Question

find the exact value of cos(pi/2) - csc(-pi/2)

Answer 1

This problem draws upon your knowledge of the values of the sine and cosine function at certain "easy" angles. cos pi/2 = 0 and sin pi/2 = 1. Can you use this info to evaluate the expression you've shared?

Answer 2

pi = 180 degrees

cos(pi/2)= cos90=0
-csc(-pi/2)=-csc(-90)= --csc90=csc90=1/sin90. and sin90=1, so 1/sin90=1

0+1=1

Answer 3

Out of curiosity, how does one get to be a white Malawian?

Answer 4

i am white livin in Malawi i was born here

Answer 5

could you guys help me with one more?

Answer 6

sure

Answer 7

1/cos^2(40) - 1/cot(40)

Answer 8

\[\color{blue}{ \frac{1}{\cos^2(40)} } -\color{red}{ \frac{1}{\cot(40)} } \]

hmm, let me think

Answer 9

As soon as I saw that "cot (40)," I thought of converting it to (cos (40)) / (sin (40)).
Also, note that there are at least two common equivalents to (cos 40)^2 (by that I mean trig identiies).

Answer 10

yeah and since it has the one on top its turns to a tan function

Answer 11

You'll probably find this question to be a lot easier if you'd please modify the fractions so as to have a common denominator (LCD)

Answer 12

well,
we know that
\[\color{green} { \cot(40)= \frac{\cos(40) }{\sin(40)} } \]

so it would be
\[\huge\color{blue}{ \frac{1}{\cos^2(40)} - } \huge\color{blue}{ \frac{1}{\frac{\cos(40) }{\sin(40)} } }\]
\[\huge\color{blue}{ \frac{1}{\cos^2(40)} - } \huge\color{blue}{ \frac{\sin(40)}{\cos(40) } }\]

can you continue on your own from here?

Answer 13

no ive got that i just dont know what to do next

Answer 14

\[\huge\color{blue}{ \frac{1}{\cos^2(40)} - } \huge\color{blue}{ \frac{\sin(40) \huge\color{red}{ \times \cos(40) } }{\cos(40) \huge\color{red}{ \times \cos(40) } } }\]

Answer 15

multiply what I wrote in red, and add.

Answer 16

yeah got that so then you have 1-cos(40)sin(40)/cos^2(40) and from there i cant do anything

Answer 17

\[\huge\color{blue}{ \frac{1-\sin(40)\cos(40)}{\cos^2(40)} } \]
right?

Answer 18

yeah but the answer should either be 1 or 0

Answer 19

No that was wrong, write that one as cos(40) sec(40)

Answer 20

\[\huge\color{blue}{ \frac{ \huge\color{red}{ \cos(40)\sec(40) } -\sin(40)\cos(40)}{\cos^2(40)} } \]

see what I substituted for '1" in red?

Answer 21

\[\huge\color{blue}{ \frac{ \huge\color{blue}{ \cos(40)\sec(40) } -\sin(40)\cos(40)}{\cos^2(40)} }\]

so far so good?

Answer 22

Are you still here?

Answer 23

sorry my dad needed me how can you do that?

Answer 24

Oh, you don't have to be sorry, I definitely understand that.

Answer 25

nevermind got it

Answer 26

yeah? what's the answer then?

Answer 27

yeah so then you have that and you can cancel out a cos so itd be sec(40)-sin(40)/cos(40)

Answer 28

yeah? what's the answer then?

Answer 29

ah i have no idea

Answer 30

Ok, then I'll show you.

Answer 31

you can flip the sec to 1/cos but that only leaves you with a sin

Answer 32

\[\huge\color{blue}{ \frac{\cos(40)\sec(40)-\cos(40)\sin(40)}{\cos^2(40)} } \]
factoring out of cos(40 on the top.
\[\huge\color{blue}{ \frac{\cos(40)~~(~~\sec(40)-\sin(40)~~)}{\cos^2(40)} } \]

Answer 33

yeah i got that it leaves you with sec-sin/cos

Answer 34

\[\huge\color{blue}{ \frac{~(~~\sec(40)-\sin(40)~~)}{\cos(40)} } \]\[\huge\color{blue}{ \frac{\sec(40)}{\cos(40)} - } \huge\color{blue}{ \frac{\sin(40)}{\cos(40)} } \]

Answer 35

genius so then the sec/cos cancels out and you are left with the sin/cos

Answer 36

\[\huge\color{blue}{ \frac{\sec(40)}{\cos(40)} - } \huge\color{blue}{\tan(40) } \]

\[\huge\color{blue}{ \sec^2(40) - } \huge\color{blue}{\tan(40) } \]

Answer 37

I think this would be the answer.

Answer 38

ok yeah but its gotta be 1 or 0

Answer 39

ah i see what you did and i think your right

Answer 40

Yeah, it can only be one, if you didn't type the question correctly, I just went over my work, and made sure.

Answer 41

yeah but to be one it would have to be sec^2-tan^2

Answer 42

YES

Answer 43

but we only have a tan not tan^2

Answer 44

so, it would remain as it is. if it was 45 not 40, then it would be simplify-able.

Answer 45

yeah i know and thats why im confused ok well thanks

Answer 46

You welcome!