Ask your own question, for FREE!
Mathematics 14 Online
OpenStudy (anonymous):

Find an exact value. cosine of negative seven pi divided by twelve

OpenStudy (anonymous):

\[\cos(-7 \pi/12)=\cos(7 \pi/12)=\cos(\frac{ 6 \pi+\pi }{ 12 })=\cos(\frac{ \pi }{ 2 }+\frac{ \pi }{ 12 })=-\sin(\frac{ \pi }{ 12 })\]and we know that:\[\sin(\alpha)=\sqrt{\frac{ 1-\cos(2 \alpha) }{ 2 }}\rightarrow -\sin(\pi/12)=-\sqrt{\frac{ 1-\cos(\pi/6) }{ 2 }}\]and \[\cos(\pi/6)=\cos(30º)=\sqrt{3}/2\]then\[\cos(-7\pi/12)=-\sqrt{\frac{ 1-\frac{ \sqrt{3} }{ 2 } }{ 2 }}=-\sqrt{\frac{ 2-\sqrt{3} }{ 4 }}=-\frac{ 1 }{ 2 }\sqrt{2-\sqrt{3}}\]

OpenStudy (anonymous):

that just looks like a bunch of weird stuff is there any way I can turn the eqautions on?

OpenStudy (anonymous):

if you find it weird i can withdraw it but it is the way to go. If you find something easier, let me know. This is about calculating the cosine of a weird angle using known values of pupular angle like 30º

OpenStudy (anonymous):

no I mean like the eqautoion system isn't working I just see a bhunch of symbols one day I just couldn't see eqautions anymore

OpenStudy (anonymous):

\[\frac{ 7 }{ 5}\] like when I do that it looks funny

OpenStudy (anonymous):

with my calculator: cos(-7pi/12)=-0.258819045

OpenStudy (anonymous):

yea but my answers look like this square root of six plus square root of two quantity square root of six minus square root of two divided by four square root of two minus square root of six quantity square root of two minus square root of six divided by four

OpenStudy (anonymous):

if none of the choices is negative, none is valid

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Latest Questions
Arriyanalol: i need help
2 days ago 7 Replies 1 Medal
Arriyanalol: CONGRATS to me
2 days ago 8 Replies 2 Medals
Legomaster346: What colleges are good that are near Arab Alabama?
1 week ago 5 Replies 0 Medals
Arriyanalol: i need help
1 week ago 6 Replies 2 Medals
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!