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Adding $0$ to any number always results in the number itself.

$a+0=a$

Because of this, $0$ is called the Additive Identity.

Consider a number $a.$ By the Reflexive Property of Equality, $a$ is equal to itself.

$a=a $

Let $b$ be another number. If $b$ is added to and subtracted from the left-hand side of the above equation, the equality still holds true.
$a=a⇔a+b−b=a $

Finally, $b−b$ is equal to $0.$
$a+b−b=a⇔a+0=a✓ $

It has been shown that $a+0=a.$ By the Commutative Property of Addition, $0+a$ is also equal to $a.$
Any number multiplied by $1$ is equal to the number itself.

$a⋅1=a$

Because of this, the number $1$ is called the Multiplicative Identity.

Consider a number $a.$ By the definition of multiplication, $a$ multiplied by another number $n$ can be written as $n$ times the addition of $a.$

$a⋅n=ntimesa+a+…+a $

If $n=1,$ the sum has only one term.
$a⋅1=1timea $

Therefore, $a⋅1$ is equal to $a.$ Also, by the Commutative Property of Multiplication, $1⋅a=a.$ For any angle $θ,$ the following trigonometric identities hold true.

$sin_{2}θ+cos_{2}θ=1$

$1+tan_{2}θ=sec_{2}θ$

$1+cot_{2}θ=csc_{2}θ$

Consider a right triangle with a hypotenuse of $1.$

Since $cosθ$ represents a side length, it is **not** $0.$ Therefore, by diving both sides of the above equation by $cos_{2}θ,$ the second identity can be obtained.
The second identity was obtained.
Since $sinθ$ represents a side length, it is **not** $0.$ Therefore, by dividing both sides of $sin_{2}θ+cos_{2}θ=1$ by $sin_{2}θ,$ the third identity can be proven.
Finally, the third identity was obtained.

By recalling the sine and cosine ratios, the lengths of the opposite and adjacent sides to $∠θ$ can be expressed in terms of the angle.

Definition | Substitute | Simplify | |
---|---|---|---|

$sinθ$ | $HypotenuseLength ofoppositeside to∠θ $ | $1opp $ | $opp$ |

$cosθ$ | $HypotenuseLength ofadjacentside to∠θ $ | $1adj $ | $adj$ |

It can be seen that if the hypotenuse of a right triangle is $1,$ the sine of an acute angle is equal to the length of its opposite side. Similarly, the cosine of the angle is equal to the length of its adjacent side.

By the Pythagorean Theorem, the sum of the squares of the legs of a right triangle is equal to the square of the hypotenuse. Therefore, for the above triangle, the sum of the squares of $sinθ$ and $cosθ$ is equal to the square of $1.$

$sin_{2}θ+cos_{2}θ⇓sin_{2}θ+cos_{2}θ =1_{2}=1 $

$sin_{2}θ+cos_{2}θ=1$

DivEqn

$LHS/cos_{2}θ=RHS/cos_{2}θ$

$cos_{2}θsin_{2}θ+cos_{2}θ =cos_{2}θ1 $

Simplify

WriteSumFrac

Write as a sum of fractions

$cos_{2}θsin_{2}θ +cos_{2}θcos_{2}θ =cos_{2}θ1 $

QuotOne

$aa =1$

$cos_{2}θsin_{2}θ +1=cos_{2}θ1 $

WritePow

Write as a power

$cos_{2}θsin_{2}θ +1=cos_{2}θ1_{2} $

$b_{m}a_{m} =(ba )_{m}$

$(cosθsinθ )_{2}+1=(cosθ1 )_{2}$

$cosθsinθ =tanθ$

$tan_{2}θ+1=(cosθ1 )_{2}$

$cosθ1 =secθ$

$tan_{2}θ+1=sec_{2}θ$

CommutativePropAdd

Commutative Property of Addition

$1+tan_{2}θ=sec_{2}θ$

$1+tan_{2}θ=sec_{2}θ$

$sin_{2}θ+cos_{2}θ=1$

DivEqn

$LHS/sin_{2}θ=RHS/sin_{2}θ$

$sin_{2}θsin_{2}θ+cos_{2}θ =sin_{2}θ1 $

Simplify

WriteSumFrac

Write as a sum of fractions

$sin_{2}θsin_{2}θ +sin_{2}θcos_{2}θ =sin_{2}θ1 $

QuotOne

$aa =1$

$1+sin_{2}θcos_{2}θ =sin_{2}θ1 $

WritePow

Write as a power

$1+sin_{2}θcos_{2}θ =sin_{2}θ1_{2} $

$b_{m}a_{m} =(ba )_{m}$

$1+(sinθcosθ )_{2}=(sinθ1 )_{2}$

$sinθcosθ =cotθ$

$1+cot_{2}θ=(sinθ1 )_{2}$

$sinθ1 =cscθ$

$1+cot_{2}θ=csc_{2}θ$

$1+cot_{2}θ=csc_{2}θ$

The first identity can be shown using the unit circle and the Pythagorean Theorem. Consider a point $(x,y)$ on the unit circle in the first quadrant, corresponding to the angle $θ.$ A right triangle can be constructed with $θ.$

By the Pythagorean Theorem, the sum of the squares of $x$ and $y$ equals $1.$$x_{2}+y_{2}=1 $

In fact, this is true not only for points in the first quadrant, but for $cos_{2}θ+sin_{2}θ=1$

When a binomial is squared, the resulting expression is a perfect square trinomial.

$(a+b)_{2}=a_{2}+2ab+b_{2}(a−b)_{2}=a_{2}−2ab+b_{2}$

For simplicity, depending on the sign of the binomial, these two identities can be expressed as one.

$(a±b)_{2}=a_{2}±2ab+b_{2}$

This rule will be first proven for $(a+b)_{2}$ and then for $(a−b)_{2}.$

$(a+b)_{2}$

PowToProdTwoFac

$a_{2}=a⋅a$

$(a+b)(a+b)$

Multiply parentheses

Distr

Distribute $(a+b)$

$a(a+b)+b(a+b)$

Distr

Distribute $a$

$a_{2}+ab+b(a+b)$

Distr

Distribute $b$

$a_{2}+ab+ba+b_{2}$

CommutativePropMult

Commutative Property of Multiplication

$a_{2}+ab+ab+b_{2}$

AddTerms

Add terms

$a_{2}+2ab+b_{2}$

$(a+b)_{2}$

PowToProdTwoFac

$a_{2}=a⋅a$

$(a−b)(a−b)$

Multiply parentheses

Distr

Distribute $(a−b)$

$a(a−b)−b(a−b)$

Distr

Distribute $a$

$a_{2}−ab−b(a−b)$

Distr

Distribute $-b$

$a_{2}−ab−ba+b_{2}$

CommutativePropMult

Commutative Property of Multiplication

$a_{2}−ab−ab+b_{2}$

SubTerms

Subtract terms

$a_{2}−2ab+b_{2}$