trigonmetric identities

Question

Use the proper trigonmetric identities to solve each part: Given that sinO=2/5, with pie/2<O<pie, find the exact values of: a.) cosO=       b.)tanO=      c.) sin2(O)=    d.)cos2(O)   e.)tan2(O)     f.)sin O/2   g.) cos(O/2)   h.)tan(O/2)    

Answer 1

Lizzy,

I realize this is probably too late to help your homework grade, but just in case you still need explanation, here you go.

 

I don't see how to use identities to get part a), but after that it's straight forward.

 

Given that the sin = 2/5, and sin = opp / hyp (or y/r),  then we can imagine a right triangle with the vertical leg length 2, and the hypotenuse of length 5. Using this picture, we apply Pythagorean theorem to say:

 

                                                               x = sqrt(5² - 2²) = sqrt(21)

 

So now we have the picture of a right triangle with x = sqrt(21),  y = 2,  and r = 5.

 

a) Since cos = adj / hyp (or x/r), the cos = -sqrt(21)

                                                                     5

 

b) tan = sin/cos     so   tan = (2/5)/(sqrt(21)/5) or 2/sqrt(21) = -2sqrt(21)/21

 

c) sin(2O) = 2sin(O)cos(O) = 2(2/5)(-sqrt(21)/5) = -4sqrt(21)/25

 

d) cos(2O) = 1 - 2sin²(O) = 1 - 2(2/5)² = 1 - 8/25 = 17/25

 

e) tan(2O) = 2tan(O)/(1-tan²(O)) = 2(-2sqrt(21)/21)/(1-(-2/sqrt(21))²) = (-4sqrt(21)/21)/(1-(4/21))

                = (-4sqrt(21)/21)/(17/21) = -4sqrt(21)/17

 

f) sin(O/2) = +/- sqrt[(1-cos(O))/2] = +/- sqrt[(1-(-sqrt(21)/5))/2] = +/- sqrt[(5+sqrt(21))/10]

                   I'm sorry I can't show the algebraic steps any better in this format

 

g) cos(O/2) = +/- sqrt[(1+cos(O))/2] = +/- sqrt[(5-sqrt(21))/10]

 

h) tan(O/2) = sin(O)/(1+cos(O)) = (2/5)/(1+(-sqrt(21)/5)) = (5+sqrt(21))/2