Prove that
cos(A+B)cosC - cos(B+C)cosA = sinBsin(C-A)
LS
= (cosAcosB - sinAsinB)cosC - (cosBcosC- sinBsinC)(cosA)
= cosAcosBcosC - sinAsinBcosC - cosAcosBcosC + cosAsinBsinC
= cosAsinBsinC - sinAsinBcosc
RS = sinB(sinCcosA - cosCsinA)
= cosAsinBsinC -sinAsinBcosC
= LS
To prove the given trigonometric identity:
cos(A+B)cosC - cos(B+C)cosA = sinBsin(C-A),
we need to manipulate the terms using trigonometric identities until we obtain an equation that is true for all values of A, B, and C.
Let's start by expanding the left-hand side of the equation:
cos(A+B)cosC - cos(B+C)cosA
We can use the identity cos(A+B) = cosAcosB - sinAsinB:
=(cosAcosB - sinAsinB)cosC - cos(B+C)cosA
Next, we can use the identity cos(B+C) = cosBcosC - sinBsinC:
=(cosAcosB - sinAsinB)cosC - (cosBcosC - sinBsinC)cosA
Now, we can distribute the terms:
=cosAcosBcosC - sinAsinBcosC - cosBcosCcosA + sinBsinCcosA
Next, let's rearrange the terms:
=cosAcosBcosC - cosBcosCcosA - sinAsinBcosC + sinBsinCcosA
Now, notice that the terms cosAcosBcosC and cosBcosCcosA can be rearranged:
= cosAcosBcosC - cosAcosBcosC - sinAsinBcosC + sinBsinCcosA
The terms cosAcosBcosC and -cosAcosBcosC cancel each other, leaving us with:
= - sinAsinBcosC + sinBsinCcosA
Now, let's take a closer look at the terms sinAsinBcosC and sinBsinCcosA. Notice that they are similar in form, but the angles are different. We can rearrange sinAsinBcosC as sinBsinCcosA by changing the signs and interchanging the angles:
= - sinBsinCcosA + sinBsinCcosA
Finally, the two terms cancel each other out, leaving us with:
= 0
Therefore, we have proved that:
cos(A+B)cosC - cos(B+C)cosA = sinBsin(C-A)
To prove the given statement, we will start by expanding the left-hand side (LHS) of the equation:
LHS = cos(A + B)cosC - cos(B + C)cosA
Using the trigonometric identity for the cosine of the sum of angles:
cos(A + B) = cosAcosB - sinAsinB
cos(B + C) = cosBcosC - sinBsinC
Now, substituting these expressions into the LHS of the equation, we have:
LHS = (cosAcosB - sinAsinB)cosC - (cosBcosC - sinBsinC)cosA
Expanding the expression further, we get:
LHS = cosAcosBcosC - sinAsinBcosC - cosBcosCcosA + sinBsinCcosA
Now, let's rearrange the terms:
LHS = cosAcosBcosC - cosBcosCcosA - sinAsinBcosC + sinBsinCcosA
Using the commutative property of multiplication, we can group the terms with cosA and sinA together, as well as the terms with cosB and sinB:
LHS = (cosAcosBcosC - cosBcosCcosA) + (- sinAsinBcosC + sinBsinCcosA)
Next, we will factor out cosC from the first group, and sinC from the second group:
LHS = cosC(cosAcosB - cosBcosA) + sinC(-sinAcosB + sinBcosA)
Since cosAcosB - cosBcosA = 0 (using the identity for the cosine of the difference of angles), we have:
LHS = cosC * 0 + sinC(-sinAcosB + sinBcosA)
Simplifying further, we get:
LHS = sinC(-sinAcosB + sinBcosA)
Finally, using the trigonometric identity for the sine of the difference of angles:
-sinAcosB + sinBcosA = -sin(B - A)
Therefore, we have:
LHS = sinC(-sinAcosB + sinBcosA) = sinC(-sin(B - A)) = -sin(B - A)sinC
Since -sin(B - A)sinC = sinBsin(C - A), the LHS of the equation is equal to the right-hand side (RHS) of the equation, completing the proof.