Questions, are all in picture, please solve

Mathematics

1 answer:

Andrews [41]2 years ago

3 0

The function represents a <em>cosine</em> graph with axis at y = - 1, period of 6, and amplitude of 2.5.

<h3>How to analyze sinusoidal functions</h3>

In this question we have a <em>sinusoidal</em> function, of which we are supposed to find the following variables based on given picture:

Equation of the axis - Horizontal that represents the mean of the bounds of the function.
Period - Horizontal distance needed between two maxima or two minima.
Amplitude - Mean of the difference of the bounds of the function.
Type of sinusoidal function - The function represents either a sine or a cosine if and only if trigonometric function is continuous and bounded between - 1 and 1.

Then, we have the following results:

Equation of the axis: y = - 1
Period: 6
Amplitude: 2.5
The graph may be represented by a cosine with no <em>angular</em> phase and a sine with <em>angular</em> phase, based on the following trigonometric expression:

cos θ = sin (θ + π/2)

To learn more on sinusoidal functions: brainly.com/question/12060967

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Prove whether the following are identities 2tanh 1/2x / 1−tanh^2 1/2 x = sinh x

Tanzania [10]

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Also, recall the identity

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Then

$\dfrac{2\tanh\left(\frac x2\right)}{1 - \tanh^2\left(\frac x2\right)} = \dfrac{2\tanh\left(\frac x2\right)}{\mathrm{sech}^2\left(\frac x2\right)} \\\\ \dfrac{2\tanh\left(\frac x2\right)}{1 - \tanh^2\left(\frac x2\right)} = 2\tanh\left(\dfrac x2\right)\cosh^2\left(\dfrac x2\right) \\\\ \dfrac{2\tanh\left(\frac x2\right)}{1 - \tanh^2\left(\frac x2\right)} = 2\sinh\left(\dfrac x2\right)\cosh\left(\dfrac x2\right) \\\\\dfrac{2\tanh\left(\frac x2\right)}{1 - \tanh^2\left(\frac x2\right)} = \sinh(x)$