Given cosθ = 3/5, find the five other trigonometric function values. 15

[tex]cos(\theta )=\cfrac{\stackrel{adjacent}{3}}{\underset{hypotenuse}{5}}\qquad \impliedby \textit{let's now find the \underline{opposite side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \sqrt{5^2-3^2}=b\implies \sqrt{25-9}=b\implies \sqrt{16}=b\implies 4=b \\\\[-0.35em] ~\dotfill[/tex]

[tex]sin(\theta )=\cfrac{\stackrel{opposite}{4}}{\underset{hypotenuse}{5}}\qquad \qquad tan(\theta )=\cfrac{\stackrel{opposite}{4}}{\underset{adjacent}{3}}\qquad \qquad cot=\cfrac{\stackrel{adjacent}{3}}{\underset{opposite}{4}} \\\\\\ sec(\theta )=\cfrac{\stackrel{hypotenuse}{5}}{\underset{adjacent}{3}}\qquad \qquad csc(\theta )=\cfrac{\stackrel{hypotenuse}{5}}{\underset{opposite}{4}}[/tex]

The values of the five other trigonometric functions value for the same input θ for which cos(θ) = 3/5 are:

Trigonometric ratios for a right angled triangle are from the perspective of a particular non-right angle.

In a right angled triangle, two such angles are there which are not right angled(not of 90 degrees).

The slant side is called hypotenuse.

From the considered angle, the side opposite to it is called perpendicular, and the remaining side will be called base.

From that angle (suppose its measure is θ),

[tex]\sin(\theta) = \dfrac{\text{Length of perpendicular}}{\text{Length of Hypotenuse}}\\\cos(\theta) = \dfrac{\text{Length of Base }}{\text{Length of Hypotenuse}}\\\\\tan(\theta) = \dfrac{\text{Length of perpendicular}}{\text{Length of base}}\\\\\cot(\theta) = \dfrac{\text{Length of base}}{\text{Length of perpendicular}}\\\\\sec(\theta) = \dfrac{\text{Length of Hypotenuse}}{\text{Length of base}}\\\\\csc(\theta) = \dfrac{\text{Length of Hypotenuse}}{\text{Length of perpendicular}}\\[/tex]

If ABC is a triangle with AC as the hypotenuse and angle B with 90 degrees then we have:

[tex]|AC|^2 = |AB|^2 + |BC|^2[/tex]

where |AB| = length of line segment AB. (AB and BC are rest of the two sides of that triangle ABC, AC being the hypotenuse).

We're given that:

cosθ = 3/5 for some angle θ

Since we've got:

[tex]\cos(\theta) = \dfrac{\text{Length of Base }}{\text{Length of Hypotenuse}}[/tex]

Therefore, we have:

[tex]\dfrac{3}{5} = \dfrac{\text{Length of Base }}{\text{Length of Hypotenuse}}[/tex]

Let we consider a right angled triangle in which there is hypotenuse of length 5 units and base (from the perspective of one of its non-right angle) of 3 units (as shown in the image attached below).

(we couldve taken 3x and 5x instead of 3 and 5, for any positive real number value of 'x' as when we would take their ratio, that common factor 'x' would get cancelled out. We can think of 3 and 5 as the special case of 3x and 5x when x = 1)

Then, from that perspective, let the perpendicular be of the length 'p' units, then as per the pythagoras theorem, we get:

[tex]p^2 + 3^2 = 5^2\\p = \sqrt{25 - 9} = \sqrt{16} = 4 \: \rm units[/tex]

(took only the positive root to remove the square term because the value of p denotes length, which is a non-negative quantity).

Thus, we have:

From the perspective of the angle θ:

Thus, using these values, and the definition of the trigonometric ratios, we get:

Thus, the values of the five other trigonometric functions value for the same input θ for which cos(θ) = 3/5 are:

Learn more about Pythagoras theorem here:

https://brainly.com/question/12105522

Learn more about trigonometric ratios here:

https://brainly.com/question/22599614

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