Asked by A
                Find its area. y = sec^2(x),    y = 8 cos(x),    −π/3 ≤ x ≤ π/3
            
            
        Answers
                    Answered by
            Reiny
            
    Notice that at the intersection of the two curves, x = π/3 
and we have symmetry, so we can take the area from 0 to π/3, then double
area = 2∫ (8cosx - sec^2 x) dx from 0 to π/3
= 2[8sinx - tanx] from 0 to π/3
= 2(8sin π/3 - tan π/3 - (8sin 0 - tan 0 )
= ......
your turn
    
and we have symmetry, so we can take the area from 0 to π/3, then double
area = 2∫ (8cosx - sec^2 x) dx from 0 to π/3
= 2[8sinx - tanx] from 0 to π/3
= 2(8sin π/3 - tan π/3 - (8sin 0 - tan 0 )
= ......
your turn
                    Answered by
            Reiny
            
    Let me know what you get, they probably want an "exact" answer.
    
                    Answered by
            A
            
    For sinπ/3, I got 16√(3)/2 but for tanπ/3, I don't know how to find the answer for it. I know at tanπ/3, sin=3/√(2) and cos= 1/√(2). Should that give me √(3)/2? So I got 16√(3)/2 -2√(3)/√(2) is the wrong answer
    
                    Answered by
            Steve
            
    tan π/3 = √3
tan π/6 = 1/√3
One of those "standard" values.
learn it, love it.
    
tan π/6 = 1/√3
One of those "standard" values.
learn it, love it.
                    Answered by
            Anonymous
            
    like pi/3 = 60 degrees and pi/6 = 30 degrees :)
    
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