Question

find the exact value
Given that cos theta = 15/17 with theta quadrant 1 , find sin 2theta and cos 2theta
Plzz help

Answer 1

knowing SOHCAHTOA find the rest.

Answer 2

@shorouq91, you can do it.

Answer 3

If cosT=15/17 and T is in first quadrant then sinT >0 and sin^2T+cos^2T=1 so sin T=8/17. Thus from the fact that sin2a=2sinacosa and cos2a=1-2sin^2a you write :
\[\sin2T=2*cosT*sinT=2*15/17*8/17=2*40/289=80/289\]
\[\cos2T=1-2*\sin^2T=1-2*(5/17)^2=1-50/289=239/289\]

Answer 4

im trying but i cant :(

Answer 5

Hope it helps ;]

Answer 6

yes it is
thank you so much @Lolavitz

Answer 7

who did u get 8

Answer 8

Holy shiet, i forgot it is 15 not 5. So it is 2*15*8/17^2=250/289 My bad, thanks for noticing

Answer 9

i'm confused now

Answer 10

I made a mistake in calculations in multiplying. But the 8/17 i got from the fact that
\[\cos^2T+\sin^2T=1\] which is real for every real number T. Thus we can write \[\sin^2T=1-\cos^2T+\] thus \[\sin^2T=1-225/289=289/289-225/289=(289-225)/289=64/289\] Thus sinT equals either -8/15 or 8/15 and since we know that T is in first quadrant so sinT >0 so sinT=8/15.

Answer 11

\(\bf cos(\theta)=\cfrac{15}{17}\implies \cfrac{\textit{adjacent side}}{hypotenuse}\implies \cfrac{a}{c}\qquad thus\\ \quad \\
a = 15\qquad c= 17\qquad b=\square ?\qquad \textit{using pythagorean theorem}\\ \quad \\
c^2=a^2+b^2\implies \pm\sqrt{c^2-a^2}=b\)

we're told that the angle is in the 1st Quadrant, in the 1st Quadrant, recall "x" and "y" or "cosine" and "sine" are positive, check your Unit Circle, thus the given cosine function is positive

and thus the sine function will also be positive

Answer 12

so once you find what the \(\bf sin(\theta)\) is, then you can find what \(\bf sin(2\theta), cos(2\theta)\) are

Answer 13

recall that \(\bf cos(2\theta)\implies 2cos^2(\theta)-1\qquad \qquad sin(2\theta)=2sin(\theta)cos(\theta)\)

Answer 14

so you're given the cosine , now you'd just need to find the sine function for the angle, so you can plug them in the Double-Angle trig identities

Answer 15

i cant do i

Answer 16

hmmm can't find \(\bf sin(\theta)\quad ?\)

Answer 17

:(

Answer 18

\(\bf cos(\theta)=\cfrac{15}{17}\implies \cfrac{\textit{adjacent side}}{hypotenuse}\implies \cfrac{a}{c}\qquad thus\\ \quad \\
a = 15\qquad c= 17\qquad b=\square ?\qquad \textit{using pythagorean theorem}\\ \quad \\
c^2=a^2+b^2\implies \pm\color{red}{\sqrt{c^2-a^2}=b}\)

keep in mind that \(\bf sin(\theta)=\cfrac{b}{c}\)

Answer 19

8/17

Answer 20

yeap

Answer 21

sin^2theta - cos^2 theta =1

Answer 22

now you know what the cosine and sine values are thus \(\bf cos(2\theta)\implies 2cos^2(\theta)-1\qquad \qquad sin(2\theta)=2sin(\theta)cos(\theta)\)

Answer 23

hmmm are you after the \(\bf cos^2(\theta) \ or \ cos(2\theta)\quad ?\)

Answer 24

im rally confused :(

Answer 25

1-225/289=
289/289-225/289=
(289-225)/289=64/289

Answer 26

well... what part confuses you?

Answer 27

Sho ( your nickname is too long i will just call you that ,you don't mind do you ? ) the equasion is sin^2theta + cos^2 theta =1. You add the values not substract them. From that you learn the value of sin^2theta, from which you know the value of sintheta. Now that you know both sinthetha and costhetha, you use their values while using the formula that jdoe gave you :
cos(2θ)⟹2cos^2(θ)−1
sin(2θ)=2sin(θ)cos(θ)

Answer 28

i cant do it

Answer 29

what do i have to do after when i got sin 8/17

Answer 30

well, are after \(\bf cos^2(\theta) \ or \ cos(2\theta)\quad ?\)

Answer 31

i dont know which way i have to start

Answer 32

You know the values of cosT and sin T. There are said formulas :
cos(2θ)=2cos^2(θ)−1
sin(2θ)=2sin(θ)cos(θ)
Now you just substitue the functions with their values .

Answer 33

cos 2 theta =2 cos^2 theta
15/17*2=2(15/17)^2

Answer 34

No, cos2T=2*cos^2T-1. You forgot to substract the 1.

Answer 35

30/17=161/289

Answer 36

cos2T=2*cos^2T-1=2*(15/17)^2-1=239/289
and
sin2T=2*cosT*sinT=2*(8/17)*(15/17)=240/289

Answer 37

cos 2 T = 161/289