Question

need help with # 57 and 61
i'll post the pic
please and thank y

Answer 1

you*

Answer 2

i dont really understand your question
what did you get for #55 for example,

Answer 3

ah ok,

Answer 4

SO we want the tangent of t , in terms of sines only

\[\tan(t)=\frac{\sin (t)}{\cos(t)}\]
to get rid of the cosine ,
solve the trig identity

\[\sin^2+\cos^2=1\]for cosine\[\quad\vdots\\\cos=\]

substitute this result into
\[\tan(t)=\frac{\sin (t)}{\cos(t)}\]

Answer 5

the answer from the book is different :O

Answer 6

sin t/ square root of 1-sin^2 t

Answer 7

wait , i think you have take into consideration that we are in the fourth quadrant

Answer 8

okay

Answer 9

use this \[\begin{array}{r|l}S&A\\\hline T&C\end{array}\]

sin,cos and tan
are All positive in quadrant I
only Sin is positive in quadrant II
only tangent is positive in quadrant III
only cosine is positive in quadrant IV

Answer 10

its still confusing but thanks for trying to explain it to me :)

Answer 11

is sin t/ √( 1-sin^2 t)

the answer you got or the answer in the book?

Answer 12

yes

Answer 13

both?

Answer 14

that's the answer for # 57

Answer 15

yes, do you get it?

Answer 16

nope

Answer 17

\[\sin^2+\cos^2=1\\
\cos^2=1-\sin^2\\\cos=\sqrt{1-\sin^2}\]

Answer 18

i got that part

Answer 19

substitute this result into \[\tan(t)=\frac{\sin (t)}{\cos(t)}\]

Answer 20

thats the part i dont get

Answer 21

so we have just showed that \[\cos(t)=\sqrt{1-\sin^2(t)}\]


so
\[\tan(t)=\frac{\sin (t)}{\cos(t)}=\frac{\sin(t)}{?}\]

Answer 22

now i get it!!

Answer 23

good good

Answer 24

Do you think you can do 61?

Answer 25

thanks