Question

suppose for an angle theta in a right triangle cos theta = C. Sketch and label this triangle, and then use it to write the other five trig functions of theta in terms of C.

Answer 1

Answer:

$sin\theta = \sqrt{1-C^2}$

$tan\theta = \dfrac{\sqrt{1-C^2}}{C}$

$cot\theta = \dfrac{C}{\sqrt{1-C^2}}$

$sec\theta = \dfrac{1}{C}}$

$cosec\theta = \dfrac{1}{\sqrt{1-C^2}}$

Step-by-step explanation:

Given that:

$\theta$ is an angle in a right angled triangle.

and $cos\theta = C$

To find:

To draw the triangle and write other five trigonometric functions in terms of C.

Solution:

We know that cosine of an angle is given by the formula:

$cosx =\dfrac{Base}{Hypotenuse}$

Here, we are given that $cos\theta = C$ OR

$cos\theta = \dfrac{C}{1}$

i.e. Base = C and Hypotenuse of triangle = 1

Please refer to the right angled triangle as per given statements.

$\triangle PQR$, with base PR = C units

and hypotenuse, QP = 1 unit

$\angle R$ is the right angle.

Let us use pythagorean theorem to find the value of perpendicular.

According to pythagorean theorem:

$\text{Hypotenuse}^{2} = \text{Base}^{2} + \text{Perpendicular}^{2}$

$1^{2} = C^{2} + QR^{2}\\\Rightarrow QR = \sqrt {1-C^2}$

$sin\theta = \dfrac{Perpendicular}{Hypotenuse}\\\Rightarrow sin\theta = \dfrac{\sqrt{1-C^2}}{1}\\\Rightarrow sin\theta = \sqrt{1-C^2}$

$tan\theta = \dfrac{Perpendicular}{Base}\\\Rightarrow tan\theta = \dfrac{\sqrt{1-C^2}}{C}$

$cot\theta = \dfrac{Base}{Perpendicular}\\\Rightarrow cot\theta = \dfrac{C}{\sqrt{1-C^2}}$

$sec\theta = \dfrac{Hypotenuse}{Base}\\\Rightarrow sec\theta = \dfrac{1}{C}}$

$cosec\theta = \dfrac{Hypotenuse}{Perpendicular}\\\Rightarrow cosec\theta = \dfrac{1}{\sqrt{1-C^2}}$