Trigonometric Values As a Function of x...1

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David Cohen Textbook Questions.
Enjoy.

IMG_20220204_134648.jpg
 
sin^2(theta)=(2x)^2
sin^2(theta)=4x^2

cos^2(theta)=3^2
cos^2(theta)=9

tan^2(theta)=sin^2(theta)/cos^2(theta)=4x^2/9
 
The legs have lengths 2x and 3 so, by the Pythagorean formula, the hypotenuse has length sqrt(4x^2+ 9).

sin(theta)= 2x/sqrt(4x^2+ 9) so sin^2(theta)= 4x^2/(4x^2+ 9)

cos(theta)= 3/sqrt(4x^2+ 9) so cos^2(theta)= 9/(4x^2+ 9)

tan(theta)= 2x/3 so tan^2(theta)= 4x^2/9.
 
The legs have lengths 2x and 3 so, by the Pythagorean formula, the hypotenuse has length sqrt(4x^2+ 9).

sin(theta)= 2x/sqrt(4x^2+ 9) so sin^2(theta)= 4x^2/(4x^2+ 9)

cos(theta)= 3/sqrt(4x^2+ 9) so cos^2(theta)= 9/(4x^2+ 9)

tan(theta)= 2x/3 so tan^2(theta)= 4x^2/9.

Cool.
 

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