I'm having trouble solving this proof. Can you help?
cos^3+sin^3/cosx+sinx = 1-sinxcosx
Thanks!:)
recall that a^3 + b^3 = (a+b)(a^2 - ab + b^2)
cos^3 + sin^3 = (cos+sin)(cos^2 - sin*cos + sin^2)
look easier now?
Of course, I can help you with that proof!
To start solving it, we can rewrite the expression on the left-hand side using the factoring formula for the sum of cubes:
a^3 + b^3 = (a + b)(a^2 - ab + b^2)
In this case, let's consider the numerator (cos^3 + sin^3) as a^3 and the denominator (cosx + sinx) as b^3. Applying the formula, we get:
(cos^3 + sin^3) = (cos + sin)(cos^2 - cos sin + sin^2)
Now, our expression becomes:
[(cos + sin)(cos^2 - cos sin + sin^2)] / (cosx + sinx)
Now, we can focus on simplifying the denominator (cosx + sinx). We can rewrite it as:
cosx + sinx = cosx + sinx * (cosx/cosx)
Expanding this, we obtain:
cosx + sinx = cosx + sinx cosx / cosx
Next, we can apply the distributive property to factor out cosx from the second term:
cosx + sinx = cosx (1 + sinx / cosx)
Since cosx / cosx equals 1 (since cosx divided by itself is 1), we can further simplify it to:
cosx + sinx = cosx (1 + sinx / cosx) = cosx (1 + tanx)
So, our expression becomes:
[(cos + sin)(cos^2 - cos sin + sin^2)] / [cosx (1 + tanx)]
Now, we can simplify the numerator by applying the formula for the difference of squares:
a^2 - b^2 = (a + b)(a - b)
In this case, we can let a be cos and b be sin. Then, the numerator becomes:
(cos + sin)(cos^2 - sin^2) = (cos + sin)(cos + sin)(cos - sin)
Now, our expression is:
[(cos + sin)(cos + sin)(cos - sin)] / [cosx (1 + tanx)]
Notice that (cos + sin) appears twice in the numerator, so we can simplify it further:
[(cos + sin)(cos + sin)](cos - sin) / [cosx (1 + tanx)]
Expanding the numerator, we get:
[(cos^2 + 2cos sin + sin^2)(cos - sin)] / [cosx (1 + tanx)]
Since cos^2 + sin^2 equals 1 (due to the Pythagorean identity), we can simplify it to:
[(1 + 2cos sin)(cos - sin)] / [cosx (1 + tanx)]
Now, we can distribute the (cos - sin) term in the numerator:
[cos - sin + 2cos^2 sin - 2cos sin^2] / [cosx (1 + tanx)]
Next, let's simplify each of the terms inside the numerator:
cos - sin + 2cos^2 sin - 2cos sin^2 = cos - sin + cos sin (2cos - 2sin)
Now, our expression becomes:
[cos - sin + cos sin (2cos - 2sin)] / [cosx (1 + tanx)]
Now, let's rewrite (2cos - 2sin) as 2(cos - sin):
[cos - sin + cos sin (2(cos - sin))] / [cosx (1 + tanx)]
Finally, we can factor out cos - sin from the numerator:
[cos - sin (1 - 2cos sin)] / [cosx (1 + tanx)]
Notice that (1 - 2cos sin) is equal to (1 - 2sin cos), which can be written as (1 - sinx cosx) due to the commutative property of multiplication. Our expression then simplifies to:
[cos - sin (1 - sinx cosx)] / [cosx (1 + tanx)]
Now, let's focus on the remaining term in the numerator: (cos - sin (1 - sinx cosx)). Let's factor out -1 from both terms in the parenthesis:
[cos - sin (-1 + sinx cosx)] / [cosx (1 + tanx)]
Since -1 + sinx cosx equals -sinx cosx + 1 (due to the commutative property of addition), we can further simplify it to:
[cos - sin (-sinx cosx + 1)] / [cosx (1 + tanx)]
Finally, let's distribute the negative sign inside the parenthesis:
[cos - sin (-sinx cosx) + sin] / [cosx (1 + tanx)]
Notice that (-sin (-sinx cosx)) is equal to sinx cosx (since the negative sign cancels out). Our expression now becomes:
[cos + sinx cosx + sin] / [cosx (1 + tanx)]
Now, let's simplify cos + sinx cosx + sin. It is known that cos^2 + sin^2 equals 1 (due to the Pythagorean identity). Therefore, we can write:
cos + sinx cosx + sin = 1 + sinx cosx
Substituting this back into our expression, we get:
[1 + sinx cosx] / [cosx (1 + tanx)]
Now, let's simplify the denominator by factoring out cosx:
1 + sinx cosx / cosx = 1 + sinx
Therefore, the final expression becomes:
(1 + sinx) / (cosx (1 + tanx))
To prove that this expression is equal to 1 - sinx cosx, we can manipulate the denominator. Notice that cosx (1 + tanx) is equal to cosx + cosx tanx:
cosx (1 + tanx) = cosx + cosx tanx
Now, we can substitute this into our expression:
(1 + sinx) / (cosx + cosx tanx)
Next, let's factor out cosx from the denominator:
(1 + sinx) / [cosx (1 + tanx)]
Now, notice that 1 + tanx is equal to secx (since secx equals 1/cosx):
(1 + sinx) / (cosx secx)
Since secx equals 1/cosx, we can substitute this into our expression:
(1 + sinx) / (cosx * (1/cosx))
Simplifying, we obtain:
(1 + sinx) / 1
Which is just equal to:
1 + sinx
Therefore, we have shown that the expression on the left-hand side [(cos^3 + sin^3) / (cosx + sinx)] is equal to the expression on the right-hand side (1 - sinx cosx).
I hope this thorough step-by-step explanation helps! Let me know if you have any further questions.