Question 1087132: Can you please explan why R cos(θ-a)= R sin(θ+a)? They said it was since cos (θ )= sin (90-θ ) but I still don't understand.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! i don't believe this is correct.
R * cos(T - a) = R * cos(T + a) becomes cos(T - a) = sin(T + a) when you divide both sides of the equation by R.
an example would be:
assume T = 50 and a = 10
the equation says that R * cos(40) = R * sin(60).
cos(40) = .7660444431
sin(60) = .8660254038
they're not the same.
it doesn't work because the angles are not complementary.
60 + 40 = 100, not 90.
the sum of the angles must be equal to 90 if the angles are complementary.
however, .....
cos(T - a) = sin(90 - T + a) will work because the angles are complementary.
if the angles are complementary, then (T - a) + (90 - T + a) must be equal to 90 degrees.
add them together and you get T - a + 90 - T + a is equal to 90.
so, unless i'm missing something .....,
R * cos(T - a) is not equal to R * sin(T + a).
however, .....
R * cos(T - a) is equal to R * sin(90 - T + a)
put it into degrees and use your calculator to find sine and cosine and you'll see what i mean.
note that sin(90 - T + a) is equal to sin(90 - (T - a))
you get cos(T - a) = sin(90 - (T - a)
the angles are complementary, and have to be in order for the equation to be correct.
cos(T-a) = sin(T+a) doesn't work.
cos(T-a) = sin(90-T+a) does work.
that's what i think.
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