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# Using Trigonometric Identities

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### Direct messages

Taking a closer look at how the trigonometric functions are defined, it is possible to find a range of identities relating the different trigonometric functions. These are highly useful when rewriting and manipulating trigonometric expressions.

## Reciprocal Identities

The trigonometric ratios cosecant, secant, and cotangent are reciprocals of sine, cosine, and tangent, respectively. Therefore, they can be defined using their respective reciprocal.

### Proof

Consider a right triangle with the three sides labeled with respect to an acute angle

Next, the sine, cosine, tangent, cosecant, secant, and cotangent ratios are written.
The reciprocal of the sine ratio will now be calculated.
Solve for
It has been found that which is the reciprocal of is equal to Therefore, by the Transitive Property of Equality, is equal to
By following a similar procedure, it can be proven that and that

## Tangent and Cotangent Identities

Both tangent and cotangent can be expressed as the ratios of sine and cosine.

### Proof

Tangent Identity

Tangent of the angle is defined as the ratio of the lengths of the opposite side a and the adjacent side b.

By manipulating the right-hand side, it can be expressed as sine over cosine instead. The proof of the identity is complete now.

### Proof

Cotangent Identity

Cotangent of the angle is defined as the ratio of the lengths of the adjacent side a and the opposite side b.

By manipulating the right-hand side, it can be expressed as sine over cosine instead.

## Pythagorean Identities

Sine and cosine values always follow a useful relation called the Pythagorean Identity. The sum of the squared sine and cosine values is equal to 1, regardless of the angle.

This 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 A right triangle can be constructed with

By the Pythagorean Theorem, it follows that
x2+y2=1.
In fact, this is true for every point on the Unit Circle, not only for points in the first quadrant. Recall that, for points (x,y) on the Unit Circle corresponding to the angle
Substituting these into the previous equality gives the Pythagorean Identity, short of rearranging the terms:
Note that represents and similarly for Dividing both sides by either or leads to two variations of the Pythagorean Identity.

## Cofunction Identities

Studying angles and their trigonometric values in a right triangle more closely reveals further relationships between sine and cosine, and between tangent and cotangent.

### Rule

The angles in a right triangle, expressed in radians, is the unknown and The third angle comes from that the sum of the angles in a triangle is π.

Cosine of can now be expressed using the angle's adjacent side and the hypotenuse:
Sine of the opposite angle can be similarly expressed as
As the right-hand sides are equal, so must the left-hand sides be, leading to the identity.

This identity is true for all angles, not just those possible to construct in a right triangle.

Using similar reasoning, the following two identities can also be found.

## Negative Angle Identities

The function has odd symmetry, and has even symmetry, which can be seen from their graphs. Thus, the corresponding identities must hold true.

Using these two identities, it can be shown that has odd symmetry.

### Rule

By expressing using sine and cosine, this identity can be shown.

## Find the trigonometric values

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Given that and find the values of and without determining

Show Solution expand_more

To begin, we can find the value of using the Pythagorean Identity. By substituting we get an equation that is solvable for the value of

Notice that there are two values for From the prompt, we know that must be between 0 and corresponding to the first quadrant on the Unit Circle. For these angles, is positive. Thus, the negative solution can be discarded.
Using the sine and cosine values, can be found.
The remaining values are the reciprocals of the ones we've found so far. Thus, we can swap the numerator with the denominator for the fractions to find the desired values.
As the tangent value isn't expressed as a fraction, it can simply be substituted.
We have now found all the desired values.