One More Step Of Simplication

For simplicity sake, let $A = \cos \theta$ and $B= \sin\theta$ , then we want to find $(A+B)^2 + (A-B)^2$ .

Let's expand and simplify the expressions $(A+B)^2$ and $(A-B)^2$ first.

$(A+B)^2 = A^2 + B^2 + 2AB \\ (A-B)^2 = A^2+B^2-2AB$

Thus,

$\begin{aligned} (A+B)^2 + (A-B)^2 &=& (A^2+B^2 + 2AB) + (A^2 + B^2-2AB) \\ &=& (A^2+B^2 + \cancel{2AB}) + (A^2 + B^2-\cancel{2AB}) \\ &=& 2(A^2+B^2) \end{aligned}$

Subsituting back $A = \cos \theta$ and $B= \sin\theta$ gives $2(A^2+B^2) = 2(\cos^2 \theta + \sin^2\theta )$ .

By Pythagorean identities , we have $\cos^2 \theta + \sin^2\theta = 1$ .

Hence our answer is $2(1) =\boxed2$ .

Using the FOIL method for these given $(\cos (\theta) + \sin(\theta))^2 +(\cos(\theta) -\sin (\theta))^2$

So

$\displaystyle{\begin{aligned}(\cos(\theta) + \sin(\theta))^2 +(\cos(\theta)-\sin (\theta))^2 &=[\cos^2(\theta) +2 \cos(\theta)\sin(\theta) + \sin(\theta)] +[\cos^2(\theta) - 2\cos(\theta) \sin(\theta) +\sin^2(\theta)] \quad\quad\quad\quad\quad\quad\quad\quad{\text{I use brackets to avoid confusion}} \\&= 2\cos^2(\theta) +2\sin^2(\theta)\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad{\text{Factor out}} \\&=2(\sin^2(\theta)+\cos^2(\theta)) \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad{\textrm{The Pythagorean identity } \cos^2(\theta)+\sin^2(\theta) = 1 }\\&= 2 \cdot 1 \\&=2 \end{aligned}}$

$\large \text{ADIOS!!!}$

Acc to identity, $(a+b)^{2} + (a-b)^{2} = 2(a^{2} + b^{2})$

And $cos^{2} \theta + sin^{2} \theta = 1$

Therefore,

$(cos \theta + sin \theta)^{2} + (cos \theta - sin \theta)^{2} = 2(cos^{2} \theta + sin^{2} \theta) = 2(1) = 2$