## mathsux4real Group Title 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 2 years ago 2 years ago

1. ajprincess Group Title

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

2. mathsux4real Group Title

So 2(-0.28)(0.276)

3. mathsux4real Group Title

:/ It's not matching up

4. ajprincess Group Title

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

5. mathsux4real Group Title

69/250

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.

7. ajprincess Group Title

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

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.

9. mathsux4real Group Title

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

10. ajprincess Group Title

U r welcome.:)

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

thanks! you are always welcome friend.

13. ajprincess Group Title

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