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olasank [31]
1 year ago
9

Maths trig functions please help

Mathematics
1 answer:
Nikitich [7]1 year ago
3 0

1. According to the plots, the curves intersect when x=90^\circ and x=360^\circ.

We can confirm this algebraically.

\sin(x) + 2 = -\cos(x) + 3

\sin(x) + \cos(x) = 1

\sqrt2 \sin\left(x + 45^\circ\right) = 1

\sin\left(x + 45^\circ\right) = \dfrac1{\sqrt2}

x + 45^\circ = \sin^{-1}\left(\dfrac1{\sqrt2}\right) + 360^\circ n \text{ or } x + 45^\circ = 180^\circ - \sin^{-1}\left(\dfrac1{\sqrt2}\right) + 360^\circ n

(where n is an integer)

x + 45^\circ = 45^\circ + 360^\circ n \text{ or } x + 45^\circ = 135^\circ + 360^\circ n

x = 360^\circ n \text{ or } x = 90^\circ + 360^\circ n

We get the two solutions we found in the interval [0°, 360°] with n=1 in the first case, and n=0 in the second case.

2. We have \sin(x)=0 when x \in \{0^\circ, \pm 180^\circ, \pm360^\circ, \ldots\}. For the given plot domain [0°, 360°], this happens when 180^\circ < x < 360^\circ.

3. The domain for both equations is all real numbers in general, but considering the given plot, you could argue the domains would be [0°, 360°].

\sin(x) is bounded between -1 and 1, so \sin(x) + 2 is bounded between -1 + 2 = 1 and 1 + 2 = 3, and its range is [1, 3].

Likewise, -\cos(x) is bounded between -1 and 1, so that -\cos(x)+3 is bounded between -1 + 3 = 2 and 1 + 3 = 4, so its range would be [2, 4].

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