Question

find the values that satisfy the trigonometric expression.

Answer 1

u have to solve the equation
\(\large\tt \begin{align} \color{black}{
\sin \theta + \tan( -\theta) =0\\~\\
\sin \theta - \tan \theta =0\\~\\
\sin \theta - \dfrac{\sin \theta }{\cos \theta }=0\\~\\
\sin \theta(1 - \dfrac{1 }{\cos \theta })=0\\~\\
\sin \theta(1 - \dfrac{1 }{\cos \theta })=0\\~\\
}\end{align}\)


see if that makes sense

Answer 2

no not really

Answer 3

which step r u not getting

Answer 4

note that

\(tan (-\theta)=-tan \theta\\
and\\
tan \theta =\dfrac{sin \theta}{cos \theta}\)

Answer 5

lol i see now ok

Answer 6

so now u will have two solutions


\(\large\tt \begin{align} \color{black}{

\sin \theta(1 - \dfrac{1 }{\cos \theta })=0\\~\\

\sin \theta=0~~or~~\dfrac{1 }{\cos \theta }=0\\~\\
\theta=\sin^{-1}0~~or~~(1-\dfrac{1 }{\cos \theta })=0\\~\\
\theta=\sin^{-1}0~~or~~(\dfrac{1 }{\cos \theta })=1\\~\\
\theta=\sin^{-1}0~~or~~\cos \theta =1\\~\\
\theta=\sin^{-1}0~~or~~\theta =\cos ^{-1}1\\~\\
}\end{align}\)

see if thats ok

Answer 7

this one


\(\large\tt \begin{align} \color{black}{

\sin \theta(1 - \dfrac{1 }{\cos \theta })=0\\~\\

\sin \theta=0~~or~~(1-\dfrac{1 }{\cos \theta })=0\\~\\
\theta=\sin^{-1}0~~or~~(1-\dfrac{1 }{\cos \theta })=0\\~\\
\theta=\sin^{-1}0~~or~~(\dfrac{1 }{\cos \theta })=1\\~\\
\theta=\sin^{-1}0~~or~~\cos \theta =1\\~\\
\theta=\sin^{-1}0~~or~~\theta =\cos ^{-1}1\\~\\
}\end{align}\)

Answer 8

ok

Answer 9

so can u solve further ?

Answer 10

it should be easy now
note that
\(0\leq \theta <2\pi\)

Answer 11

tbh i dont really know how to do it any farther this was a sample problem in my book and i am totally lost

Answer 12

ok u need to study unit circle for this

Answer 13

\(\sin 0^{\circ} =0 \\and\\

\sin \pi=0\)

similarly


\(\cos 0^{\circ} = 1\\and\\

\sin \pi=1\)

refer this
http://www.mathsisfun.com/geometry/unit-circle.html

Answer 14

sry the last step should be

\(cos \pi=1\) not \(sin \pi=1\)

Answer 15

ok im following

Answer 16

so according to the condition

unit circle is from

\(0\rightarrow 2\pi\) same as ur

theta restrictions

\(0\leq \theta <2\pi\)

so the values which satisfy the answer are

\(0\\and \\\pi\)

Answer 17

thanks for explaining it :)

Answer 18

yw