Question

Last problem I need help with! Please! Using the given data, find the values of sine, cosine, and tangent of 2 A and the quadrant in which 2 A terminates: angle A in Quadrant II, cos A = -7/25
cos 2 A =
a. 527/625
b. -336/625
c. 336/625
d. -527/625

Answer 1

\(\cos2A=2\cos A\sin A\)
Find the value of sinA using
\(\sin A=\sqrt{1-\cos^2A}\).
Can u do nw?

Answer 2

So 2(-0.28)(0.276)

Answer 3

:/ It's not matching up

Answer 4

Find them in the fraction form. Vat do u get for sinA?

Answer 5

69/250

Answer 6

cos=Adjacent/Hypotenuse and sin=Opposite/Hypotenuse and tan=Opposite/ajacent so here you have given just adjacent and hypotenuse and you can fine opposite by Pythagourus Theorum.

Answer 7

\(\sin A=\sqrt{1-\cos^2A}\)
\(\sin A=\sqrt{1-(\large\frac{-7}{25})^2}\)
\(\sin A=\sqrt{1-\large\frac{49}{625}}\)
\(\sin A=\sqrt{\large\frac{625-49}{625}}\)
\(\sin A=\sqrt{\large\frac{576}{625}}\)
\(\sin A=\large\frac{24}{25}\)
Nw can u find cos2A?

Answer 8

its hard method friend. find sin by sin=perp/hyp we need perp so from cos=base/hyp here base=-7 and hyp=25 so now use Pythagoras theorem to find perp.

Answer 9

Yes! -336/625. Thank you so much for step by step explanation! @ajprincess Thank you to @muhammad9t5 as well. I appreciate the time and effort my friends

Answer 10

U r welcome.:)

Answer 11

hyp^2=base^+perp^2
25^2=-7^2+perp^2
625=49+perp^2
625-49=perp^2
576=perp^2
24=perp

Answer 12

thanks! you are always welcome friend.

Answer 13

I agree vth u @muhammad9t5. Ur method is easy.

Answer 14

yup. @ajprincess please fan me.

Answer 15

I fanned u. i fan all those who fan me.:)