Question

Trig Identities
Cos^2 x [(pi/2) - x] /cos x

Answer 1

what would you like to know?

Answer 2

I would like to know how to simplify the expression by using the fundamental identities

Answer 3

hey, is it \(\dfrac{cos^2(\pi/2 -x)}{cos x}\)??

Answer 4

\[\cos ^{2}x[(\pi/2)-x] /cosx\]

Answer 5

so, why cos^2 x and cos x from numerator and denominator respectively cannot cancel out?

Answer 6

to get cos x (pi/2 -x)?

Answer 7

cos(x - pi/2) = cos x * cos pi/2 + sin x * sin pi/2
cos(x - pi/2) = sin x

Answer 8

cos(x - pi/2) = cos(pi/2 - x) because cosine is an even function.

= sin(x) because co-functions of complements are equal.

Answer 9

can cos^2x cross out with cos x even though they dont have the same exponent?

Answer 10

1 sec let me think about that.

Answer 11

i really don't know about that.

Answer 12

cos^2x can cancel with cos x
but do we have same angle for cos here ??

is your numerator cos x (pi/2 -x)

or just cos (pi/2-x)
??

Answer 13

\(\large \dfrac{\cos^2x}{\cos x} = \dfrac{\cos x \times \cancel {\cos x} }{\cancel{\cos x}}= \cos x\)