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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
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

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ajprincess Group TitleBest ResponseYou've already chosen the best response.1
\(\cos2A=2\cos A\sin A\) Find the value of sinA using \(\sin A=\sqrt{1\cos^2A}\). Can u do nw?
 2 years ago

mathsux4real Group TitleBest ResponseYou've already chosen the best response.0
So 2(0.28)(0.276)
 2 years ago

mathsux4real Group TitleBest ResponseYou've already chosen the best response.0
:/ It's not matching up
 2 years ago

ajprincess Group TitleBest ResponseYou've already chosen the best response.1
Find them in the fraction form. Vat do u get for sinA?
 2 years ago

mathsux4real Group TitleBest ResponseYou've already chosen the best response.0
69/250
 2 years ago

muhammad9t5 Group TitleBest ResponseYou've already chosen the best response.1
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.
 2 years ago

ajprincess Group TitleBest ResponseYou've already chosen the best response.1
\(\sin A=\sqrt{1\cos^2A}\) \(\sin A=\sqrt{1(\large\frac{7}{25})^2}\) \(\sin A=\sqrt{1\large\frac{49}{625}}\) \(\sin A=\sqrt{\large\frac{62549}{625}}\) \(\sin A=\sqrt{\large\frac{576}{625}}\) \(\sin A=\large\frac{24}{25}\) Nw can u find cos2A?
 2 years ago

muhammad9t5 Group TitleBest ResponseYou've already chosen the best response.1
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.
 2 years ago

mathsux4real Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

ajprincess Group TitleBest ResponseYou've already chosen the best response.1
U r welcome.:)
 2 years ago

muhammad9t5 Group TitleBest ResponseYou've already chosen the best response.1
hyp^2=base^+perp^2 25^2=7^2+perp^2 625=49+perp^2 62549=perp^2 576=perp^2 24=perp
 2 years ago

muhammad9t5 Group TitleBest ResponseYou've already chosen the best response.1
thanks! you are always welcome friend.
 2 years ago

ajprincess Group TitleBest ResponseYou've already chosen the best response.1
I agree vth u @muhammad9t5. Ur method is easy.
 2 years ago

muhammad9t5 Group TitleBest ResponseYou've already chosen the best response.1
yup. @ajprincess please fan me.
 2 years ago

ajprincess Group TitleBest ResponseYou've already chosen the best response.1
I fanned u. i fan all those who fan me.:)
 2 years ago
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