Double-angle and half-angle identities

J

Joanna

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In each case find sinx, cosx, tanx, cscx, secx, cotx.
47) sin(2x)= 5/13 and 0<2x<45* (degrees)

49) cos(x/2)= -1/4 and 90*<(x/2)<135*

52) sin(x/2)= 1/5 and (x/2) is in quadrant II


Verify that each equation is an identity:

61) 2sin^2(u/2)=sin^2u/(1+cosu)

64) sec^2(x/2)=(2secx + 2)/ (secx + 2 + cosx)
 
I'm just asking for help, if you can't help me then why are you wasting your time posting something stupid, don't you have more important things to do?
 
Joanna said:
I'm just asking for help
Click to expand...

Could you be more specific, please?

How may we help you to figure these out? In other words, what kind of help do you need, where are you stuck? What have you thought about or tried thus far?
 
Hello, Joanna!

You misinterpreted mmm4444bot's sarcasm.

You posted FIVE problems with no additional information, such as:
. . Do you know the double- and half-angle identities?
. . What have you done so far?
. . Exactly where and why are you stuck?

So what you did was post five problems
. . without even a "Please help me".

And you gave the unspoken command, "Solve them for me ... now!"

Can you imagine how that makes my day?


Oh, is THAT all she wants?
(Maybe she'd like some white truffles with saffron sauce?)
 

Sorry I'm new to this so I wasn't sure what to do or what to expect, but thank you for willing to help me, I really appreciate
The thing I need most help is to help me get the cosx or sinx from sin2x=5/13 or how to get sin or cos from cos(x/2)= -1/4 I know you need to use one of the double angle and half angle identity but I really have no idea how to start the problem, I'm pretty sure I would know what to do when I get the answer for sinx or cosx but Im really blanking on how to use the formulas to get there-> my book does not have any examples of those particular problems and I'm lost ( I know how to go from cos2x and find sinx, cosx, tanx...)
Double angle identities
SinA=2sinAcosA
CosA= cos^2A-sin^2A
= 1-2sin^2A
= 2cos^2A-1
Tan2A= (2tanA)/1-tan^2A

I'm not sure how to write the radical signal in the half-angle identities, so hopefully you guys can look them up :)
 
Hello, Joanna!

I'll get you started . . .

\(\displaystyle \text{Find }\sin x,\:\cos x,\:\tan x,\:\csc x,\:\sec x,\:\cot x\)

\(\displaystyle 47)\;\;\sin(2x) = \frac{5}{13}\:\text{ and }\:0\,<\,2x\,<\,45^o\)
Click to expand...

First of all, the angle is in the first quadrant, where all trig functions are positive.
. . So we don't have to worry about the plus-or-minus.

\(\displaystyle \text{We are given: }\:\sin(2x) \,=\,\frac{5}{13} \,=\,\dfrac{opp}{hyp}\)

\(\displaystyle 2x\) is an angle in a right triangle with: \(\displaystyle opp = 5,\:hyp = 13\)
. . Pythagorus tells us that: \(\displaystyle adj = 12\)
\(\displaystyle \text{Hence: }\:\cos(2x) \,=\,\dfrac{adj}{hyp} \,=\,\frac{12}{13}\)


\(\displaystyle \text{Half-angle Identity: } \:\sin x \:=\:\pm\sqrt{\dfrac{1-\cos(2x)}{2}}\)

\(\displaystyle \text{So we have: }\:\sin x \;=\;\sqrt{\dfrac{1-\frac{12}{13}}{2}} \;=\; \sqrt{\dfrac{\frac{1}{13}}{2}} \;=\;\sqrt{\dfrac{1}{26}}\)

. . . . . . . . . . \(\displaystyle \sin x \:=\:\dfrac{1}{\sqrt{26}} \:=\:\dfrac{opp}{hyp}\)


\(\displaystyle x\) is an angle in a right triangle with: \(\displaystyle opp = 1,\:hyp = \sqrt{26}\)
. . Pythagorus tells us that: .\(\displaystyle adj = 5\)
\(\displaystyle \text{We have the three sides of the right triangle: }\:\begin{Bmatrix}opp &=& 1 \\ adj &=& 5 \\ hyp &=& \sqrt{26}\end{Bmatrix}\)


\(\displaystyle \text{Now we can write the six trig functions:}\)

. . . . . \(\displaystyle \begin{bmatrix} \sin x &=& \dfrac{1}{\sqrt{26}} && \csc x &=& \sqrt{26} \\ \\ \cos x &=& \dfrac{5}{\sqrt{26}} && \sec x &=& \dfrac{\sqrt{26}}{5} \\ \\ \tan x &=& \dfrac{1}{5} && \cot x &=& 5 \end{bmatrix}\)
 
Thank you so much that makes a lot of sense now, but for the problem with cos(x/2) would I be able to use the half angle identity but substitute the cos(2x) with cos(x/2) is that ok? Or do i need to use another identity ?
 
