, 26.12.2019organicmemez

# Which of the following trigonometric functions, restricted to 0 < x < π, have values greater than 1? a.) f(x) = sin x b.) f(x) = cos x c.) f(x) = cot x d.) f(x) = sec x

We will evaluate each function separately in the given interval:
0 <x <π
f (x) = sin x
f (0) = sin 0 = 0
f (π) = sin π = 0

f (x) = cos x
f (0) = cos 0 = 1
f (π) = cos π = -1

f (x) = cot x
f (0) = cot 0 = Cos0 / sin0 = 1/0 = inf
f (π) = cot π = Cosπ / sinπ = -1 / 0 = -inf
We can observe that the function f (x) = cot x tends to infinity when it approaches zero.

f (x) = sec x
f (π / 2) = sec π / 2 = 1 / cos π / 2 = 1/0 = inf
We can observe that this function has values that tend towards infinity when it approaches π / 2

this function has values greater than 1 in 0 <x <π
f (x) = sec x

cot x and sec x

Step-by-step explanation:

Given are 4 trignometric functions defined in the interval [0,\pi]

We are asked to identity the functions which can take values more than 1.

We have

have values always between -1 and 1 for any value of x

Hence sinx and cosx cannot take values greater than 1.

This value being ratio of cos to sin will become greater than 1 whenever cos >sin value.

SO this function can take values >1

This function being reciprocal of cos function always can take values outside

(-1,1)

Hence this value can be greater than 1

The answer is (c) f(x) = cos x and f(x) = sec x.

Step-by-step explanation: We are given four trigonometric functions out of we are to select those which have values greater than for 0 < x < π.

Since the values of sine and cosine functions lies between -1 and 1. So, they cannot go beyond 1 and hence option (a) and (b) are incorrect.

Since

so we will have

So, option (c) is correct.

And, we know that

so we will have

So, this option is also correct.

Thus, (c) and (d) are the correct ones.

I think the answer is both C and D. A and B are incorrect because sin and cos never go above 1 but both cot and sec go above one within that domain.

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