given tan(x)=1/square root of 3 and sec(x) < 0, find sin(x) and cos(x)

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hala718's profile pic

Posted on

Given that tanx = 1/sqrt(3)

We need to find sinx and cosx

We know that tanx = sinx /cosx

==> sinx/cosx = 1/sqrt(3)

Cross multiply:

==> cosx = sqrt(3)* sinx...........(1)

Also, we know that sin^2 x + cos^2 x = 1...........(2)

We will substitute with (1) into (2):

==> sin^2 x + [sqrt3*sinx]^2 = 1

==> sin^2 x + 3sin^2 x = 1

==> 4sin^2 x = 1

Divide by4.

==> sin^2 x = 1/4

==> sinx = +- 1/2

==> cosx = sqrt3 sinx

==> cosx = +- sqrt3 / 2

But given that secx < 0 ==> cosx < 0

==> cosx = -sqrt3 /2

==> sinx = -1/2

giorgiana1976's profile pic

Posted on

We'll use the following identity to calculate the values of the functions sine and cosine:

1 + `tan^(2)` x = `sec^(2)` x

But tan x = 1/`sqrt(3)` => `tan^(2)` x = 1/3

1 + 1/3 = `sec^(2)` x

sec x = `+-` `sqrt(4/3)`

But sec x < 0 => sec x = -2`sqrt(3)` /3

But sec x = 1/ cos x =>  cos x = 1/sec x => cos x = -3`sqrt(3)` /6

cos x = -`sqrt(3)` /2

tan x = sin x/cos x => sin x = cos x*tan x = (-`sqrt(3)` /2)*(1/`sqrt(3)` )

sin x = -1/2

Therefore, the requested values are sin x = -1/2 and cos x = -`sqrt(3)` /2 and the angle x is located in the third quadrant, x =7 `pi` /6.

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