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# Pythagorean Identities

Pythagorean Identities are identities that are derived from the Pythagorean theorem. There are three important Pythagorean Identities.

sin2 x + cos2 x = 1

tan2 x + 1 = sec2 x

cot2x + 1 = csc2 x

We can derive all these trigonometric identities from any one of the others using simple algebraic equations.

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## Pythagorean Identities Proof

Proof of Pythagorean identities is given below:

Identity 1: sin$^2$x + cos$^2$x = 1

Proof:

In a Right-Angle Triangle ABC, let AC be the hypotenuse, AB be the opposite side, and BC be the adjacent side.

Then, sin x = $\frac{\text{opp}}{\text{hyp}}$ => $\frac{AB}{AC}$

cos x = $\frac{\text{adj}}{\text{hyp}}$ => $\frac{BC}{AC}$

Now,

sin2x + cos2x = $\frac{AB^2}{AC^2} + \frac{BC^2}{AC^2}$

= $\frac{AB^2 + BC^2}{AC^2}$

= $\frac{AC^2}{AC^2}$
[Using Pythagorean theorem ]

= 1

Therefore, sin2x + cos2x = 1.

Hence proved.

Identity 2: tan2x + 1 = sec2x

Proof:

Let us consider Identity 1, sin2x + cos2x = 1

Dividing the equation by cos2x,

$\frac{\sin^2x}{\cos^2x} + \frac{\cos^2x}{\cos^2x} = \frac{1}{\cos^2x}$

tan2x + 1 = sec2x

Hence proved.

Identity 3: 1 + cot2x = csc2x

Proof:

Let us consider Identity1, sin2x + cos2x = 1

Dividing the equation by sin2x,

$\frac{\sin^2x}{\sin^2x} + \frac{\cos^2x}{\sin^2x} = \frac{1}{\sin^2x}$

1 + cot2x = csc2x

Hence proved.

## Solving Pythagorean Identities

Given below are some of the examples in solving pythagorean identities.

### Solved Examples

Question 1:

In a triangle ABC, we have that sin X = $\frac{4}{5}$, what is the value of cos x?

Solution:

We have the basic trigonometric identity:

sin2 x + cos2 x = 1

In our case, we know sin x = $\frac{4}{5}$, so we replace in the previous equation to get:

cos2 x = 1 - $(\frac{4}{5})^2$

cos2 x = 1 - $\frac{16}{25}$ = $\frac{9}{25}$

cos x = $\frac{3}{5}$.

Question 2:

If csc x = $\frac{5}{3}$ and tan x = $\frac{3}{4}$, find the values of remaining trigonometric functions, using Pythagorean identity.

Solution:

Given:
csc x = $\frac{5}{3}$ and tan x = $\frac{3}{4}$

Step 1: We know, cot$^2$x + 1 = csc$^2$x

cot$^2$x + 1 = $(\frac{5}{3})^2$

cot$^2$x + 1 = $\frac{25}{9}$

cot$^2$x = $\frac{25}{9}$ - 1 = $\frac{16}{9}$

cot x = $\frac{4}{3}$

Step 2: tan$^2$x + 1 = sec$^2$x

$(\frac{3}{4})^2$ + 1 = sec$^2$x

$\frac{9}{16}$ + 1 = sec$^2$x

$\frac{25}{16}$ = sec$^2$x

or sec x = $\frac{5}{4}$

Step 3: sin x = $\frac{1}{csc\ x}$ = $\frac{1}{\frac{5}{3}}$ = $\frac{3}{5}$

=> sin x = $\frac{3}{5}$

Step 4: sin$^2$x + cos$^2$x = 1

$(\frac{3}{5})^2$ + cos$^2$x = 1

$\frac{9}{25}$ + cos$^2$x = 1

cos$^2$x = 1 - $\frac{9}{25}$ = $\frac{16}{25}$

cos x = $\frac{4}{5}$

## Pythagorean Identities Problems

Below are some solved examples using Pythagorean identities.

### Solved Examples

Question 1: Simplify the expression (1 - cos2x) (csc x) to a single trigonometric function.
Solution:

We know sin2x + cos2x = 1, 1 - cos2x = sin2x and cscx = $\frac{1}{\sin x}$

So, the function becomes, (sin2 x)$\frac{1}{\sin x}$ = sin x

=> (1 - cos2x) (csc x) = sinx

Question 2: Solve by using trigonometric functions : $\frac{\csc a \cos^2 a}{1 + \csc a}$
Solution:
$\frac{\csc a \cos^2 a}{1 + \csc a}$ =$\cos^2a$ $\frac{\frac{1}{\sin a}}{1 + \frac{1}{\sin a}}$ [ csc (a) = $\frac{1}{\sin(a)}$ ]

= $\frac{\cos^2a}{\sin a + 1}$

= $\frac{1 - \sin^2a}{\sin a + 1}$ [ Here, $cos^2a = 1 - sin^2a$ ]

= $\frac{(1 - \sin a)(1 + \sin a)}{\sin a + 1}$

= 1 - $\sin a$

Question 3: In a triangle ABC, we have that cos Y = $\frac{5}{13}$, what is the values of sin Y?
Solution:

We have the basic trigonometric identity, Sin2 A + cos2 A = 1

In our case, we know cos Y = $\frac{5}{13}$

sin2 Y = 1 - $(\frac{5}{13})^2$

sin2 Y = 1 - $\frac{25}{169}$ = $\frac{144}{169}$

sin Y = $\pm$ $\sqrt{\frac{144}{169}}$ = -$\frac{12}{13}$.

(Because cos Y is +ve, lies in lV quadrant. Value of sin Y is -ve in this quadrant.
)

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