SOLUTION: Find the value of the five other trig functions, given sinx=square root -2/2 and cosx<0 also give degree and radian measure of x

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Question 198049: Find the value of the five other trig functions, given sinx=square root -2/2 and cosx<0 also give degree and radian measure of x
Answer by Edwin McCravy(6938) About Me  (Show Source):
You can put this solution on YOUR website!
Find the value of the five other trig functions, given
sinX=-sqrt%282%29%2F2 and cosX%3C0 also
give degree and radian measure of X

I will use a capital X to represent the angle, and a 
small x to represent the x-coordinate in the graph.
Don't get them confused!

Since
sin%28X%29=%28-sqrt%282%29%29%2F2

and that is a negative number, this tells us that 
the angle X is in either QIII of QIV.

Since
cosx%3C0

that tells us that the cosine is a negative number,
and we know that the cosine is negative in QII and QIII,

so therefore we know that angle X is in QIII, so we
draw it:

drawing%28400%2C400%2C-3%2C3%2C-3%2C3%2C%0D%0Agraph%28400%2C400%2C-3%2C3%2C-3%2C3%29%2C+triangle%28-sqrt%282%29%2C-sqrt%282%29%2C-sqrt%282%29%2C0%2C0%2C0%29%2C%0D%0Alocate%28-.7%2C.3%2Cx%29%2C+locate%28-1.6%2C-.6%2Cy%29%2Clocate%28-.7%2C-.7%2Cr%29%2C%0D%0Arectangle%28-sqrt%282%29%2B.2%2C-.2%2C-sqrt%282%29%2C0%29%2Cblue%28arc%280%2C0%2C.8%2C-.8%2C0%2C225%29+%29%2C%0D%0Alocate%28-.2%2C.6%2CX%29%0D%0A%0D%0A%29

Since the sine is y%2Fr, we take the
numerator of the given sine, which is -sqrt%282%29 to be the 
value of y, and the denominator 2 to be the value of r.

drawing%28400%2C400%2C-3%2C3%2C-3%2C3%2C%0D%0Agraph%28400%2C400%2C-3%2C3%2C-3%2C3%29%2C+triangle%28-sqrt%282%29%2C-sqrt%282%29%2C-sqrt%282%29%2C0%2C0%2C0%29%2C%0D%0Alocate%28-.7%2C.3%2Cx%29%2C+locate%28-2%2C-.6%2Cy=sqrt%282%29%29%2Clocate%28-.6%2C-.6%2Cr=2%29%2C%0D%0Arectangle%28-sqrt%282%29%2B.2%2C-.2%2C-sqrt%282%29%2C0%29%2Cblue%28arc%280%2C0%2C.8%2C-.8%2C0%2C225%29+%29%2C%0D%0Alocate%28-.2%2C.6%2CX%29%0D%0A%0D%0A%0D%0A%29

Next we calculate the value of x (small x) using the Pythagorean
theorem:

x%5E2%2By%5E2=r%5E2
x%5E2%2B%28-sqrt%282%29%29%5E2=%282%29%5E2
x%5E2%2B2=4
x%5E2=2
x=%27%27%2B-sqrt%282%29

We know to take the velue of x as negative, because
x goes left from the origin, so
x=-sqrt%282%29

and we label it thusly:

drawing%28400%2C400%2C-3%2C3%2C-3%2C3%2C%0D%0Agraph%28400%2C400%2C-3%2C3%2C-3%2C3%29%2C+triangle%28-sqrt%282%29%2C-sqrt%282%29%2C-sqrt%282%29%2C0%2C0%2C0%29%2C%0D%0Alocate%28-1.1%2C.3%2Cx=-sqrt%282%29%29%2C+locate%28-2%2C-.6%2Cy=sqrt%282%29%29%2Clocate%28-.6%2C-.6%2Cr=2%29%2C%0D%0Arectangle%28-sqrt%282%29%2B.2%2C-.2%2C-sqrt%282%29%2C0%29%2Cblue%28arc%280%2C0%2C.8%2C-.8%2C0%2C225%29+%29%2C%0D%0Alocate%28-.2%2C.6%2CX%29+%29

Now we get the other 5 trigonometric functions by using
the definitions:

cos%28X%29=x%2Fr=%28-sqrt%282%29%29%2F2=-sqrt%282%29%2F2

sec%28X%29=r%2Fx=2%2F%28-sqrt%282%29%29=-2%2Fsqrt%282%29=%28-2sqrt%282%29%29%2F%28sqrt%282%29sqrt%282%29%29=%28-2sqrt%282%29%29%2F2=-sqrt%282%29

csc%28X%29=r%2Fy=2%2F%28-sqrt%282%29%29=-2%2Fsqrt%282%29=%28-2sqrt%282%29%29%2F%28sqrt%282%29sqrt%282%29%29=%28-2sqrt%282%29%29%2F2=-sqrt%282%29

tan%28X%29=y%2Fx=%28-sqrt%282%29%29%2F%28-sqrt%282%29%29=1

cot%28X%29=r%2Fx=%28-sqrt%282%29%29%2F%28-sqrt%282%29%29=1

Since the right triangle has two legs equal it is isosceles
are the reference angle is therefore 45°.

Since the angle is in QIII, we must add 180° to its
reference angle, so

the angle X in degree measure is 45°+180° or 225°

In radian measure, we multiply by pi%2F180%5Eo

%28225%5Eo%29%28pi%2F180%5Eo%29=5pi%2F4+


Edwin