MHB Solve sin(x)-1=cos(x): Step-by-Step Guide

[Elissa89]

So the problem is sin(x)-1=cos(x) and I don't know how to do this one.

 

[I like Serena]

Which formulas do you have available to add/subtract sine and/or cosine?

 

[Elissa89]

Which formulas do you have available to add/subtract sine and/or cosine?

I don't know what you mean.

 

[I like Serena]

I don't know what you mean.

Well... erm... I'm a bit at a loss of the formulas you can use or not...

See for instance the wiki page of Trigonometric Identities for a list of such formulas...
This may be a bit overwhelming, but one of the formulas in that page is:
$$a\sin x+b\cos x=c\sin(x+\varphi)$$
where $c = \sqrt{a^2 + b^2}$ and $\varphi = \operatorname{atan2} \left( b, a \right)$.

To be fair, there's a good chance that you haven't been taught this formula... but what have you been taught?
Or what are you otherwise supposed to know and be able to apply?

 

[Greg]

I don't think it has to be too complicated. :)

Square both sides of the given equation (post back if you don't know how to do that) and use the identity $\cos^2(x)=1-\sin^2(x)$. Simplify and solve the resulting equation, then check your results with the given equation.

 

[Elissa89]

I don't think it has to be too complicated. :)

Square both sides of the given equation (post back if you don't know how to do that) and use the identity $\cos^2(x)=1-\sin^2(x)$. Simplify and solve the resulting equation, then check your results with the given equation.

I did that but I'm still lost. My professor emailed me back, said to square both sides the squared cos can be turned into sines using the pythagorean theorem identity. Which doesn't make sense to me because the pythagorean theorem identity still has cosines in it?

Sorry I'm replying so late.

 

[I like Serena]

I did that but I'm still lost. My professor emailed me back, said to square both sides the squared cos can be turned into sines using the pythagorean theorem identity. Which doesn't make sense to me because the pythagorean theorem identity still has cosines in it?

Sorry I'm replying so late.

Let's start with squaring both sides as your professor said:
$$\sin(x)-1=\cos(x) \\
(\sin(x)-1)^2=\cos^2(x) \\
\sin^2 x - 2\sin x + 1 = \cos^2x
$$
Now we can turn the $\cos^2x$ into sines by using $\cos^2x=1-\sin^2x$, as greg1313 suggested, can't we?
Then there will be no cosines left. (Thinking)