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
\(\tan\big(\frac{7\pi}{12}\big)=\tan\big( \frac{\pi}{3}+\frac{\pi}{4} \big)\)
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° = \(\frac{\sqrt6+\sqrt2}{4}\)