Question

if $cosA=\frac{2}{5}$ and $tanA\ \textless \ 0$ then $sin(2A)=?$

Please help
Absurd answers will not be tolerated

Answer 1

Answer:

$\displaystyle \sin(2A)=\frac{-4\sqrt{21}}{25}$

Step-by-step explanation:

We are given that:

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

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:

$o=\sqrt{5^2-2^2}=\sqrt{21}$

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:

$\displaystyle \sin(2A)=2\Big(-\frac{\sqrt{21}}{5}\Big)\Big(\frac{2}{5}\Big)$

Simplify. Therefore:

$\displaystyle \sin(2A)=\frac{-4\sqrt{21}}{25}$

Answer 2

Answer:

sin 2A=2sin Acos A

=2×2/5×√21/5=±4√21/25

Step-by-step 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