Math  /  Trigonometry

QuestionWhat is the exact value for cot(165)\cot \left(165^{\circ}\right) ?

Studdy Solution

STEP 1

1. The cotangent function is the reciprocal of the tangent function.
2. The angle 165165^\circ can be expressed using known angles for which trigonometric values are known.
3. We can use the angle subtraction identity for tangent to find the exact value.

STEP 2

1. Express 165165^\circ in terms of known angles.
2. Use the tangent subtraction identity to find tan(165)\tan(165^\circ).
3. Calculate cot(165)\cot(165^\circ) as the reciprocal of tan(165)\tan(165^\circ).

STEP 3

Express 165165^\circ as a combination of known angles. We can write:
165=18015 165^\circ = 180^\circ - 15^\circ
This allows us to use the identity for tangent of a difference.

STEP 4

Use the tangent subtraction identity:
tan(180θ)=tan(θ)\tan(180^\circ - \theta) = -\tan(\theta)
Thus,
tan(165)=tan(18015)=tan(15)\tan(165^\circ) = \tan(180^\circ - 15^\circ) = -\tan(15^\circ)

STEP 5

Find tan(15)\tan(15^\circ) using the tangent subtraction formula:
tan(15)=tan(4530)=tan(45)tan(30)1+tan(45)tan(30)\tan(15^\circ) = \tan(45^\circ - 30^\circ) = \frac{\tan(45^\circ) - \tan(30^\circ)}{1 + \tan(45^\circ)\tan(30^\circ)}
Using known values tan(45)=1\tan(45^\circ) = 1 and tan(30)=13\tan(30^\circ) = \frac{1}{\sqrt{3}}, we have:
tan(15)=1131+113=313+1\tan(15^\circ) = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}
Rationalize the denominator:
tan(15)=(31)(31)(3+1)(31)=323+12=23\tan(15^\circ) = \frac{(\sqrt{3} - 1)(\sqrt{3} - 1)}{(\sqrt{3} + 1)(\sqrt{3} - 1)} = \frac{3 - 2\sqrt{3} + 1}{2} = 2 - \sqrt{3}
Thus,
tan(165)=(23)\tan(165^\circ) = -(2 - \sqrt{3})

STEP 6

Calculate cot(165)\cot(165^\circ) as the reciprocal of tan(165)\tan(165^\circ):
cot(165)=1tan(165)=1(23)=132\cot(165^\circ) = \frac{1}{\tan(165^\circ)} = \frac{1}{-(2 - \sqrt{3})} = \frac{1}{\sqrt{3} - 2}
Rationalize the denominator:
cot(165)=1323+23+2=3+2(32)(3+2)=3+234=3+21=(3+2)\cot(165^\circ) = \frac{1}{\sqrt{3} - 2} \cdot \frac{\sqrt{3} + 2}{\sqrt{3} + 2} = \frac{\sqrt{3} + 2}{(\sqrt{3} - 2)(\sqrt{3} + 2)} = \frac{\sqrt{3} + 2}{3 - 4} = \frac{\sqrt{3} + 2}{-1} = -(\sqrt{3} + 2)
Thus, the exact value is:
32\boxed{-\sqrt{3} - 2}

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