Question

If
`cosA=(2)/(5)` and
`tanA\ \textless \ 0` then
`sin(2A)=?`

Please help
Absurd answers will not be tolerated

Answer 1

sin 2A=2sin Acos A

=2×2/5×√21/5=±4√21/25

Explanation:

cos A=2/5

B/H=2/5

corresponding value we get

B=2

H=5

by using Pythagoras law

H²=P²+B²

5²=P²+2²

P²=25-4

p=√21

now

sin A=P/H=√21/5

now

sin 2A=2sin Acos A=2×2/5×√21/5=±4√21/25

Answer 2

`\displaystyle \sin(2A)=(-4√(21))/(25)`

Explanation:

We are given that:

`\displaystyle \cos(A)=(2)/(5)\text{ and } \tan(A)<0`

And we want to determine the value of:

`\sin(2A)`

First, since cos(A) is positive and tan(A) is negative, this means that ∠A must be in QIV.

In QIV, cosine is positive, sine is negative, and tangent is also negative.

The given ratio tells us that the adjacent side to ∠A is 2, and the hypotenuse is 5.

Then by the Pythagorean Theorem, the opposite side will be given by:

`(2)^2+o^2=(5)^2`

Solve for the opposide side:

`o=√(5^2-2^2)=√(21)`

So, with respect to ∠A, the adjacent side is 2, the opposite side is √(21), and the hypotenuse is 5.

Using the double-angle identity, we can rewrite our original expression as:

`\sin(2A)=2\sin(A)\cos(A)`

Using the above information, substitute in values. Remember that cosine is positive and that sine is negative:

`\displaystyle \sin(2A)=2\Big(-(√(21))/(5)\Big)\Big((2)/(5)\Big)`

Simplify. Therefore:

`\displaystyle \sin(2A)=(-4√(21))/(25)`