Solved on Nov 07, 2023

Find the value of $\cot(2A)$ given that $\cos A = -\frac{3}{\sqrt{10}}$ and $A$ is in the second quadrant.

STEP 1

Assumptions1. The value of $\cos A$ is given as $-\frac{3}{\sqrt{10}}$ . The angle $A$ is in the second quadrant (QII)
3. We need to find the value of $\cot (A)$

STEP 2

In the second quadrant, cosine is negative and sine is positive. Since we know the value of $\cos A$ , we can use the Pythagorean identity to find the value of $\sin A$ .
The Pythagorean identity is $\sin^2 A + \cos^2 A =1$

STEP 3

Rearrange the Pythagorean identity to solve for $\sin A$ $\sin^2 A =1 - \cos^2 A$

STEP 4

Substitute the given value of $\cos A$ into the equation $\sin^2 A =1 - \left(-\frac{3}{\sqrt{10}}\right)^2$

STEP 5

olve the equation to find $\sin^2 A$ $\sin^2 A =1 - \frac{9}{10} = \frac{1}{10}$

STEP 6

Since $\sin A$ is positive in the second quadrant, we take the positive square root to find $\sin A$ $\sin A = \sqrt{\frac{1}{10}} = \frac{1}{\sqrt{10}}$

STEP 7

Now we need to find $\cot (2A)$ , which can be expressed in terms of $\sin A$ and $\cos A$ using the double angle formula for tangent and the reciprocal identity for cotangent. The double angle formula for tangent is $\tan (2A) = \frac{2 \sin A \cos A}{1 - \sin^2 A}$ And the reciprocal identity for cotangent is $\cot x = \frac{1}{\tan x}$

STEP 8

Substitute the values of $\sin A$ and $\cos A$ into the double angle formula for tangent $\tan (2A) = \frac{2 \left(\frac{1}{\sqrt{10}}\right) \left(-\frac{3}{\sqrt{10}}\right)}{1 - \left(\frac{1}{\sqrt{10}}\right)^2}$

STEP 9

implify the equation to find $\tan (2A)$ $\tan (2A) = \frac{-2 \times3}{ -} = -\frac{6}{9} = -\frac{2}{3}$

STEP 10

Finally, use the reciprocal identity for cotangent to find $\cot (2A)$ $\cot (2A) = \frac{}{\tan (2A)}$

STEP 11

Substitute the value of $\tan (A)$ into the equation $\cot (A) = \frac{}{-\frac{}{3}}$

STEP 12

olve the equation to find $\cot (2A)$ $\cot (2A) = -\frac{}{2}$ So, $\cot (2A) = -\frac{}{2}$ .

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Find the value of $\cot(2A)$ given that $\cos A = -\frac{3}{\sqrt{10}}$ and $A$ is in the second quadrant.

STEP 1

Assumptions1. The value of $\cos A$ is given as $-\frac{3}{\sqrt{10}}$ . The angle $A$ is in the second quadrant (QII)
3. We need to find the value of $\cot (A)$

STEP 2

In the second quadrant, cosine is negative and sine is positive. Since we know the value of $\cos A$ , we can use the Pythagorean identity to find the value of $\sin A$ .
The Pythagorean identity is $\sin^2 A + \cos^2 A =1$

STEP 3

Rearrange the Pythagorean identity to solve for $\sin A$ $\sin^2 A =1 - \cos^2 A$

STEP 4

Substitute the given value of $\cos A$ into the equation $\sin^2 A =1 - \left(-\frac{3}{\sqrt{10}}\right)^2$

STEP 5

olve the equation to find $\sin^2 A$ $\sin^2 A =1 - \frac{9}{10} = \frac{1}{10}$

STEP 6

Since $\sin A$ is positive in the second quadrant, we take the positive square root to find $\sin A$ $\sin A = \sqrt{\frac{1}{10}} = \frac{1}{\sqrt{10}}$

STEP 7

STEP 8

Substitute the values of $\sin A$ and $\cos A$ into the double angle formula for tangent $\tan (2A) = \frac{2 \left(\frac{1}{\sqrt{10}}\right) \left(-\frac{3}{\sqrt{10}}\right)}{1 - \left(\frac{1}{\sqrt{10}}\right)^2}$

STEP 9

implify the equation to find $\tan (2A)$ $\tan (2A) = \frac{-2 \times3}{ -} = -\frac{6}{9} = -\frac{2}{3}$

STEP 10

Finally, use the reciprocal identity for cotangent to find $\cot (2A)$ $\cot (2A) = \frac{}{\tan (2A)}$

STEP 11

Substitute the value of $\tan (A)$ into the equation $\cot (A) = \frac{}{-\frac{}{3}}$

STEP 12

olve the equation to find $\cot (2A)$ $\cot (2A) = -\frac{}{2}$ So, $\cot (2A) = -\frac{}{2}$ .

