Trigonometric Values As a Function of x...2

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David Cohen Textbook Questions.
Enjoy.

IMG_20220204_134658.jpg
 
sin(90-β)=sin(90) cos(β) - cos(90) sin(β).......sin(90)=1, cos(90)=0
sin(90-β)=1* cos(β) - 0* sin(β)

sin(90-β)= cos(β)

cos(90-β)=sin(90) sin(β) + cos(90) cos(β)
cos(90-β)=1*) sin(β) +0* cos(β)
cos(90-β)=sin(β)

tan(90-β)= cos(β)/sin(β)
 
I don't see any reason to use anything as complicated as trig identities here.

90- beta is the angle opposite beta so the "opposite side" has length 3 and the hypotenuse has length 4x. sin(90- beta)=3/(4x).

Cosine is "near side over hypotenuse". The near side to angle 90- beta is unlabeled but by the Pythagorean formula its length is sqrt(16x^2- 9). cos(90-beta)= sqrt(16x^2- 9)/(4x).

Tangent is "opposite side over near side" so tan(90- beta)= 3/sqrt(16x^2- 9).
 
sin(90-β)=sin(90) cos(β) - cos(90) sin(β).......sin(90)=1, cos(90)=0
sin(90-β)=1* cos(β) - 0* sin(β)

sin(90-β)= cos(β)

cos(90-β)=sin(90) sin(β) + cos(90) cos(β)
cos(90-β)=1*) sin(β) +0* cos(β)
cos(90-β)=sin(β)

tan(90-β)= cos(β)/sin(β)

Thank you for using trigonometric identities.
 

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