Question

cosA=-3/5, pi<=A,=3pi/2 find sin2A

Answer 1

Answer:

Given that cosA=-3/5

and

$\pi \leqslant A < \frac{3 \pi}{2} \\ \implies \: A \: lies \: in \: 3rd \: quadrant$

As ,

$sin^{2} A \: + cos {}^{2} A = 1 \\ \implies \: ( \frac{ - 3}{5}) ^{2} + sin {}^{2} A = 1 \\ \implies \: sin {}^{2} A = \frac{16}{25} \\ \implies \: sinA = \pm \frac{4}{5}$

As x € 3rd quadrant

⟹ only cot and tan will be positive and other trigonometric ratios will be negative hence SinA will be negative.

So,

$sinA = \frac{ - 4}{5}$

Now using formula

$sin2A = 2sinAcosA$

$sin2A = 2 \times \frac{ - 4}{5} \times \frac{ - 3}{5} \\ = \frac{24}{25}$

Answer 2

$\underline{\green{\textsf{Answer:-}}}$

$\pink{\boxed{\blue{sin2A = \dfrac{24}{25}}}}$

$\underline{\green{\textsf{Explanation:-}}}$

Given

$cosA = \dfrac{-3}{5}$

where,

$\pi ≤ A ≤ \dfrac{3 \pi}{2}$

•°• A is in Q-lll

To Find:

sin2A

Solution

From Pythagoras theorem

Hypoteneuse² = Adj. side² + Opp. side²

5² = 3² + x²

x² = 5²-3²

x² = 25-9

x = √16

x = 4

Opp. side = 4

$sinA = \dfrac{opp.side}{Hypoteneuse}$

$sinA = \dfrac{-4}{5}$

Since, A is in Q-lll

["sin " is negative]

w.k.t,

$\boxed{sin2A = 2 sinA \: cosA}$

$sin2A = 2 \times \dfrac{-4}{5} \times \dfrac{-3}{5}$

$sin2A = 2 \times \dfrac{12}{25}$

$\blue{\boxed{\red{\green{sin\orange{2} \pink{A} }= \dfrac{24}{\blue{25}}}}}$