Is (sin(A)+cos(A))^2=1+2sin(A)cos(A) ?

The answer to whether (sin(A)+cos(A))^2=1+2sin(A)cos(A) is True

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prove (sin(A)+cos(A))^2=1+2sin(A)cos(A)

prove

(sin (A) + cos (A))^{2} = 1 + 2 sin (A) cos (A)

True

Solution steps

(sin (A) + cos (A))^{2} = 1 + 2 sin (A) cos (A)

Manipulating left side

(sin (A) + cos (A))^{2}

Expand

(cos (A) + sin (A))^{2} : cos^{2} (A) + 2 cos (A) sin (A) + sin^{2} (A)

(cos (A) + sin (A))^{2}

Apply Perfect Square Formula:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = cos (A), b = sin (A)

= cos^{2} (A) + 2 cos (A) sin (A) + sin^{2} (A)

= cos^{2} (A) + sin^{2} (A) + 2 cos (A) sin (A)

Rewrite using trig identities

cos^{2} (A) + sin^{2} (A) + 2 cos (A) sin (A)

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

= 1 + 2 cos (A) sin (A)

= 1 + 2 cos (A) sin (A)

= 1 + 2 sin (A) cos (A)

We showed that the two sides could take the same form

\Rightarrow True

\frac{cos ( b )}{sec ( b )} + \frac{sin ( b )}{csc ( b )} = csc^{2} (b) - cot^{2} (b)

2 cos^{2} (\frac{θ}{2}) - cos (θ) - 1 = 0

\frac{1}{csc ( θ ) - cot ( θ )} = csc (θ) + cot (θ)

1 + cot^{2} (- x) = csc^{2} (x)

cos (θ + \frac{π}{4}) = \frac{2}{2} (cos (θ) - sin (θ))

Is (sin(A)+cos(A))^2=1+2sin(A)cos(A) ?
The answer to whether (sin(A)+cos(A))^2=1+2sin(A)cos(A) is True