Math problem ...

MaHe · Nov 25, 2007

tanx = - sqrt(15)
90 degrees < x < 180 degrees

Calculate sinx and cosx, using the calculator is forbidden. I actually know the results (sinx = sqrt(15)/4 and cosx = -1/4), but desperately need the procedure. Please help, if you can.

castillo · Nov 25, 2007

tanx = sinx/cosx ===> sinx = -sqrt(15)*cosx
cosx^2 + sinx^2 = 1
cosx^2 + (-sqrt(15)*cosx)^2 = 1
cosx^2 + 15*cosx^2 = 1
16*cosx^2 = 1
cosx^2 = 1/16
cosx = sqrt(1/16) = 1/4
From this, you obtain sinx by "sinx^2 = 1 - cosx^2"

Edit:
As 90 < x < 180, cosx should be negative, so cosx = -1/4, not 1/4, sorry.

Veho · Nov 25, 2007

Oooh, can I try?

tan(x) = -sqrt(15)

square(tan(x)) = 15

square(sin(x))/square(cos(x)) = 15;

square(sin(x)) = 15square(cos(x))

Since square(sin(x)) + square(cos(x)) = 1

then: square(sin(x)) = 15(1-square(sin(x)))

16square(sin(x)) = 15

square(sin(x)) = 15/16

Square(cos(x)) = (1 - 15/16)

EDIT: Yeah, not done yet... oops

Square(sin(x)) = 15/16 ==> sin(x) = +/- sqrt(15)/4

square(cos(x)) = 1/16 ==> cos(x) = +/- 1/4

If x is from the second quadrant (180>x>90),

then sin(x) > 0,

cos(x) < 0

Meaning,

sin(x) = sqrt(15)/4;

cos(x) = -1/4.

xcalibur · Nov 25, 2007

ugh... trigonometric identities...
got a D because of these and integration
redoing maths Alevel this year

any tips? i tend to lose track of the whole process pretty quick

wiithepeople · Nov 25, 2007

...

my brain hurts.

MaHe · Nov 25, 2007

QUOTE(castillo @ Nov 25 2007 said: ↑

tanx = sinx/cosx ===> sinx = -sqrt(15)*cosx
cosx^2 + sinx^2 = 1
cosx^2 + (-sqrt(15)*cosx)^2 = 1
cosx^2 + 15*cosx^2 = 1
16*cosx^2 = 1
cosx^2 = 1/16
cosx = sqrt(1/16) = 1/4
From this, you obtain sinx by "sinx^2 = 1 - cosx^2"

Edit:
As 90 < x < 180, cosx should be negative, so cosx = -1/4, not 1/4, sorry.

QUOTE(veho @ Nov 25 2007 said: ↑

Oooh, can I try?

tan(x) = -sqrt(15)

square(tan(x)) = 15

square(sin(x))/square(cos(x)) = 15;

square(sin(x)) = 15square(cos(x))

SinceÂ Â square(sin(x)) + square(cos(x)) = 1

then: square(sin(x)) = 15(1-square(sin(x)))

16square(sin(x)) = 15

square(sin(x)) = 15/16

Square(cos(x)) = (1 - 15/16)

EDIT: Yeah, not done yet... oopsÂ Â Â

Square(sin(x)) = 15/16Â ==>Â sin(x) = +/-Â sqrt(15)/4

square(cos(x)) = 1/16Â ==>Â cos(x) = +/-Â 1/4

If x is from the second quadrant (180>x>90),

then sin(x) > 0,

cos(x) < 0

Meaning,

sin(x) = sqrt(15)/4;

cos(x) = -1/4.

Click to expand...

Click to expand...

Wow, thanks guys! The solution was right before my nose and I spent three hours. How the heck didn't I figure it out by myself?
Anyways, I owe you both a beer or a dosen.