Question

which trigonometric function is equivalent to f(x)=sinx ?

f(x)=cos(x-3pi/2)
f(x)=cos(x-pi/2)
f(x)=cos(-x-pi/2)  <-- i think this one,not sure
f(x)=cos(x+pi)

Answer 1

sin(x) = cos(pi/2 - x)
cos(y) = cos(-y)
sin(x) = cos(x -pi/2)
... look at unit circle , cos is x, sin is y and examine their relations... it's easy to find relation from that .

Answer 2

Answer:  The correct option is (B) $f(x)=\cos\left(x-\dfrac{\pi}{2}\right).$

Step-by-step explanation:  We are given to select the correct trigonometric function that is equivalent to the following trigonometric function :

$f(x)=\sin x.$

Option (A) :

Here, the given function is

$f(x)\\\\=\cos\left(x-3\dfrac{\pi}{2}\right)\\\\=\cos\{-\left(3\dfrac{\pi}{2}-x\right)\}\\\\=\cos \left(3\dfrac{\pi}{2}-x\right)\\\\=-\sin x\neq \sin x.$

So, this option is incorrect.

Option (B) :

Here, the given function is

$f(x)\\\\=\cos\left(x-\dfrac{\pi}{2}\right)\\\\=\cos\{-\left(\dfrac{\pi}{2}-x\right)\}\\\\=\cos \left(\dfrac{\pi}{2}-x\right)\\\\=\sin x.$

So, this option is CORRECT.

Option (C) :

Here, the given function is

$f(x)\\\\=\cos\left(-x-\dfrac{\pi}{2}\right)\\\\=\cos\{-\left(\dfrac{\pi}{2}+x\right)\}\\\\=\cos \left(\dfrac{\pi}{2}+x\right)\\\\=-\sin x\neq \sin x.$

So, this option is incorrect.

Option (D) :

Here, the given function is

$f(x)\\\\=\cos\left(x+\pi\right)\\\\=\cos\left(2\dfrac{\pi}{2}+x\right)\}\\\\=-\cos x\neq \sin x.$

So, this option is incorrect.

Thus, (B) is the correct option.