Question

a student was asked to prove cos(x+pi)=-cosx the students work follows, where was the students mistake?

Answer 1

Hello!

The answer is: The mistake was in the second expression.

The second expression should be the follow:

`cos(x+\pi)=cosxcos\pi-sinxsin\pi=-cosx`

Why?

The students are working with a cosine identity, cosine sum.

According to the theorem, cosine of a sum will be:

`cos(a+b)=cosacosb-sinasinb`

Where, for this case:

`a=x\\b=\pi`

So, substituting we have:

`cos(x+\pi)=cosxcos\pi-sinxsin\pi=cosx*(-1)-sinx*(0)\\cos(x+\pi)=-cosx-0\\cos(x+\pi)=-cosx`

So, the mistake was in the second step expression.

The expression should have been:

`cosxcos\pi-sinxsin\pi`

But why the result was correct even using a wrong expression?

The answer to that question is based on the value of sin(π) which is equal to 0.

Have a nice day!

Answer 2

The answer is the second ⇒ cosx cosπ - sinx sinπ

Explanation:

∵ cos(x + π) = cosx cosπ - sinx sinπ

∵ cosπ = -1

∵ sinπ = 0

∴ cos(x + π) = cosx(-1) - sinx(0) = -cosx - 0 = -cosx

The mistake in the rule it will not give different answer because + 0 or - 0 give us the same answer but if the measure of angle not π the answer will change.

Note: cos(x + π) means the angle in the third quadrant and the value of cos in the third quadrant must be negative