Find the exact value of each trigonometric function

Find the exact value of each trigonometric function

2. COS 135 degrees
3. TAN 930 degrees
4. COT 810 degreesÂÂ
5. sec -240 degrees

The correct answer and explanation is :

Let’s go step by step to find the exact values of the given trigonometric functions.

1. ( \cos 135^\circ )

The angle ( 135^\circ ) is in the second quadrant. The reference angle is:
[
180^\circ – 135^\circ = 45^\circ
]
Since cosine is negative in the second quadrant:
[
\cos 135^\circ = -\cos 45^\circ = -\frac{\sqrt{2}}{2}
]

2. ( \tan 930^\circ )

Since tangent has a period of ( 180^\circ ), we reduce ( 930^\circ ) by subtracting multiples of ( 360^\circ ):

[
930^\circ – 2(360^\circ) = 930^\circ – 720^\circ = 210^\circ
]

The reference angle is:

[
210^\circ – 180^\circ = 30^\circ
]

Since tangent is positive in the third quadrant:

[
\tan 210^\circ = \tan 30^\circ = \frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3}
]

Thus,

[
\tan 930^\circ = -\frac{\sqrt{3}}{3}
]

3. ( \cot 810^\circ )

Since cotangent has a period of ( 180^\circ ), we reduce ( 810^\circ ) by subtracting multiples of ( 180^\circ ):

[
810^\circ – 4(180^\circ) = 810^\circ – 720^\circ = 90^\circ
]

Since ( \cot 90^\circ = 0 ), we conclude:

[
\cot 810^\circ = 0
]

4. ( \sec (-240^\circ) )

Since secant is the reciprocal of cosine, we first find ( \cos (-240^\circ) ). Adding ( 360^\circ ) to bring it within a standard range:

[
-240^\circ + 360^\circ = 120^\circ
]

The reference angle is:

[
180^\circ – 120^\circ = 60^\circ
]

Since cosine is negative in the second quadrant:

[
\cos 120^\circ = -\cos 60^\circ = -\frac{1}{2}
]

Now, taking the reciprocal:

[
\sec (-240^\circ) = \sec 120^\circ = \frac{1}{\cos 120^\circ} = \frac{1}{-1/2} = -2
]

Final Answers:

1. ( \cos 135^\circ = -\frac{\sqrt{2}}{2} )
2. ( \tan 930^\circ = -\frac{\sqrt{3}}{3} )
3. ( \cot 810^\circ = 0 )
4. ( \sec (-240^\circ) = -2 )

Explanation (300 words)

To solve these problems, we used fundamental trigonometric concepts, including reference angles, quadrant rules, and periodicity.

1. For ( \cos 135^\circ ), we recognized that ( 135^\circ ) lies in the second quadrant where cosine is negative. Using the reference angle ( 45^\circ ), we found ( \cos 135^\circ = -\frac{\sqrt{2}}{2} ).
2. For ( \tan 930^\circ ), we reduced the angle by subtracting multiples of ( 360^\circ ) to bring it within a standard range, leading to ( 210^\circ ). Since ( 210^\circ ) is in the third quadrant, where tangent is positive, we found ( \tan 210^\circ = -\frac{\sqrt{3}}{3} ).
3. For ( \cot 810^\circ ), we reduced the angle using the periodicity of ( 180^\circ ), bringing it to ( 90^\circ ). Since ( \cot 90^\circ = 0 ), the final answer was 0.
4. For ( \sec (-240^\circ) ), we converted the negative angle to a positive equivalent by adding ( 360^\circ ), leading to ( 120^\circ ). We found ( \cos 120^\circ = -\frac{1}{2} ) and took its reciprocal to get ( \sec 120^\circ = -2 ).

Understanding quadrant signs, reference angles, and periodic properties of trigonometric functions allows us to determine exact values efficiently.