Question

how to solve this . I know how to get the inverse cos , but not the pi/2 . cos ( pi/2 + cos^-1 1/3 )

Answer 1

Start with the sum identity for cosine.

Answer 2

this one ey ? cos ( a+/-b) = cos A cos B +/- sin A sin B ) ?

Answer 3

yes

Answer 4

then , how abt the A and B ? from where can I take them ?

Answer 5

expand into\[\cos (\pi/2)\cos (\cos^{-1} (\frac{1}{3}))-\sin(\pi/2)\sin(\cos^{-1} (\frac{1}{3}))\]

Answer 6

The left side zeros out from the cosine, the left side leaves you with \[-\sin (\cos^{-1} (\frac{1}{3}))\]work that out with triangles

Answer 7

er... the right side I mean

Answer 8

i got\[\pi/6 - ( \sqrt{8}.\pi ) /6\]

Answer 9

not I

Answer 10

erm

Answer 11

cos (pi/2)=0\[-\sin(\cos^{-1} (\frac{1}{3}))=-2\sqrt{2}\]

Answer 12

hold it

Answer 13

ok
i got that triangle

Answer 14

Then the sine is opposite over hypotenuse... oops... root 8 over 3 which simplifies to 2root 2 over 3.... sorry.

Answer 15

okay ,

Answer 16

so the answer is \[-\sqrt{8}/6\] right ?

Answer 17

why over 6?

Answer 18

erm , i don't know . i'm not sure

Answer 19

because i multiply the sin \[\pi/2\times \sqrt{8}/ 3\]. like this

Answer 20

so when it is multiplied it become \[( \pi . \sqrt{8})/6\]

Answer 21

\[-\sin(\pi/2)\sin(\cos^{-1} (\frac{1}{3}))\]\[-1*\sin (\cos^{-1} (\frac{1}{3}))\]\[-\sqrt{8}/3\]\[- \frac{2\sqrt{2}}{3}\]

Answer 22

then i just cut of the \[\pi/6\]
just like that

Answer 23

\[\sin (\pi/2)=1\]

Answer 24

There are no pi's in the answer because you take the sine or cosine of every angle leaving only a ratio.

Answer 25

okay thanks

Answer 26

can you just answer this one last question ?

Answer 27

We got scattered along the way, but the steps are...
put your formula into the sum of cosines identity
evaluate each sin and cos
simplify
sure, I can answer another

Answer 28

given \[\tan ( \sec ^-1 5/3 )\] , find (i) \[\tan ( x/2 )\] and (ii)\[\tan (x )\]

Answer 29

i got the triangle already

Answer 30

part ii is easiest
from the triangle, find opposite over adjacent

Answer 31

The other require the identity for tan(x/2)

Answer 32

\[\tan(\frac{x}{2})=\frac{1-\cos x}{\sin x}\]

Answer 33

that identities is from what chapter ?

Answer 34

ew... I get my identity off of a cheat sheet from the front of my trig book... not likely to match your chapter even if I knew.

Answer 35

I actually have 3 identities for tan(x/2)

Answer 36

maybe i don't learn it yet . can you list it to me ? or maybe i just forgot

Answer 37

\[\tan(\frac{x}{2})=\frac{1-\cos x}{\sin x}=\frac{\sin x}{1+\cos x}=\pm \sqrt{\frac{1-\cos x}{1+\cos x}}\]

Answer 38

oh , okay thanks , then after we use the identities , what should we do ?

Answer 39

start with your triangle

Answer 40

find tan x=opp/adj\[=6\sqrt{6}/3=2\sqrt{6}\]

Answer 41

Then use your identity, finding the sin and cos from the triangle and inserting it into the right places in the identity.

Answer 42

\[\tan(x/2)=\frac{1-(\frac{3}{15})}{(\frac{6\sqrt{6}}{15})}\]

Answer 43

so we just leave the answer like that ?

Answer 44

No, you should simplify.

Answer 45

okay thank you so much . by the way how often you online here ? , so that I can ask you question :D

Answer 46

Not predictable, this is my first time in many months.
Hartnn is my favorite go to guy. He is here a lot. He makes me look stupid. Look for him if I'm not here.

Answer 47

okay , but if i just can't find him , where can I find you ?

Answer 48

hmm.. not sure what to tell you. If I'm not on, I'm not on.

Answer 49

okay can we make it simple , which account do you usually use ?

Answer 50

I'm EulerGroupie. I log on through the MIT site.

Answer 51

don't worry i'm not a stalker , i'm just a very math lover that didn't know much about maths

Answer 52

erm do you have skype ?
or y!m?

Answer 53

nope... my e-mail is Eulergroupie@comcast.net

Answer 54

I'm a community college math tutor that sometimes stays up late messing with this site.

Answer 55

okay , got it . oh well , i think that is all my question , just in case i need your help , i'll leave a message here

Answer 56

cool... good luck

Answer 57

thanks again , and sorry for my bad english usage

Answer 58

erm , are you still there ? i'm kinda stuck again

Answer 59

yeah

Answer 60

if the question is like this , how should i do it ? \[\csc ( 2 \sec ^{-1} 3x + \pi/2 )\]

Answer 61

reciprocal identity first
\[=\frac{1}{\sin (2\sec ^{-1}(3x)+\pi/2)}\]then use the sum of sines identity on this denominator
make the triangle from the inverse secant and use it to build the large fraction
evaluate each sine and cosine using the triangle and reduce

Answer 62

okay thanks

Answer 63

You're welcome.

Answer 64

I don't know how to insert the value for sum of sines , the sines have 2 in front of it

Answer 65

Oh my, that IS nasty. I wouldn't make the triangle until you resolve the reciprocal and then the half angle identity.

Answer 66

== , okay , I lost my way right now . how to solve the reciprocal identities ?

Answer 67

This one seems a bit extreme... I need a sec to find a way out.

Answer 68

sec=1/cos

Answer 69

take your time

Answer 70

set and angle variable equal to 2sec^-1(3x)\[\alpha=2\sec ^{-1}(3x)\]I must be missing something, this is getting terrible.

Answer 71

if we use my triangle it can be 2 teta right ?or not..