Question

Let cosA= 1/root10

and sin A < 0. Find sin (2A), cos (2A) and tan (2A).
V10
Give exact answers as fully-simplified fractions, if necessary.
sin (2A) =
cos (2A) =
tan (2A) =

Answer 1

`\sin(2A)=-(3)/(5)`

`\cos(2A)=-(4)/(5)`

`\tan (2A)=(3)/(4)`

Explanation:

Given:

`\cos A=(1)/(√(10))`

`\sin A < 0`

The cosine of an angle in a right triangle is defined as the ratio of the length of the side adjacent the angle to the length of the hypotenuse:

`\cos(x)=\sf (Adjacent\;side)/(Hypotenuse)`

Given that cos(A) = 1/√(10)​, we can create a right triangle where the side adjacent to angle A is 1, and the hypotenuse is √(10).

To find the length of the side opposite angle A, we can use the Pythagorean Theorem:

`O=\sqrt{\left(√(10)\right)^2-1^2}`

`O=√(10-1)`

`O=√(9)`

`O=3`

If cos(A) is positive and sin(A) is negative, then angle A can be found in quadrant IV of the Cartesian coordinate system. Therefore, both sin(A) and tan(A) will be negative:

`\sin(A)=\sf -(Opposide\;side)/(Hypotenuse)=-(3)/(√(10))`

`\tan(A)=(\sin(A))/(\cos(A))=(-(3)/(√(10)))/((1)/(√(10)))=-3`

To find the exact values of sin(2A), cos(2A) and tan(2A), we can use the double angle identities:

`\boxed{\begin{array}{l}\underline{\textsf{Double Angle Identities}}\\\\\sin (2x)= 2 \sin x\cos x\\\\\cos (2x)=\cos^2 x- \sin^2 x\\\\\tan (2 x)=(2\tan x)/(1 - \tan^2x)\\\\\end{array}}`

To find the exact value of sin(2A), substitute the values of sin(A) and cos(A) into the sine double angle identity:

`\sin(2A)=2\left(-(3)/(√(10))\right)\left((1)/(√(10))\right)`

`\sin(2A)=-(2\cdot 3\cdot 1)/(√(10)\cdot√(10))`

`\sin(2A)=-(6)/(10)`

`\boxed{\boxed{\sin(2A)=-(3)/(5)}}`

To find the exact value of cos(2A), substitute the values of cos(A) and sin(A) into the cosine double angle identity:

`\cos(2A)=\left((1)/(√(10))\right)^2-\left(-(3)/(√(10))\right)^2`

`\cos(2A)=(1)/(10)-(9)/(10)`

`\cos(2A)=(1-9)/(10)`

`\cos(2A)=-(8)/(10)`

`\boxed{\boxed{\cos(2A)=-(4)/(5)}}`

To find the exact value of tan(2A), substitute the value of tan(A) into the tangent double angle identity:

`\tan (2A)=(2(-3))/(1 -(-3)^2)`

`\tan (2A)=(-6)/(1 -9)`

`\tan (2A)=(-6)/(-8)`

`\boxed{\boxed{\tan (2A)=(3)/(4)}}`

Answer 2

1.
`\bf{\sin (2A) = -(3)/(5)}`

2.
`\bf{\cos (2A) = -(4)/(5)}`

3.
`\bf{\tan (2A) = (3)/(4)}`

Explanation:

Given that
`\tt{\cos A = (1)/(√(10))}` and
`\tt{\sin A < 0}`, we can find the values of
`\tt{\sin (2A)}`,
`\tt{\cos (2A)}`, and
`\tt{\tan (2A)}` using the double-angle formulas.

The double-angle formulas are:

1.
`\tt{\sin (2A) = 2 \sin A \cos A}`

2.
`\tt{\cos (2A) = \cos^2 A - \sin^2 A}`

3.
`\tt{\tan (2A) = (\sin (2A))/(\cos (2A))}`

First, we need to find
`\tt{\sin A}`. We know that
`\tt{\sin^2 A + \cos^2 A = 1}`, so we can solve for
`\tt{\sin A}`:

• 
  `\tt{\sin A = √(1 - \cos^2 A) = \sqrt{1 - \left((1)/(√(10))\right)^2} = \sqrt{1 - (1)/(10)} = \sqrt{(9)/(10)} = (3)/(√(10))}`

Since we're given that
`\tt{\sin A < 0}`, we take the negative root:

• 
  `\tt{\sin A = -(3)/(√(10))}`

Now we can find
`\tt{\sin (2A)}`,
`\tt{\cos (2A)}`, and
`\tt{\tan (2A)}`:

1.
`\tt{\sin (2A) = 2 \sin A \cos A = 2 \left(-(3)/(√(10))\right) \left((1)/(√(10))\right) = -(6)/(10) = -(3)/(5)}`

2.
`\tt{\cos (2A) = \cos^2 A - \sin^2 A = \left((1)/(√(10))\right)^2 - \left(-(3)/(√(10))\right)^2 = (1)/(10) - (9)/(10) = -(8)/(10) = -(4)/(5)}`

3.
`\tt{\tan (2A) = (\sin (2A))/(\cos (2A)) = (-(3)/(5))/(-(4)/(5)) = (3)/(4)}`

So, the exact values are:

1.
`\tt{\sin (2A) = -(3)/(5)}`

2.
`\tt{\cos (2A) = -(4)/(5)}`

3.
`\tt{\tan (2A) = (3)/(4)}`

#BTH1

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