Find the exact value of each equation between

Ответ:

moran14

10.03.2022

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.

$\cos\left(-\dfrac{13\pi}3\right)=\cos\dfrac{13\pi}3=\cos\left(\dfrac\pi3+4\pi\right)=\cos\dfrac\pi3=\boxed{\dfrac12}$

16.

$\sin\dfrac{23\pi}4=\sin\left(\dfrac{3\pi}4+5\pi\right)=\sin\left(\dfrac{3\pi}4+\pi\right)=\sin\dfrac{7\pi}4=-\dfrac1{\sqrt2}$

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

17.

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

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

18.

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

$\sin\dfrac{29\pi}6=\sin\left(\dfrac{5\pi}6+4\pi\right)=\sin\dfrac{5\pi}6=-\dfrac12$

$\cos\dfrac{29\pi}6=\cos\dfrac{5\pi}6=\dfrac{\sqrt3}2$

$\implies\tan\left(-\dfrac{29\pi}6\right)=-\dfrac{-\frac12}{\frac{\sqrt3}2}=\dfrac1{\sqrt3}$

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

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Find the exact value of each equation between $0\leq theta\leq2\pi$ 15) $cos(-\frac{13\pi }{3} )$ 16) $csc(\frac{23\pi }{4}$ ) 17) $sec-(\frac{7\pi }{2}$ ) 18) $cot(-\frac{29\pi }{6}$ )

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Find the exact value of each equation between 15) 16)) 17)) 18))

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Find the exact value of each equation between $0\leq theta\leq2\pi$ 15) $cos(-\frac{13\pi }{3} )$ 16) $csc(\frac{23\pi }{4}$ ) 17) $sec-(\frac{7\pi }{2}$ ) 18) $cot(-\frac{29\pi }{6}$ )