Calculate $\sin^5\alpha-\cos^5\alpha$ if $\sin\alpha-\cos\alpha=\frac12$ Calculate $$\sin^5\alpha-\cos^5\alpha$$ if $\sin\alpha-\cos\alpha=\dfrac12$.
The main idea in problems like this is to write the expression that we need to calculate in terms of the given one (in this case we know $\sin\alpha-\cos\alpha=\frac12$).
I don't see how to even start to work on the given expression as we cannot use $a^2-b^2=(a-b)(a+b)$ or $a^3-b^3=(a-b)(a^2+ab+b^2)$. So in other words, I can't figure out how to factor the expression (even a little).

 The given answer is $\dfrac{79}{128}$.

 A: I think we're all overthinking the problem a bit here!
Given that
$$\sin\alpha -\cos\alpha=\frac{1}{2}$$
set $x=\sin\alpha$ so that $\cos\alpha=\sqrt{1-x^2}$ and solve the quadratic for $x$. Your answer is now
$$\sin^5\alpha -\cos^5\alpha=x^5-(1-x)^\frac{5}{2}$$
A: using the hint: $x^5-y^5 = (x-y)(x^4+x^3y+x^2y^2+xy^3+y^4)$
Let's create the parts:
since $x^2+y^2=1$ we have:
$(x^2+y^2)^2 = x^4+2x^2y^2+y^4 = 1$ (2)
We have a big part of it but we are missing $xy$ so
we know $x-y = 1/2$ squaring both side: $x^2 + y^2 -2xy = 1/4$ since $x^2+y^2=1$ then $xy=3/8$ (1).
using (2)
$(x-y)(x^4+x^3y+x^2y^2+xy^3+y^4) = \frac{1}{2} (1 - x^2y^2 + xy(x^2+y^2)) = \frac{1}{2} (1 - \frac{9}{64} + \frac{3}{8}\times1) = \frac{79}{128}$
A: HINT
I would start with noticing that
\begin{align*}
x^{5} - y^{5} & = (x - y)(x^{4} + x^{3}y + x^{2}y^{2} + xy^{3} + y^{4})\\\\
\end{align*}
Then you can rearrange the inner terms in order to fit the proposed information.
A: We will use a few identities.
First is the Pythagorean Identity:
$$\sin^2\alpha +\cos^2\alpha = 1$$
Second is a simple factorization:
$$a^5 - b^5 = (a-b)(a^4 + a^3b + a^2b^2 + ab^3 + b^4) = (a-b)(a^4 + b^4 + ab(a^2+b^2+ab))$$
Third will help us write things of fourth degree in terms of second and first degree terms:
$$a^4 + b^4 = (a^2+b^2)^2-2(ab)^2$$
So we can combine them and get the following:
$$a^5 - b^5 = (a-b)((a^2+b^2)^2-2(ab)^2 + ab(a^2+ab+b^2))$$
Now, all we have to do is calculate $\sin\alpha \cos\alpha$ and plug the values in.
$$\sin\alpha - \cos\alpha = \frac{1}{2}$$
Squaring both sides, we get
$$\sin^2\alpha + \cos^2\alpha - 2\sin\alpha\cos\alpha = \frac{1}{4}$$
$$\Rightarrow 1 - 2\sin\alpha\cos\alpha = \frac{1}{4}$$
$$\Rightarrow\sin\alpha\cos\alpha = \frac{3}{8}$$
Then, if we let $a = \sin\alpha$ and $b = \cos\alpha$, we get
$$\sin^5\alpha - \cos^5\alpha = (\sin\alpha - \cos\alpha)((\sin^2\alpha + \cos^2\alpha)^2 - 2(\sin\alpha\cos\alpha)^2 \\+ \sin\alpha\cos\alpha (\sin^2\alpha + \cos^2\alpha + \sin\alpha\cos\alpha))$$
$$= \frac{1}{2}\left(1^2 - 2\left(\frac{3}{8}\right)^2 + \frac{3}{8}\left(1+\frac{3}{8}\right)\right) = \frac{79}{128}$$
A: Another approach: Solve $ \sin{z}-\cos{z}=\frac{1}{2}$
This results in $e^{iz}=\frac{\sqrt{7}-i}{2(i-1)}$
Plugging this into the complex sin and cos definition results in:
$\sin{z}=\frac{1-1\sqrt{7}}{4}$ and $\cos{z}= \frac{-1-1\sqrt{7}}{4}$
When applying this in $sin^5z-cos^5z$ all uneven powers of $\sqrt(7)$ cancel out.
What remains is: $sin^5z-cos^5z=\frac{1+70+5\cdot 49}{2^9}=\frac{79}{128}$
A: Alternative approach:
Let $~\displaystyle x = \sin(\alpha), y = \cos(\alpha) = \implies 
(x - y) = \frac{1}{2}, ~\left(x^5 - y^5\right) = \color{red}{what?}$
Tools

*

*$T_1 ~: ~(x - y)^2 = x^2 - 2xy + y^2.$


*$T_2 ~: ~(x - y)^3 = x^3 - 3x^2y + 3xy^2 - y^3.$


*$T_3 ~: ~(x - y)^5 = x^5 - 5x^4y + 10x^3y^2 - 10x^2y^3 + 5xy^4 - y^5.$
Plan of Attack 
Since $~x^2 + y^2 = 1~,$ there is sufficient information to compute $(xy)$.  Using this you can compute $x^3 - y^3$, and using this you can compute $x^5 - y^5.$

From $T_1,$ you have that
$$\frac{1}{4} = 1 - 2xy \implies xy = \frac{3}{8}. \tag1 $$

Using (1), from $T_2,$ you have that
$$\frac{1}{8} = \left(x^3 - y^3\right) - 3xy(x-y) \implies $$
$$\frac{1}{8} = \left(x^3 - y^3\right) - 3\left(\frac{3}{8}\right) \left(\frac{1}{2}\right) \implies (x^3 - y^3) = \frac{11}{16}. \tag2 $$

Using (1) and (2), from $T_3,$ you have that
$$\frac{1}{32} = \left(x^5 - y^5\right) -5(xy)(x^3 - y^3) + 10(xy)^2(x - y) \implies $$
$$\frac{1}{32} = \left(x^5 - y^5\right) - 5\left(\frac{3}{8}\right)\left(\frac{11}{16}\right) + 10\left(\frac{3}{8}\right)^2\left(\frac{1}{2}\right) \implies $$
$$\frac{4}{128} = \left(x^5 - y^5\right) - \frac{165}{128} + \frac{90}{128} \implies$$
$$\left(x^5 - y^5\right) = \frac{4 + 165 - 90}{128} = \frac{79}{128}.$$
A: $$\dfrac1{2\sqrt2}=\sin\left(\alpha-\dfrac\pi4\right)=\sin x$$ where $\alpha-\dfrac\pi4=x$
$$\sin^5\alpha-\cos^5\alpha=\sin^5\left(x+\dfrac\pi4\right)-\cos^5\left(x+\dfrac\pi4\right)=\dfrac{(\cos x+\sin x)^5-(\cos x-\sin x)^5}{(\sqrt2)^5}$$
Now,
$$(\cos x+\sin x)^5-(\cos x-\sin x)^5=2\sin^5x+2\binom53\sin^3x\cos^2x+2\binom54\sin x\cos^4x$$
Use $\cos^2x=1-\sin^2x=1-\left(\dfrac1{2\sqrt2}\right)^2=?$