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Find the value of cot⁡(2A)\cot(2A)cot(2A) given that cos⁡A=−310\cos A = -\frac{3}{\sqrt{10}}cosA=−10​3​ and AAA is in the second quadrant.

STEP 1

Assumptions1. The value of cos⁡A\cos AcosA is given as −310-\frac{3}{\sqrt{10}}−10​3​ . The angle AAA is in the second quadrant (QII)3. We need to find the value of cot⁡(A)\cot (A)cot(A)

STEP 2

In the second quadrant, cosine is negative and sine is positive. Since we know the value of cos⁡A\cos AcosA, we can use the Pythagorean identity to find the value of sin⁡A\sin AsinA.The Pythagorean identity issin⁡2A+cos⁡2A=1\sin^2 A + \cos^2 A =1sin2A+cos2A=1

STEP 3

Rearrange the Pythagorean identity to solve for sin⁡A\sin AsinAsin⁡2A=1−cos⁡2A\sin^2 A =1 - \cos^2 Asin2A=1−cos2A

STEP 4

Substitute the given value of cos⁡A\cos AcosA into the equationsin⁡2A=1−(−310)2\sin^2 A =1 - \left(-\frac{3}{\sqrt{10}}\right)^2sin2A=1−(−10​3​)2

STEP 5

olve the equation to find sin⁡2A\sin^2 Asin2Asin⁡2A=1−910=110\sin^2 A =1 - \frac{9}{10} = \frac{1}{10}sin2A=1−109​=101​

STEP 6

Since sin⁡A\sin AsinA is positive in the second quadrant, we take the positive square root to find sin⁡A\sin AsinAsin⁡A=110=110\sin A = \sqrt{\frac{1}{10}} = \frac{1}{\sqrt{10}}sinA=101​​=10​1​

STEP 7

STEP 8

STEP 9

implify the equation to find tan⁡(2A)\tan (2A)tan(2A)tan⁡(2A)=−2×3−=−69=−23\tan (2A) = \frac{-2 \times3}{ -} = -\frac{6}{9} = -\frac{2}{3}tan(2A)=−−2×3​=−96​=−32​

STEP 10

Finally, use the reciprocal identity for cotangent to find cot⁡(2A)\cot (2A)cot(2A)cot⁡(2A)=tan⁡(2A)\cot (2A) = \frac{}{\tan (2A)}cot(2A)=tan(2A)​

STEP 11

Substitute the value of tan⁡(A)\tan (A)tan(A) into the equationcot⁡(A)=−3\cot (A) = \frac{}{-\frac{}{3}}cot(A)=−3​​

STEP 12

olve the equation to find cot⁡(2A)\cot (2A)cot(2A)cot⁡(2A)=−2\cot (2A) = -\frac{}{2}cot(2A)=−2​So, cot⁡(2A)=−2\cot (2A) = -\frac{}{2}cot(2A)=−2​.

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Find the value of $\cot(2A)$ given that $\cos A = -\frac{3}{\sqrt{10}}$ and $A$ is in the second quadrant.

Assumptions1. The value of $\cos A$ is given as $-\frac{3}{\sqrt{10}}$ . The angle $A$ is in the second quadrant (QII)
3. We need to find the value of $\cot (A)$

In the second quadrant, cosine is negative and sine is positive. Since we know the value of $\cos A$ , we can use the Pythagorean identity to find the value of $\sin A$ .
The Pythagorean identity is $\sin^2 A + \cos^2 A =1$

Rearrange the Pythagorean identity to solve for $\sin A$ $\sin^2 A =1 - \cos^2 A$

Substitute the given value of $\cos A$ into the equation $\sin^2 A =1 - \left(-\frac{3}{\sqrt{10}}\right)^2$

olve the equation to find $\sin^2 A$ $\sin^2 A =1 - \frac{9}{10} = \frac{1}{10}$

Since $\sin A$ is positive in the second quadrant, we take the positive square root to find $\sin A$ $\sin A = \sqrt{\frac{1}{10}} = \frac{1}{\sqrt{10}}$

implify the equation to find $\tan (2A)$ $\tan (2A) = \frac{-2 \times3}{ -} = -\frac{6}{9} = -\frac{2}{3}$

Finally, use the reciprocal identity for cotangent to find $\cot (2A)$ $\cot (2A) = \frac{}{\tan (2A)}$

Substitute the value of $\tan (A)$ into the equation $\cot (A) = \frac{}{-\frac{}{3}}$

olve the equation to find $\cot (2A)$ $\cot (2A) = -\frac{}{2}$ So, $\cot (2A) = -\frac{}{2}$ .