Question

1-cosx=-sinx x∈[-180, 180]

I moved the -sinx over to the other side so now I have -cosx+sinx+1=0 but I'm not sure where to go from there. The answers are 0 and -90 degrees but I don't know how to get there.

Help is much appreciated, thanks!

Answer 1

so you want to solve
\[\cos x -\sin x = 1\]

Answer 2

the trick here is to use the identity \(\cos A\cos B-\sin A\sin B = \cos(A+B)\)

Answer 3

familiar with that identity ?

Answer 4

Yes, how would I apply it to cosx-sinx=1 though?

Answer 5

Yes that's the interesting part, divide \(\sqrt{2}\) through out recalling the fact that
\[\sin(45)=\cos(45)=\frac{1}{\sqrt{2}}\]

Answer 6

\[\cos x -\sin x = 1\]

dividing \(\sqrt{2}\) through out gives
\[\cos x\frac{1}{\sqrt{2}} -\sin x\frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}}\]

replace \(\frac{1}{\sqrt{2}}\) by \(\cos(45)\) or \(\sin(45)\) as needed :
\[\cos x\cos(45) -\sin x\sin(45) = \frac{1}{\sqrt{2}}\]

Answer 7

Now the left hand side is ready to use that identity, yes ?

Answer 8

Got it. Why choose the square root of 2 specifically to divide though? I'm a little confused on that part.

Answer 9

ikr, that \(\sqrt{2}\) might seem coming from nowhere, but there is a more general formula... maybe for now keep going and finish solving the equation

Answer 10

Alright thank you so much for the help, I think I know how to solve it from here (:

Answer 11

Nice :) about that \(\sqrt{2}\) :
when you have an expression like \[a\cos x+b\sin x\]

draw a right triangle with legs \(a,b\)

Answer 12

Notice that \(\color{blue}{\sin\theta=\frac{a}{\sqrt{a^2+b^2}}}\) and \(\color{purple}{\cos\theta=\frac{b}{\sqrt{a^2+b^2}}}\) , and the earlier expression can be rearranged as

\[\begin{align}a\cos x+b\sin x&=\sqrt{a^2+b^2}\left(\color{blue}{\frac{a}{\sqrt{a^2+b^2}}}\cos x+\color{purple}{\frac{b}{\sqrt{a^2+b^2}}}\sin x\right)\\~\\
&=\sqrt{a^2+b^2}\left(\color{blue}{\sin\theta} \cos x+\color{purple}{\cos\theta}\sin x\right)\\~\\
&=\sqrt{a^2+b^2}\sin(\theta+x)
\end{align}\]