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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

(1 + sin (x))^{2} + 1 - sin^{2} (x) : 2 sin (x) + 2

(1 + sin (x))^{2} + 1 - sin^{2} (x)

(1 + sin (x))^{2} : 1 + 2 sin (x) + sin^{2} (x)

Apply Perfect Square Formula:

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

a = 1, b = sin (x)

= 1^{2} + 2 \cdot 1 \cdot sin (x) + sin^{2} (x)

Simplify

1^{2} + 2 \cdot 1 \cdot sin (x) + sin^{2} (x) : 1 + 2 sin (x) + sin^{2} (x)

1^{2} + 2 \cdot 1 \cdot sin (x) + sin^{2} (x)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 + 2 \cdot 1 \cdot sin (x) + sin^{2} (x)

Multiply the numbers:

2 \cdot 1 = 2

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

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

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

Simplify

1 + 2 sin (x) + sin^{2} (x) + 1 - sin^{2} (x) : 2 sin (x) + 2

1 + 2 sin (x) + sin^{2} (x) + 1 - sin^{2} (x)

Group like terms

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

Add similar elements:

sin^{2} (x) - sin^{2} (x) = 0

= 2 sin (x) + 1 + 1

Add the numbers:

1 + 1 = 2

= 2 sin (x) + 2

= 2 sin (x) + 2

= 2 sin (x) + 2

= 2 sin (x) + 2

= 2 + 2 sin (x)

We showed that the two sides could take the same form

\Rightarrow True

2 sin^{2} (x) + sin (x) = 0

cot (6 0^{\circ}) = \frac{cos ( 6 0 ^{\circ} )}{sin ( 6 0 ^{\circ} )}

\frac{sec ( x )}{cos ( x )} = sec^{2} (x)

tan (- x) tan (\frac{π}{2} - x) = - 1

tan (z) \cdot cos (z) \cdot csc (z) = 1

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