Given sin 20°=k express the following in terms of k a) sin 200° b) cos 20° c) tan(-20°) d) sin 70°Related to trignometry.

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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a) You need to consider `200^o = 180^o + 20^o` , hence you need to use the next formula to evaluate sin `200^o`  in terms of k:

`sin (alpha + beta) = sin alpha*cos beta + sin beta*cos alpha`

Plugging `alpha = 180^o` , `beta = 20^o`  in the formula yields:

`sin 200^o = sin (180^o + 20^o) = sin 180^o*cos20^o + sin 20^o*cos 180^o`

Plugging sin `180^o = 0`  and cos`180^o = -1 ` in the formula yields:

`sin 200^o = - sin 20^o = -k`

b) You need to use the basic trigonometric formula such that:

`sin^2 20^o + cos^2 20^o = 1`

`cos^2 20^o = 1 - sin^2 20^o=gt cos 20^o = +-sqrt(1 - sin^2 20^o)`

You need to keep the positive value only since `20^o`  is in the first quadrant, hence `cos 20^o = sqrt(1 - k^2)`

c) You need to remember that tangent function is rational such that:

`tan (-20^o) = (sin(-20^o))/(cos(-20^o))`

You need to remember that sine function is odd and cosine function is even such that:

`tan (-20^o) = (-sin(20^o))/(cos(20^o)) = -k/sqrt(1 - k^2)`

d) You need to consider `70^o = 90^o- 20^o` , hence you need to use the next formula to evaluate sin `70^o`  in terms of k:

`sin (alpha- beta) = sin alpha*cos beta- sin beta*cos alpha`

Plugging `alpha = 90^o` , `beta = 20^o`  in the formula yields:

`sin 70^o = sin (90^o- 20^o) = sin 90^o*cos20^o- sin 20^o*cos 90^o`

Plugging sin `90^o =1`  and cos `90^o =0`  in the formula yields:

`sin 70^o = cos20^o = sqrt(1 - k^2)`

Hence, writing the values of trigonometric functions in terms of k yields: a) `sin 200^o = -k` ; b) `cos 20^o = sqrt(1 - k^2)` ; c)`tan (-20^o) = -k/sqrt(1 - k^2)` ; d) `sin 70^o =sqrt(1 - k^2).`

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