Right Triangle Trigonometry" If sin 20 =  a  and cos 20 =  b  then what would be the correct value of sin 70 in terms of a and/or b?

Answer 1

sin20 = a

cos20 = b

sin70 = cos(90-70)= cos20 = b

sin70 = b

check the answer

use a calculator

sin20 = .342

cos20 = .940

sin70 = .940

or another method with other trig identities

sin30 = 1/2

cos30= (1/2)sqr3

use double angle formulas

sin2x =2sinxcosx

cos2x=cos^2(x)-sin^2(x)

sin70 = sin(40+30) = sin(2(20)+30)

= b^2/2 -a^2/2 +absqr3

sin(40+30) = sin40cos30+cos40sin30

=sin40(sqr3/2) + cos40(1/2)

=(2sin20cos20)(sqr3/2) + (cos^2(20)-sin^2(20))(1/2)

=2ab(sqr3/2) + (b^2 -a^2)/2

= b^2/2 -a^2/2 + absqr3

=(.94)^2/2 -(.342)^2/2 + (.94)(.342)sqr3

=(.883- .117)/2 + .557

=,383 + .557

= .940

it works

so

b = b^2/2 -a^2/2 +absqr3

that gives you solutions of sin70

in terms of b alone

and in terms of both a and b together

as well as the numerical value without either a or b

now there's another solution, in terms of a alone

sin70 = f(a) = ?

take b= (b^2-a^2)/2 +absqr3

rewrite in quadratic form

use the quadratic formula and solve for b solely in terms of a

b = 1-asqr3 + 2sqr(a^2-asqr3+1)

and you now have 4 solutions for sin70,

all = .940 when you plug in numerical values for a and b, a=.342, b= .940

Answer 2

The two acute angles in a right triangle are always complementary. The Sin of the first acute angle is the Cos of it's complement therefore the Sin 70 degrees will be the same as the Cos 20 degrees, which is b.