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Find the Exact value of each equation between
0\leq theta\leq2\pi

15)
cos(-(13\pi )/(3) )

16)
csc((23\pi )/(4))

17)
sec-((7\pi )/(2))

18)
cot(-(29\pi )/(6))

1 Answer

3 votes

Use the fact that the co/sine functions are
2\pi-periodic and that the tangent function is
\pi-periodic. Also, recall that
\cos x is even (so that
\cos(-x)=\cos x) and
\sin x is odd (so that
\sin(-x)=-\sin x.

15.


\cos\left(-\frac{13\pi}3\right)=\cos\frac{13\pi}3=\cos\left(\frac\pi3+4\pi\right)=\cos\frac\pi3=\boxed{\frac12}

16.


\sin\frac{23\pi}4=\sin\left(\frac{3\pi}4+5\pi\right)=\sin\left(\frac{3\pi}4+\pi\right)=\sin\frac{7\pi}4=-\frac1{\sqrt2}


\implies\csc\frac{23\pi}4=\boxed{-\sqrt2}

17.


\cos\left(-\frac{7\pi}2\right)=\cos\frac{7\pi}2=\cos\left(\frac\pi2+3\pi\right)=\cos\left(\frac\pi2+\pi\right)=\cos\frac{3\pi}2=0


\implies\sec\left(-\frac{7\pi}2\right)=\boxed{\text{undefined}}

18.


\tan\left(-\frac{29\pi}6\right)=\frac{\sin\left(-\frac{29\pi}6\right)}{\cos\left(-\frac{29\pi}6\right)}=-\frac{\sin\frac{29\pi}6}{\cos\frac{29\pi}6}


\sin\frac{29\pi}6=\sin\left(\frac{5\pi}6+4\pi\right)=\sin\frac{5\pi}6=-\frac12


\cos\frac{29\pi}6=\cos\frac{5\pi}6=\frac{\sqrt3}2


\implies\tan\left(-\frac{29\pi}6\right)=-\frac{-\frac12}{\frac{\sqrt3}2}=\frac1{\sqrt3}


\implies\cot\left(-\frac{29\pi}6\right)=\boxed{\sqrt3}

User Stokke
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