Determine without the use of a calculator

Question

sin( arctan(-(2/3)^1/2) + arccos(-2/5^1/2) )

Answer 1

First of all -(2/3)^(1/2) = -sqrt(6)/3 and -2/sqrt(5) = -2*sqrt(5) / 5

First term:

tangent is -sqrt(6)/3.....

tangent is negative in quadrants 2 and 4

in quadrant 2, x=-3 and y = sqrt(6) and in quadrant 4, x=3 and y = -sqrt(6)

in either case, the hypotenuse is sqrt(15) by pythagorean

so the sine of that same angle is sqrt(6)/sqrt(15) = sqrt(6/15) = sqrt(2/5)

= sqrt(10) / 5

Second term:

cosine is -2*Sqrt(5)/ 5

cosine is negative in quadrants 2 and 4

so either x=-2*Sqrt(5) and the hypotenuse is 5 in quadrant 2 or

x = 2*Sqrt(5) and the hypotenuse is 5 in quadrant 4.

This forces y=sqrt(5) in quadrant 2 and y=-sqrt(5) in quadrant 5

IN either case, we need the angle whose cosine is 2*sqrt(5) /5 .

Unless this is a type and you meant to pass this angle to another trig function

(I see you have an extra STRAY parenthesis out there) you must estiamte 2*Sqrt(5)/5

without a calculator and then use the trig table to find the angle

The trig functions are : sine = sqrt(5)/5 , cos = -2*sqrt(5)/5 as given and tangent = -1/2 in quadrant 2

in quadrant 4, sine = -sqrt(5)/5, cos = (2/5)*Sqrt(5) and tangent = =1/2

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IF YOU WANT THE ACTUAL ANGLE, READ ON !!!!!!!

To estimate the angle:

2 < sqrt(5)< 2.3 because 2.3*2.3 = 5.29

Using the bisection method to approximate this angle even more....

it is about 2.235

so the angle is about 2*2.235/5 = 0.894

Per the table the angle is about 26.619