HHHEEELLLPPP!!
I have a trig identity problem that I am in desprite need of help.I need help proving that
sin(("pie"/2) + x)= cos(x). I think I need to use the addition formula but I'm not sure:confused:

It's pi.

Pleas don't make fun of me and help I have been working on this and oother problems for 2 months and the spelling of pi is not my first worry.So can you help me or not :(

I will help if you promise not to spell {\pi} as 'pie' anymore;) .

You are correct, use the addition formula for sin.


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

Let a=\frac{\pi}{2} and b=x. Plug them in and you should see it.

OK so I distributed like you said to do but I don't know how to evaluate the sin(pi/2) and cos(pi/2)

Run them through a calculator. But you should know that sin(\frac{\pi}{2})=1 and cos(\frac{\pi}{2})=0

Remember, \frac{\pi}{2} \;\ radians corresponds to 90 degrees.

Doesn't sin(90)=0 and cos(90)=1?

See now?

omg duh I'm a retard lol I got it... now I'm stuck on another (sinx+cosx)^2=1+sin2x...
I got it to (six+cosx)(sinx+cosx)=1+2sinxcosx
then I distributed to get sin^2x+2sinxcosx+cos^x=1+2sinxcosx

Now what??

there are the steps to follow to solve that problem
(sinx+cosx)^2=(sinx)^2+2sinxcosx+(cosx)^2 (1)
(cosx)^2+(sinx)^2=1 that is a rule (2)
(2) in (1) will give (sinx+cosx)^2=((cosx)^2+(sinx)^2)+2sinxcosx
(sinx+cosx)^2= 1+2sinxcosx
note:(a+b)^2=a^2+2ab+b^2 I hope that will help you get it
sin2x=2sinxcosx
(sinx+cosx)^2=1+sin2x

omg duh im a retard lol i got it.... now im stuck on another (sinx+cosx)^2=1+sin2x......
i got it to (six+cosx)(sinx+cosx)=1+2sinxcosx
then i distributed to get sin^2x+2sinxcosx+cos^x=1+2sinxcosx

Now what ??????

Now what? You're pretty much done. Doesn't sin^{2}(x)+cos^{2}(x)=1?

sin2x=2sinxcosx
(cox+sinx)^2=1+2sinxcosx
=1+sin2x