Question

Given cos(theta)=3*radical*7/8, find sin and tan

Answer 1

\[\cos \theta =3\sqrt{7}\div8\]

Answer 2

if cos = 3root7 over 8, and cos is adjacent over hypotenuse, you can assume that the adjacent is 3root 7, and the hypo. is 8. Use the pythagorean to find the 3rd side of the triangle. i.e. : 8squared= sqrt( (3root7)squared + (OTHER LEG)2 )

Answer 3

then, with that value, you have all three sides of the triangle and you can plug those numbers into the formula for sin and tan : opposite/hypotenuse and opposite/adjacent.

Answer 4

thanks a lot, do you mind helping me with a few more??

Answer 5

go ahead an post! I can help for about 10 more minutes.

Answer 6

and mark one of these as a good answer so your question won't still say "needs a good answer"

Answer 7

tan \[293\pi/6\]

Answer 8

is that a question?

Answer 9

oh ha my b never mind.

Answer 10

ok well 293/6 is many, many revolutions around the unit circle. so just get the next closest multiple of 12 (2 times 6) which is 288. 288pi/6 would imply going around the circle 24 times, so subtracting 288 from 293 just means you are forgetting about those 24 revolutions and starting, hypothetically, at 0. so 293-288 is 5, which means your new angle is 5pi/6. using your knowledge of the unit circle, find tan of that!

Answer 11

haha, its all good

Answer 12

thanks a million

Answer 13

Find \[\csc 13\pi/6\]

Answer 14

this is kinda the same concept. since 13 is more than 2 times 6, you are already going around the circle. so subtracting 12 gives you 1, so find csc (which is 1/sin or just hypotenuse/opposite) of pi/6

Answer 15

\[\sqrt{\cos ^{2}(\theta/2)}\]
How do you rewrite the expression in nonradical form without using absolute values
0<theta<pi

Answer 16

well if you have anything 2 under a radical you can just pull it out. for instance, \[\sqrt{2^2}\] is just 2. so you can just pull out the cos^2 and it should just be cos(theta/2)