Question

Find sin & tan if cos=2/3 & cot>0

Answer 1

\[\sin ^{2} + \cos ^{2} =1\]

Answer 2

cos(x) = adjacent / hypotenuse = 2 / 3
BC = sqrt(3^2 - 2^2) = sqrt(9-4) = sqrt(5)
tan(x) = opposite / adjacent = ?

Answer 3

tan = sin /cos

Answer 4

what would BC be?

Answer 5

to find what BC is, you gotta use Pythagoreans Theorem. \[a^2 + b^2 = c^2\] so plug in the numbers you have and solve for little b

Answer 6

So little b is =5

Answer 7

not quite, you'd get \[b^2 = 5\] but we want b not b squared

Answer 8

Do you see the formula I gave above sin ^2 + cos^2 = 1. You know cos so you can get sin.

Answer 9

and how would i set up my equation to find sin?

Answer 10

sin^2 + (2/3) ^2 = 1 thus 1 - (2/3)^2 = sin^ 2.

Answer 11

yeah i still don't get it

Answer 12

\[\sin ^{2} + (2/3) ^{2} =1...... So..... 1- (2/3)^{2}=\sin ^{2}\]

Answer 13

Figure what 2/3 squared is and subtract it from 1.

Answer 14

I got .4444444

Answer 15

Need to keep in fractions ..but .4444 = 4/9 so we subtract that from 1 and we get what?

Answer 16

14/25

Answer 17

What @nelsonjedi is saying is there is a theorem that states that \[\sin^2(x) + \cos^2(x) = 1\] we know that \[\cos(x) = \frac{2}{3}\] but the equation i stated above only has \[\cos^2(x)\] so we need to change our cos to cos^2 and you do that by squaring both sides\[(\cos(x))^2 = (\frac{2}{3})^2\] which gives us \[\cos^2(x) = \frac{2^2}{3^2} = \frac{4}{9}\] so \[\cos^2(x) = \frac{4}{9}\] now we can plug that into our equation like this \[\sin^2(x) + \frac{4}{9} = 1\] solve for sin^2(x) \[\sin^2(x) = 1 - \frac{4}{9}\] \[\sin^2(x) = \frac{5}{9}\] but we want sin(x) not sin^2(x) so take the square root on both sides \[\sqrt{\sin^2(x)} = \sqrt{\frac{5}{9}}\] which equals \[\sin(x) = \frac{\sqrt{5}}{\sqrt{9}} = \frac{\sqrt{5}}{3}\]

Answer 18

Thank you Tyler...

Answer 19

Now that you have sin you can figure tan. Which is sin/cos = tan

Answer 20

So I'd put 5/9 divided by 2/3?

Answer 21

\[\tan(x) = \frac{\sin(x)}{\cos(x)}\] what is sin(x) and cos(x) ?

Answer 22

sin=5/9
cos=2/3
Right?

Answer 23

The cos is right but look again at sin(x). Its the last thing i typed in my long reply ^^ up there

Answer 24

sin= sqrt(5)/3
cos=2/3

Answer 25

good, now you can divide those two numbers :)

Answer 26

But I don't have a calculator that'll give me a specific answer

Answer 27

so it'll give me a decimal

Answer 28

You dont need a calculator to do it unless it wants it in a decimal form

Answer 29

do i convert what i get to a fraction?

Answer 30

\[\frac{\frac{\sqrt{5}}{3}}{\frac{2}{3}}\] when dividing with fractions you take the fraction on the bottom, flip it upside down and then multiply so you would do \[\frac{\sqrt{5}}{3} * \frac{3}{2}\] which equals \[\frac{3\sqrt{5}}{6}\] you can divide the top and bottom number by 2 to reduce the fraction so your final answer is \[\frac{\sqrt{5}}{2}\]

Answer 31

I mean divide the top and bottom number by 3

Answer 32

not 2

Answer 33

so sqrt 5/3 divided by 3?

Answer 34

No, when i was doing the simplifying above i said divide by 2 and i meant 3. But the final answer that I got above is correct \[\frac{\sqrt{5}}{2}\]

Answer 35

Thanks :)