1. ## trigonometry

Find the values of x , valid between -pi and pi . I am confused because of the range .

sin x =cos x

My attempt .

tan x =1

so x will be 45 .. yes this is when x is between 0 and pi but how bout -pi and 0 ?? thanks

2. Originally Posted by thereddevils
Find the values of x , valid between -pi and pi . I am confused because of the range .

sin x =cos x

My attempt .

tan x =1

so x will be 45 .. yes this is when x is between 0 and pi but how bout -pi and 0 ?? thanks
you should take the angle when sin equal cos it is 45
and 45-180=-135 in the third quarter since sin and cos are negative

3. Originally Posted by Amer
you should take the angle when sin equal cos it is 45
and 45-180=-135 in the third quarter since sin and cos are negative

Thanks Amer , i am not sure if what i did is correct .

tan x =1

x=45

tan (-x) = 1

-x=135 ( since quadrant 3 is positive )

x=135

Can i do like this ? Is my thought process correct >

4. Originally Posted by thereddevils
Find the values of x , valid between -pi and pi . I am confused because of the range .

sin x =cos x

My attempt .

tan x =1

so x will be 45 .. yes this is when x is between 0 and pi but how bout -pi and 0 ?? thanks
Since the domain is given in radians the answer should also be given in radians (not degrees).

$\displaystyle x = \frac{\pi}{4} + n \pi$ where n is an integer. Now substitute different values of n to find all of the solutions that lie in the solution domain.

5. sin x = cos x,

sin x/cos x = 1,

tan x = 1,

x = pi/4 + pi (k), where k is an integer, for domain -pi to pi,

let k = -1 and 0

(1) k = 0,

x = pi/4,

(2) k = -1,

x = pi/4 - pi = -(3/4)pi

see graph, ist quadrant sin x and cos x are positive, in the 3rd quadrant, both are negative.

6. If you need a different way to show why
tan(x) = 1, draw the triangle with angle x with the opposite and adjacent sides equal to 1 [in the first quadrant]

The hypotenuse is then sqrt(2), and you should recognize that the triangle is a special traingle most likely focused on in your class, where the angle of x is pi/4 radians (the reference angle)

7. $\displaystyle \sin x=\cos x\implies \sqrt{2}\sin \left( x-\frac{\pi }{4} \right)=0\implies x-\frac{\pi }{4}=k\pi ,\,k\in \mathbb{Z}.$