<u>Answer</u>:
25) If f(x) = sin ([1/3] x) , find f(π/2).
Substitute with x = π/2
f(π/2) = sin ([1/3] *[π/2]) = sin (π/6) = 1/2
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26) If f(x) = cos (2x) , find f(3π/4).
Substitute with x = 3π/4
f(3π/4) = cos ( 2 * 3π/4) = cos ( 3π/2) = zero.
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27) If f(x) = sin (2x) + cos (3x) , find f(π/4)
Substitute with x = π/4
f(π/4) = sin ( 2 * π/4) + cos ( 3 * π/4) =
= sin ( π/2) + cos ( 3π/4)
= 1 - 1/(√2)
= 0.293
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28) f(x) = tan (5x) - sin (2x) , find f(π/6).
Substitute with x = π/6
So, f(π/6) = tan (5 * π/6) - sin ( 2 * π/6)
= tan (5π/6) - sin (π/3)
= -1/(√3) - (√3)/2
= -1.443
To the nearest tenth : 32.7
To the nearest hundredth : 32.70
Whole Number : 33
Hope this helps!
The equation of line PQ which is the perpendicular bisector of AB is
y=1.33x + 9.98
None of these are correct, if it is asking for a number (x) divided by 45 the correct answer would be x/45
There are two noncoterminal angles which share that sine. They will be supplementary. So two different possibilities for the cosine and the tangent, indicated by the plus or minus.
We already know one possibility is x=45 degrees so the cosine is also
and we prepend the plus or minus to reflect the supplementary angle.
We could have gotten that even if we didn't know the angle,
![\cos x =\pm \sqrt{1- \sin^2 x}=\pm \sqrt{1 - ( \sqrt{2}/2)^2} = \pm \sqrt{1 - \frac 1 2} = \pm \sqrt{ \frac 1 2 } = \pm \sqrt{2}/2](https://tex.z-dn.net/?f=%5Ccos%20x%20%3D%5Cpm%20%5Csqrt%7B1-%20%5Csin%5E2%20x%7D%3D%5Cpm%20%5Csqrt%7B1%20-%20%28%20%5Csqrt%7B2%7D%2F2%29%5E2%7D%20%3D%20%5Cpm%20%5Csqrt%7B1%20-%20%5Cfrac%20%201%202%7D%20%3D%20%5Cpm%20%5Csqrt%7B%20%5Cfrac%201%202%20%7D%20%3D%20%5Cpm%20%5Csqrt%7B2%7D%2F2%20)
So the tangent is the sine divided by the cosine, which are the same except sometimes opposite signs.
![\tan x = \pm 1](https://tex.z-dn.net/?f=%20%5Ctan%20x%20%3D%20%5Cpm%201%20)