Question

Cos (0) = square root 2/2 and 3pi/2<0<2 pi , evaluate sin(0) and tan (0)

Answer 1

The first one is B.... -sqrt(2)/ 2

The Second one is -1

Explanation:

Answer 2

`\sin\theta=-(\sqrt2)/(2)}` and
`\tan\theta=-1`

Explanation:

Given :
`\cos\theta=(\sqrt2)/(2)` and
`(3\pi)/(2)<\theta<2`

To find : Evaluate
`\sin\theta` and
`\tan\theta` ?

Solution :

As
`(3\pi)/(2)<\theta<2` means
`\theta` belong to the third or fourth quadrant .

So,
`\sin\theta` is negative and
`\tan\theta` is positive in third and negative in fourth.

Using the formula of trigonometric,

`\sin^2\theta+\cos^2\theta=1`

`\sin\theta=√(1-\cos^2\theta)`

Substitute
`\cos\theta=(\sqrt2)/(2)` ,

`\sin\theta=\pm\sqrt{1-((\sqrt2)/(2))^2}`

`\sin\theta=\pm\sqrt{1-(2)/(4)}`

`\sin\theta=\pm\sqrt{1-(1)/(2)}`

`\sin\theta=\pm\sqrt{(1)/(2)}`

`\sin\theta=-(\sqrt2)/(2)}`

Now using another formula,

`\tan\theta=(\sin\theta)/(\cos\theta)`

`\tan\theta=(-(\sqrt2)/(2))/((\sqrt2)/(2))`

`\tan\theta=-1`

Therefore,
`\sin\theta=-(\sqrt2)/(2)}` and
`\tan\theta=-1`.