Verify the identity. tan x + (pi/2)= -cot x
tan(x+(pi/2))=-cot(x) ?
yes
You could use that tan(t)=sin(t)/cos(t) then use the sum identity for sine and cosine
i know the identities but i'm not sure how to solve it
This has to do with phase shifts: \[\sin(x+\frac{\pi}{2}) = \cos(x)\] \[\cos(x+\frac{\pi}{2}) = -\sin(x)\] The two shifts above can help you prove your identity.
I've always hated memorizing stupid pointless "shortcuts" for trigonometry, so here's the easiest way to do it in my opinion. All you need to know is what the common values of sin and cos are (for 0,30,60,90,180 degrees) and sin(a+b) with cos(a+b) and you've pretty much slashed through 80% of things some teachers might insist that you need to memorize.
\[\frac{\sin(x+\frac{\pi}{2})}{\cos(x+\frac{\pi}{2})}\] use the sum identity for both sin and co
Go to the definition of tan: tan(x) = sin(x)/cos(x) So if you have tan(a+b) it's sin(a+b)/cos(a+b).
Now instead of b you have pi/2. Replace sin(x+b) with sin(x+pi/2) which is sin(x)*cos(pi/2)-sin(pi/2)*cos(x). Since cos(pi/2) (or cos(90), whatever you prefer) is 0, the above thing equals -cos(x). Do the same for cos(x+pi/2) and you'll get sin(x).
okay, i think i understand. thanks all!
can anyone help with this problem too? it's my last one Verify the identity. (1- sin x)/(cos x) = (cos x)/(1+ sin x)
do i cross multiply?
@AngusV
If you cross multiply you'll see that you get to the fundamental equation of trigonoetry - the one with: sin(x)^2+cos(x)^2=1. So just reverse engineer it a bit.
\[\frac{1-\sin(x)}{\cos(x)} \cdot \frac{1+\sin(x)}{1+\sin(x)}\] multiply top and bottom by top's conjugate for the left hand side expression
I don't think you can move something over from one side to another when proving something is something else
that is you can take the left hand side and show it is the same as the right hand side or you can take the right hand side and show it is the left hand side
okay, and then what would i do? i'm still lost
have you tried doing what i suggest above
\[\frac{1-\sin(x)}{\cos(x)} \cdot \frac{1+\sin(x)}{1+\sin(x)}=?\]
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