Question

PLEASE HELP!!!!!!!!
give the exact value of
sin(cos^−1(e^(ln(3√/2)

Answer 1

eq editor please

Answer 2

\[\sin (\cos^{-1} (e ^{\ln(\sqrt{3}/2)})\]

Answer 3

\(\bf \huge sin \left(cos^{-1} \left(e^{ln(\frac{\sqrt{3}}{2})}\right)\right)\)

Answer 4

hint:\[e ^{\ln \left( \frac{ \sqrt{3} }{ 2 } \right)}=\frac{ \sqrt{3} }{ 2 }\]

Answer 5

ahh

Answer 6

@pgpilot326 thanks i got that part i just dont know how to do the sin cos inverse thing....

Answer 7

what angle gives yo a cos of \[\frac{ \sqrt{3} }{ 2 }?\]

Answer 8

@pgpilot326 pi/6 i think

Answer 9

Remember that the inverse trig functions will give you angles and take Lengths for input. Trig functions take angles for input and give you lengths! You can use this to find the angle from the unit circle as this is one that exists on it. :D

Did I give you a link to my trig sheet last time ses11?

Answer 10

@agentc0re yes thank you! it is very helpful!! so my final answer will be an angle right?

Answer 11

yep for 0<angle<pi. what are the restrictions on the range of cos ^-1?

Answer 12

you're being asked for the sin(), thus the sine value

Answer 13

so find sin of pi/6 and your done.

Answer 14

oh ok thank you all!!!

Answer 15

@pgpilot326 wait sorry... one more question how does the cosine change to the inverse? (does it change the answer pi/6?)

Answer 16

pi/6 is not the answer to the whole question, just this part:
\[\cos^{-1} \left( \frac{ \sqrt{3} }{ 2 } \right)=\frac{ \pi }{ 6 }\]
you still need to find
\[\sin \left( \frac{ \pi }{ 6 } \right)\]
that's the answer

Answer 17

ok thanks i was just wondering if the restrictions changed the pi/6 for the first half of the question but i figured it out and the final answer should be 1/2 I think

Answer 18

i'm pretty sure that
\[0 \le \cos^{-1} x \le \]
that covers all of the values of cos without repeating.

Answer 19

what happened to the pi?