+9 Let
θ
be an angle in quadrant
III
such that
=sinθ−78
.
Find the exact values of
secθ
and
cotθ
.
+118658 Hi avdb33,
I think something is missing from your question.
There sould be something on both sides of the equal sign. :)
also, do you mean
sin(theta)-78
or
sin(theta-78) ( I assume the angles are in degrees.)
+37093 OK....remember sin^2 + cos^2 =1
(-7/8)^2 + cos^2 =1
cos^2 = 1- 49/64 = 16/64
cos = - 4/8 (third quadrant)
Secant = 1/cos = -8/4 = - 2
Cot = cos/sine = (-4/8) / (-7/8) = - 4/8 x - 8/7 = 4/7
+118658 Let θ be an angle in quadrant III such that sinθ=−7/8
.
Find the exact values of secθ and cotθ
In the third quadrant cos is negative so sec is negative too (sec=1/cos)
and
In the third quadrant tan is positive so cot is positive too (cot=1/tan)
Draw a right angled triangle and put theta as one of the acute angles.
sin theta =7/8 you can ignor the negative sign for this.
so
put 7 on the opposite side to theta and 8 on the hypotenuse.
Use pythagoras's theorum to work out the adjacent side
\(opp=7\\ hyp=8\\ adj=\sqrt{64-49}=\sqrt{15}\)
\(sec(\theta)= -\frac{ hyp}{adj} =-\frac{8}{\sqrt{15}}=\frac{-8\sqrt{15}}{15}\\ cot(\theta)= \frac{adj}{ opp} =+\frac{\sqrt{15}}{7}\\\)
