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Chapter 4: Problem 17

Evaluate (if possible) the sine, cosine, and tangent at the real number. $$t=-\frac{7 \pi}{4}$$

Short Answer

Expert verified
The sine of \(t = -\frac{7 \pi}{4}\) is \(\frac{\sqrt{2}}{2}\), the cosine of \(t = -\frac{7 \pi}{4}\) is \(\frac{\sqrt{2}}{2}\) and the tangent of \(t = -\frac{7 \pi}{4}\) is 1.

Step by step solution

01

Calculate sine

Recall that the sine function has a period of \(2\pi\). Therefore, \(\sin(-\frac{7 \pi}{4})\) is equivalent to \(\sin(\frac{ \pi}{4})\) since \(-\frac{7 \pi}{4} + 2\pi = \frac{ \pi}{4}\). Sin(\(\frac{ \pi}{4}\)) = \(\frac{\sqrt{2}}{2}\)
02

Calculate cosine

The cosine function also has a period of \(2\pi\). Therefore, \(\cos(-\frac{7 \pi}{4})\) is equivalent to \(\cos(\frac{ \pi}{4})\) since \(-\frac{7 \pi}{4} + 2\pi = \frac{ \pi}{4}\). So, \(\cos(\frac{ \pi}{4})\) = \(\frac{\sqrt{2}}{2}\)
03

Calculate tangent

Tangent is the ratio of sine to cosine. So, \(\tan(-\frac{7 \pi}{4}) = \tan(\frac{ \pi}{4})\), since tangent has period \( \pi\), which means \(-\frac{7 \pi}{4} + 2\pi = \frac{ \pi}{4}\). So, \(\tan(\frac{\pi}{4}) = 1\) because \(\frac{\sin(\frac{\pi}{4})}{\cos(\frac{\pi}{4})} = \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1\).

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