Joanna said:
Thank you so much that makes a lot of sense now, but for the problem with cos(x/2) would I be able to use the half angle identity but substitute the cos(2x) with cos(x/2) is that ok? Or do i need to use another identity ?
Click to expand...

No you do not need another identity.

But show us your work - I am not excatly sure how you plan to substitute.
 
So for cos(x/2)= -1/4. 90*<x/2<135 QII

So I used the formula that you showed me previously but instead of cos(2x) I put in the cos(x/2):

Sinx= + or - square root of (1-cos(x/2)/2)
Sinx= + or - square root of (1- (-1/4)/2) = + or - square root of (5/8)
So then I can multiply top and bottom by square root of 8 which would equal square root of 40/8
But does that equal 2squareroot10/ 8 which equals square root 10/4
And I'm not sure it it's positive or negative because cos(x/2) is in QII but
90<x/2<135 would you multiply both sides by 2 and get 180<x<270 which would be in QIII ?? Ahh so confusing :/
 
Joanna said:
So for cos(x/2)= -1/4. 90*<x/2<135 QII

So I used the formula that you showed me previously but instead of cos(2x) I put in the cos(x/2):

Sinx= + or - square root of (1-cos(x/2)/2)
Sinx= + or - square root of (1- (-1/4)/2) = + or - square root of (5/8)
So then I can multiply top and bottom by square root of 8 which would equal square root of 40/8
But does that equal 2squareroot10/ 8 which equals square root 10/4
And I'm not sure it it's positive or negative because cos(x/2) is in QII but
90<x/2<135 would you multiply both sides by 2 and get 180<x<270 which would be in QIII ?? Ahh so confusing :/
Click to expand...

I was afraid you were going to do that with substitution. That would be incorrect.

The substitution you should do is:

substitute

x → Θ/2 and 2x → Θ

then the original equation:

sin(x) = ±√[{1-cos(2x)}/2]

becomes:

sin(Θ/2) = ±√[{1-cos(Θ)}/2]

or if you want to have in terms of "x" - you should get:

sin(x/2) = ±√[{1-cos(x)}/2]

convince yourself that the equation above is correct.
 
So I can plug in -1/4 for cos(x/2) and solve for cosA right?
This is what I did:
(-1/4)= + or - square root of (1+cosx) /(2)
Square both sides : 1/16 = (1+cosx)/(2)
Then multiply both sides by 2 : 1/8 = 1+cosx
-cosx= 1-(1/8)
-cosx= 7/8
Cosx=-7/8

Is this correct?
 
Joanna said:
So I can plug in -1/4 for cos(x/2) and solve for cosA right?
This is what I did:
(-1/4)= + or - square root of (1+cosx) /(2)
Square both sides : 1/16 = (1+cosx)/(4) <<<< You need to fix rest of it accordingly
Then multiply both sides by 2 : 1/8 = 1+cosx
-cosx= 1-(1/8)
-cosx= 7/8
Cosx=-7/8

Is this correct?
Click to expand...

.
 
I'm having a problem with figuring out this proof : 2sin^2(u/2)=sin^2u/(1+cosu)
I would think you should start from the left side but I'm not sure what to do with the squared root of sin(u/2) and the 2 in front of it also, because sin(u/2) is the half angle identity and I want to write it out but I do not know where does the squared root and the 2 appropriately fits into the equation?? Help please just help me start the proof I know I'll be capable of completing it but just the initial step is very difficult for me
 
Joanna said:
I'm having a problem with figuring out this proof : 2sin^2(u/2)=sin^2u/(1+cosu)
I would think you should start from the left side but I'm not sure what to do with the squared root of sin(u/2) and the 2 in front of it also, because sin(u/2) is the half angle identity and I want to write it out but I do not know where does the squared root and the 2 appropriately fits into the equation?? Help please just help me start the proof I know I'll be capable of completing it but just the initial step is very difficult for me
Click to expand...

Hint:

\(\displaystyle 2*sin^2(\frac{u}{2}) \ = 1 \ - \ cos(u)\)
 
​
thank you do much that helped a lot and next time I will start a new problem in a new thread- sorry about that :)
 
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