Question

Let t = cosA + sinA prove that t^2 = 1 + sin2A

Answer 1

$\bf \begin{array}{llll} \textit{Pythagorean Identities} \\\\ sin^2(\theta)+cos^2(\theta)=1 \end{array}\qquad \qquad \qquad \begin{array}{llll} \textit{Double Angle Identities} \\\\ sin(2\theta)=2sin(\theta)cos(\theta) \end{array}\\\\ -------------------------------\\\\ \begin{cases} t=cos(A)+sin(A)\\ t^2=[cos(A)+sin(A)]^2 \end{cases}\implies \stackrel{FOIL}{cos^2(A)+2cos(A)sin(A)+sin^2(A)} \\\\\\ \boxed{sin^2(A)+cos^2(A)}+2cos(A)sin(A)\implies \boxed{1}+2sin(A)$

Answer 2

$t= \cos A + \sin A$

$t^2=(\cos A + \sin A)^2 = \cos^2 A + 2 \cos A \sin A + \sin ^2 A$

$t^2= \cos^2 A+ \sin ^2 A + 2 \cos A \sin A$

$t^2= 1 + 2 \cos A \sin A \quad\checkmark$

$t^2= 1 + \sin 2A \quad\checkmark$

We use the Pythagorean Theorem $\cos^2 A + \sin ^2 A = 1$ and the double angle formula for sine, $\sin 2A = 2 \sin A \cos A$