I do not understand how to do this

Question

Using the following values, how many other trigonometric values of angles between 0 and 180 degrees can you find (without using a calculator)? Don't forget about complementary angles, co-terminal angles and other trig rules. Show work and justification for each value.

sin(5°)=0.0872; cos(45°)=0.7071; sin(60°)=0.8660; cos(60°)=0.5

You need to produce at least 10 different functions and use at least 5 different identities. Show your work and use reasoning.

Answer 1

You need to pick at least 5 identities and replace the angle value of -5 degrees and solve for the other angle.

For instance, sin (-theta)= - sin(theta) or the sine of the negative angle = the negative of the sine of the same positive angle, so,

sin (5 deg)=.0872, which was given, so -.0872 = sin(-5 deg), which means that by knowing the sine of 5 degrees, you also know the sine of negative 5 degrees.

From the Pythagorean theorem, sin² (theta)+ cos² (theta)= 1 so cos(5 deg) = square root of (1 - sin² (5 deg)) = square root of (1 -(.0872)^2) = sqrt (.99239616) = .9962, so you now also know the cosine of 5 degrees.

Answer 2

To calculate (without a calculator) construct right triangles with each angle

45 degrees is a 1-1-sqr2 right triangle. 1=opposite side=adjacent side

sum their squares to get the square of the hypotenuse = 2

take the square root of 2 = the hypotenuse length.

cosine of 45 degrees is adjacent side over hypotenuse = 1/sqr2 = sqr2/2 = about 1.414/2 = 0.707

secant 45 degrees = 1/cos45 = sqr2/1 = sqr2 = 1.414

sin45 = cos45 = 0.707

cosecant45 = 1/sin45 = 1.414

tan45 = opposite side over adjacent side = 1/1 = 1

cotan45 = tan45 = 1

60 degrees involves a 1-sqr3- 2 right triangle, with adjacent side=1, opposite side = sqr3 and

hypotenuse = 2

cos60=adjacent over hypotenuse = 1/2 = 0.5

sin60 = opposite side over hypotenuse = sqr3/2 = 1.732/2 = 0.866

sec 45 = 1/cos45 = 1//1/2 = 2

csc60 = 1/sin60 =2/sqr3

Those are standard right triangles, for 30, 45, 60, 90, 120, 135, 150, 180

all involving the same 1-sqr3-2 and 1-1-sqr2 right triangles, but with some negative signs between 90 and 180 degrees for tangent, cotangent, cosine, and secant. sines and cosecants are still positive in both quadrant I and II.

to get 5 degrees fiunctons you'd need a triple angle formula with 5 x3 =15 and a double angle formula for 15x2 =30

just google double angle formula and triple trigonometric formula