Question

find the exact values of sin2u, cos2u, tan2u using cotu = sqrt2, pi < u < 3pi/2

Answer 1

$$
cot(u) = \sqrt{2} \implies cot(u) = \cfrac{cos(u)}{sin(u)}\\
\text{for the section between } \pi < u < \cfrac{3\pi}{2} \implies 3rd \ quadrant\\
\text{sine and cosine are both negative on 3rd quadrant, so}\\
cot(u) = \cfrac{cos(u)}{sin(u)} = \cfrac{-\sqrt{2}}{-1}\\
$$

Answer 2

so, that'd give you a cosine of \(-\sqrt{2}\) and a sine of -1
from there you can use that to expand the "double angle trig identities" and get their value

Answer 3

ok you are given angle is in 3rd quadrant (sin is neg, cos is neg)

start with triangle, then you will need the double angle formulas

from the triangle you obtain
\[\sin u = \frac{1}{\sqrt{3}}\]
\[\cos u = \frac{\sqrt{2}}{\sqrt{3}}\]
\[\tan u = \frac{1}{\sqrt{2}}\]

Answer 4

oops i forgot about neg signs

Answer 5

@jdoe0001 , careful with assigning cos and sine , they have to be within -1 and 1
sqrt2 > 1

Answer 6

http://www.sosmath.com/trig/Trig5/trig5/img7.gif

Answer 7

oohh yeah, my bad
$$
cot(u) = \sqrt{2} \implies cot(u) = \cfrac{adjacent}{opposite}\\
\text{for the section between } \pi < u < \cfrac{3\pi}{2} \implies 3rd \ quadrant\\
\text{"y" and "x" are both negative on 3rd quadrant, so}\\
cot(u) = \cfrac{adjacent}{opposite} = \cfrac{-\sqrt{2}}{-1}\\
$$

Answer 8

So cosine is negative in Quadrant 3? But sqrt2 > 1, so that means that cosine can't equal - sqrt2.

Answer 9

yes, as properly indicated by dumbcow

Answer 10

so, we're just dealing with adj/opp sides for this case

Answer 11

\[\cot u=\sqrt{2},\tan u=\frac{ 1 }{ \sqrt{2} }\]
\[\sin 2u=\frac{ 2\tan u }{ 1+\tan ^{2}u }=\frac{ 2*\frac{ 1 }{ \sqrt{2} } }{1-\frac{ 1 }{ 2 } }\]
\[=\frac{ \frac{ 2 }{\sqrt{2}} }{\frac{ 1 }{2 } }=2\sqrt{2}\]
\[similarly \cos 2u=\frac{ 1-\tan ^{2}u }{1+\tan ^{2}u } ,\tan 2 u=\frac{ 2\tan u }{ 1-\tan ^{2}u }\]
Plug the values of tan u ,you get the required values.

Answer 12

I used 2sinucosu for sin2u and got 2sqrt2/3.

Answer 13

How did you go from 1 + tan^2u to 1 - (1/2). Isn't is 1 + (1/2)?

Answer 14

@rascal , you are correct , that was a typo i believe
either formula will work, however 2sincos is much more familiar

Answer 15

Ok thanks!

Answer 16

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

\[\cos 2u=\frac{ 1-\tan ^{2}u }{1+\tan ^{2}u }\]