Trig identities (is my method correct?)

[supernova1203]

Homework Statement

Prove the following identities

31c) sin($\frac{\pi}{2}$+x)=cosx

Homework Equations

sin²x+cos²x=1

The Attempt at a Solution

The idea here is to prove the identity by making LS=RS

so here is what i have done, but I am not sure if it is the right way, since the book shows it went about it in a different way.
My method:

sin($\frac{\pi}{2}$+x)=cosx

1+sinx=cosx

now we use the pythagreon identity

sin²x+cos²x=1

If we move the cos to the right side, we are left with sinx=-cosx

and we use another identity

cos-x=cosx

therefore LS=RS and we have proved the identity? Book method:sin($\frac{\pi}{2}$+x)=cosx

LS=sin($\frac{\pi}{2}$+x) RS=cosx

=sin$\frac{\pi}{2}$cosx+cosx+cos$\frac{\pi}{2}$sinx

=(1)cosx+(0)sinx

=cosx

LS=RS therefore sin($\frac{\pi}{2}$+x)=cosx
Is my method correct? Also at the point where they do "=sin$\frac{\pi}{2}$cosx+cosx+cos$\frac{\pi}{2}$sinx" Are they still just dealing with LS only?

thanks!

Also if someone could please explain how the book got to the solution it did, what thought process does one have to use to get to the solution the way book did? Thanks

 

[jedishrfu]

The algebra in your method is not quite right:

sin(pi/2 + x) =/= sin(pi/2) + sin(x) _________ (i.e. =/= 1 + sin(x) )

with sin(pi/2) = 1

the sine of the sum of two angles is:

sin(a + b) = sin(a)cos(b) + cos(a)sin(b)

 

[kevinj888]

Homework Statement

Prove the following identities

31c) sin($\frac{\pi}{2}$+x)=cosx

Homework Equations

sin²x+cos²x=1

The Attempt at a Solution

The idea here is to prove the identity by making LS=RS

so here is what i have done, but I am not sure if it is the right way, since the book shows it went about it in a different way.



My method:

sin($\frac{\pi}{2}$+x)=cosx

1+sinx=cosx

now we use the pythagreon identity

sin²x+cos²x=1

If we move the cos to the right side, we are left with sinx=-cosx

and we use another identity

cos-x=cosx

therefore LS=RS and we have proved the identity?





Book method:


sin($\frac{\pi}{2}$+x)=cosx

LS=sin($\frac{\pi}{2}$+x) RS=cosx

=sin$\frac{\pi}{2}$cosx+cosx+cos$\frac{\pi}{2}$sinx

=(1)cosx+(0)sinx

=cosx

LS=RS therefore sin($\frac{\pi}{2}$+x)=cosx



Is my method correct? Also at the point where they do "=sin$\frac{\pi}{2}$cosx+cosx+cos$\frac{\pi}{2}$sinx" Are they still just dealing with LS only?

thanks!

Also if someone could please explain how the book got to the solution it did, what thought process does one have to use to get to the solution the way book did? Thanks

The mistake is in this line :
1+sinx=cosx
From where you got this equation?

 

[verty]

The book's method is not what you should be using, you should be using the CAST rule and the rule about http://www.regentsprep.org/Regents/math/algtrig/ATT6/cofunctions.htm.

 

[Mark44]

The most straightforward approach to this problem is to use the identity for the sine of a sum of angles: sin(A + B) = sinA*cosB + cosA*sinB. You don't have it listed in your relevant identities/equations, but should.

The mistake you made is thinking that sin(A + B) "distributes" over a sum to result in sinA + sinB. That is definitely NOT true.

 

[sherrellbc]

The mistake is in this line :
1+sinx=cosx
From where you got this equation?

I was wondering the same thing.
Perhaps, you meant to write:
$1 - sin^2(x) = cos^2(x)$

Otherwise I am not sure.

Also,
$sin(A+B) = sin(A)cos(B) + sin(B)cos(A)$

 

[Mark44]

The mistake is in this line :
1+sinx=cosx
From where you got this equation?

I was wondering the same thing.
Perhaps, you meant to write:
$1 - sin^2(x) = cos^2(x)$

I believe this is what the OP was doing:
$sin(\pi/2 + x) = cos(x)$
$sin(\pi/2) + sin(x) = cos(x)$
1 + sin(x) = cos(x)

IOW, the OP was to distribute in a situation for which it wasn't applicable.