqrt3)\) .

\( \tan\big( \frac{\pi}{3}+\frac{\pi}{4} \big)=\frac{ (\sqrt3 +1)(1+\sqrt3)}{(1-\sqrt3)(1+\sqrt3)}\\~\\ \tan\big( \frac{\pi}{3}+\frac{\pi}{4} \big)=\frac{ 2\sqrt3+4}{-2}\\~\\ \tan\big( \frac{\pi}{3}+\frac{\pi}{4} \big)=-\sqrt3-2\\~\\ \tan\frac{7\pi}{12}=-\sqrt3-2\\~\\ \text{________________________}\)

 

sin( 105° )  =  sin( 45° + 60° )

 

And the sum of two angles formula for sin is:

 

sin(α + β)  =  sin α cos β + cos α sin β         so....

 

sin(45° + 60°)  =  sin 45° cos 60° + cos 45° sin 60°    

 

sin(45° + 60°)  =  \(\big(\frac{\sqrt2}{2}\big)\big(\frac{\sqrt3}{2}\big)+\big(\frac{\sqrt2}{2}\big)\big(\frac12\big)\)

 

sin(45° + 60°)  =  \(\frac{\sqrt6}{4}+\frac{\sqrt2}{4}\)

 

sin(45° + 60°)  =  \(\frac{\sqrt6+\sqrt2}{4}\)

 

sin 105°   =   \(\frac{\sqrt6+\sqrt2}{4}\)

hectictar  May 15, 2018

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