We will learn how to find the trigonometrical ratios of any angle using the following step-by-step procedure.
Step I: To find the trigonometrical ratios of angles (n ∙ 90° ± θ); where n is an integer and θ is a positive acute angle, we will follow the below procedure.
First we need to determine the sign of the given trigonometrical ratio. Now to determine the sign of the given trigonometrical ratio we need to find the quadrant in which the angle (n ∙ 90° + θ) or (n ∙ 90° - θ) lies.
Now, using the rule “All, sin, tan, cos” we will find the sign of the given trigonometrical ratio.Therefore,
(i) All trigonometrical ratios are positive if the given angle (n ∙ 90° + θ) or (n .90° + θ) lies in the I quadrant (first quadrant);
(ii) Only sin and csc
ratios is positive if the given angle (n ∙
90° + θ) or (n ∙ 90° - θ) lies in the II quadrant (second quadrant);
(iii) Only tan and cot ratios is positive if the given angle (n ∙ 90° + θ) or (n ∙ 90° - θ) lies in the III quadrant (third quadrant);
(iv) Only cos and sec ratios is positive if the given angle (n ∙ 90° + θ) or (n ∙ 90° - θ) lies in the IV quadrant (fourth quadrant).
Step II: Now determine whether n is an even or odd integer.
(i) If n is an even integer the form of the given trigonometrical ratio will remain the same i.e.,
sin (n ∙ 90° + θ) = sin θ
sin (n ∙ 90° - θ) = - sin θ; cos (n ∙ 90° + θ) = cos θ; cos (n ∙ 90° - θ) = - cos θ; tan (n ∙ 90° + θ) = tan θ; tan (n ∙ 90° - θ) = - tan θ. |
csc (n ∙ 90° + θ) = csc θ
csc (n ∙ 90° - θ) = - csc θ; sec (n ∙ 90° + θ) = sec θ; sec (n ∙ 90° - θ) = - sec θ; cot (n ∙ 90° + θ) = cot θ; cot (n ∙ 90° - θ) = - cot θ. |
(ii) If n is an odd integer then the form of the given trigonometrical ratio is altered i.e.,
sin changes to cos;
i.e., sin (n ∙ 90° + θ) = cos θ or, sin (n ∙ 90° - θ) = - cos θ |
csc changes to sec;
i.e., csc (n ∙ 90° + θ) = sec θ or, csc (n ∙ 90° - θ) = - sec θ |
cos changes to sin;
i.e., cos (n ∙ 90° + θ) = sin θ or, cos (n ∙ 90° - θ) = - sin θ |
sec changes to csc;
i.e., sec (n ∙ 90° + θ) = csc θ or, sec (n ∙ 90° - θ) = - csc θ |
tan changes to cot;
i.e., tan (n ∙ 90° + θ) = cot θ or, tan (n ∙ 90° - θ) = - cot θ |
cot changes to tan;
i.e., cot (n ∙ 90° + θ) = tan θ or, cot (n ∙ 90° - θ) = - tan θ |
● Trigonometric Functions
11 and 12 Grade Math
From Trigonometrical Ratios of any Angle to HOME PAGE
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.
Sep 15, 24 04:57 PM
Sep 15, 24 04:08 PM
Sep 15, 24 03:16 PM
Sep 14, 24 04:31 PM
Sep 14, 24 03:39 PM
New! Comments
Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.