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Elissa89

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

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In summary: 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)

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Elissa89

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

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I like Serena

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

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Elissa89

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Klaas van Aarsen said:Which formulas do you have available to add/subtract sine and/or cosine?

I don't know what you mean.

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Elissa89 said: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?

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Greg

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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.

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Elissa89

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greg1313 said:

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.

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Elissa89 said: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)

The first step in solving this equation is to use the Pythagorean identity, sin^2(x) + cos^2(x) = 1, to rewrite the equation as sin(x) - 1 = 1 - sin^2(x). This will allow us to solve for sin(x) in terms of cos(x).

Using the equation from the first step, we can solve for sin(x) by adding 1 to both sides of the equation, giving us sin(x) = 1 - sin^2(x) + 1. Then, we can factor out the sin(x) term to get sin(x)(1 + sin(x)) = 2. Finally, we can divide both sides by (1 + sin(x)) to get sin(x) = 2/(1 + sin(x)).

After solving for sin(x) in terms of cos(x), we can substitute this expression into the original equation, sin(x) - 1 = cos(x). This will give us an equation only in terms of cos(x) which we can then solve for using algebraic methods.

To solve for cos(x), we can rearrange the equation from the previous step to get cos(x) = sin(x) - 1. Then, using the half-angle formula, cos(x) = ±√[(1 - cos(x))/2], we can solve for cos(x) in terms of sin(x). Finally, we can substitute this expression back into the equation to find the values of x that satisfy the original equation.

Yes, when using the half-angle formula in the previous step, we must keep in mind that we are finding solutions for cos(x), so we need to consider the restrictions on the domain of cos(x). This includes avoiding values that would make the expression under the square root negative or undefined, such as when sin(x) = -1 or cos(x) = 1.

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