Please solve this trigonometric identity proof problem. I have completed 20; this is the hardest one. Many thanks
(sin^3(x)-cos^3(x))/sin(x)-cos(x)=1+sin(x)cos(x)
recall that
(a^3-b^3) = (a-b)(a^2+ab+b^2)
sin^2+cos^2 = 1
and it all drops right out.
To solve this trigonometric identity proof problem, we will start from the left-hand side (LHS) of the equation and simplify it until we reach the right-hand side (RHS). Here is the step-by-step solution:
LHS: (sin^3(x) - cos^3(x))/(sin(x) - cos(x))
Step 1: Recognize that the numerator and denominator are both cubes (sin^3(x) and cos^3(x)). We can use the formula for the difference of cubes to simplify this expression:
(a^3 - b^3) = (a - b)(a^2 + ab + b^2)
Applying this formula, we rewrite the numerator as:
sin^3(x) - cos^3(x) = (sin(x) - cos(x))(sin^2(x) + sin(x)cos(x) + cos^2(x))
So, the LHS becomes:
LHS: (sin(x) - cos(x))(sin^2(x) + sin(x)cos(x) + cos^2(x))/(sin(x) - cos(x))
Step 2: Cancel out the common factor of (sin(x) - cos(x)) in the numerator and denominator:
LHS: sin^2(x) + sin(x)cos(x) + cos^2(x)
Step 3: Recall the Pythagorean identity, which states that sin^2(x) + cos^2(x) = 1. Replace sin^2(x) + cos^2(x) in the equation with 1:
LHS: 1 + sin(x)cos(x)
We have now reached the RHS of the equation, which confirms that the identity holds:
RHS: 1 + sin(x)cos(x)
Therefore, we have successfully proven the trigonometric identity:
(sin^3(x) - cos^3(x))/(sin(x) - cos(x)) = 1 + sin(x)cos(x)