SOLUTION: if f(x)=3sinx + 2cos2x, the value of (Pie/2) is
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Question 77324: if f(x)=3sinx + 2cos2x, the value of (Pie/2) is
Found 2 solutions by jim_thompson5910, chitra:
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
If you're trying to evaluate:
)
for the function
=3sinx + 2cos2x )
then:
=3sin(\pi/2) + 2cos(2(\pi/2)) )
Since =3*1=3)
we can reduce the first term
=3 + 2cos(2(\pi/2)) )
=3 + 2cos(\not{2}(\pi/\not{2})) )
notice these terms cancel
=3 + 2cos(\pi) )
Since =2(-1)=-2)
Which reduces to
=3 + (-2) )
Which finally reduces to
=3-2=1 )
So =1 )
Answer by chitra(359) (Show Source): You can put this solution on YOUR website!
The given function is:
f(x) = 3sinx + 2cos2x
To find the value of f(pi/2)
Substituting for x, we get:
f(Pi/2) = 3 Sin (pi/2) + 2 Cos(2(pi/2)) -------------(1)
We know that the value of
Sin(pi/2) = 1 and cos (pi) = -1
Hence, the equation becomes...
f(pi/2) = 3 (1) + 2 (-1)
f(pi/2) = 3 - 2
f(pi/2) = 1
Hence, the solution.
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