Question

If sinA=4/5 solve sin2A, cos2A and tan2A​

Answer 1

Step-by-step explanation:

1) if m(∠A)∈[0;90°), then

$cos(A)=\sqrt{1-sin^2A} =\frac{3}{5};$

$sin2A=2sinAcosA=2*\frac{3}{5} *\frac{4}{5}=\frac{24}{25};$

$cos2A=cos^2A-sin^2A=\frac{9}{25}-\frac{16}{25}=-\frac{7}{25};$

$tan2A=\frac{sin2A}{cos2A}=-\frac{\frac{24}{25}}{\frac{7}{25}}=-\frac{24}{7}.$

2) if m∠A∈[90°;180°), then

cos(A)=-0.6;

sin2A=-0.96;

cos2A=-0.28;

tan2A=-24/7.