Violet Woodward

2022-07-28

In A and B, cosine is given..find sine and tangent if x liesin the specific interval.
A.) $\mathrm{cos}\left(x\right)=1/3x\in \left[\pi /2,0\right]$
B.) $\mathrm{cos}\left(x\right)=-5/13x\in \left[\pi /2,\pi \right]$
C.) $s=\mathrm{sec}\left(\pi x/2\right)$...Find the period. What symmetries does the function have?
D.) $\mathrm{cos}\left(x+\pi /4\right)-1$. What is the period?

eyiliweyouc

(A) since $\left[0,\pi /2\right]$ comes in first interval so sine and tan are positive
so;$\mathrm{sin}\left(x\right)=\sqrt{1-{\mathrm{cos}}^{2}\left(x\right)}=\left(\sqrt{9-{x}^{2}}\right)/3$
so;$\mathrm{tan}\left(x\right)=\mathrm{sin}\left(x\right)/\mathrm{cos}\left(x\right)=\left(\sqrt{9-{x}^{2}}\right)/x$
(B) since$\left[\pi /2,\pi \right]$ comes in second interval sosine is positive and tan is negative.
so; $\mathrm{sin}\left(x\right)=\sqrt{1-{\mathrm{cos}}^{2}\left(x\right)}=\left(\sqrt{169-25{x}^{2}}\right)/13$
so;$\mathrm{tan}\left(x\right)=\mathrm{sin}\left(x\right)/\mathrm{cos}\left(x\right)=\left(\sqrt{169-25{x}^{2}}\right)/-5x$
(c)the period of $\mathrm{sec}\left(\pi x/2\right)=2$
(d)the period of $\mathrm{tan}\left(x+\pi /4\right)=\pi /2$

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